Integrand size = 21, antiderivative size = 285 \[ \int \frac {\sin ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {3 a \left (a^6-25 a^4 b^2+35 a^2 b^4-3 b^6\right ) x}{8 \left (a^2+b^2\right )^5}+\frac {3 a^2 b \left (a^4-5 a^2 b^2+2 b^4\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right )^5 d}-\frac {a^4 b}{2 \left (a^2+b^2\right )^3 d (a+b \tan (c+d x))^2}-\frac {2 a^3 b \left (a^2-2 b^2\right )}{\left (a^2+b^2\right )^4 d (a+b \tan (c+d x))}+\frac {\cos ^4(c+d x) \left (b \left (3 a^2-b^2\right )+a \left (a^2-3 b^2\right ) \tan (c+d x)\right )}{4 \left (a^2+b^2\right )^3 d}-\frac {a \cos ^2(c+d x) \left (24 a b \left (a^2-b^2\right )+\left (5 a^4-34 a^2 b^2+9 b^4\right ) \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^4 d} \] Output:
3/8*a*(a^6-25*a^4*b^2+35*a^2*b^4-3*b^6)*x/(a^2+b^2)^5+3*a^2*b*(a^4-5*a^2*b ^2+2*b^4)*ln(a*cos(d*x+c)+b*sin(d*x+c))/(a^2+b^2)^5/d-1/2*a^4*b/(a^2+b^2)^ 3/d/(a+b*tan(d*x+c))^2-2*a^3*b*(a^2-2*b^2)/(a^2+b^2)^4/d/(a+b*tan(d*x+c))+ 1/4*cos(d*x+c)^4*(b*(3*a^2-b^2)+a*(a^2-3*b^2)*tan(d*x+c))/(a^2+b^2)^3/d-1/ 8*a*cos(d*x+c)^2*(24*a*b*(a^2-b^2)+(5*a^4-34*a^2*b^2+9*b^4)*tan(d*x+c))/(a ^2+b^2)^4/d
Time = 6.30 (sec) , antiderivative size = 521, normalized size of antiderivative = 1.83 \[ \int \frac {\sin ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {b \left (-\frac {a^3 \left (a^2-5 b^2\right ) \arctan (\tan (c+d x))}{b \left (a^2+b^2\right )^4}+\frac {3 a \left (a^2-3 b^2\right ) \arctan (\tan (c+d x))}{8 b \left (a^2+b^2\right )^3}-\frac {3 a^2 (a-b) (a+b) \cos ^2(c+d x)}{\left (a^2+b^2\right )^4}+\frac {\left (3 a^2-b^2\right ) \cos ^4(c+d x)}{4 \left (a^2+b^2\right )^3}-\frac {a^2 \left (3 a^4-15 a^2 b^2+6 b^4-\frac {a^5-13 a^3 b^2+10 a b^4}{\sqrt {-b^2}}\right ) \log \left (\sqrt {-b^2}-b \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^5}+\frac {3 a^2 \left (a^4-5 a^2 b^2+2 b^4\right ) \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^5}-\frac {a^2 \left (3 a^4-15 a^2 b^2+6 b^4+\frac {a^5-13 a^3 b^2+10 a b^4}{\sqrt {-b^2}}\right ) \log \left (\sqrt {-b^2}+b \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^5}-\frac {a^3 \left (a^2-5 b^2\right ) \cos (c+d x) \sin (c+d x)}{b \left (a^2+b^2\right )^4}+\frac {3 a \left (a^2-3 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b \left (a^2+b^2\right )^3}+\frac {a \left (a^2-3 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{4 b \left (a^2+b^2\right )^3}-\frac {a^4}{2 \left (a^2+b^2\right )^3 (a+b \tan (c+d x))^2}-\frac {2 a^3 \left (a^2-2 b^2\right )}{\left (a^2+b^2\right )^4 (a+b \tan (c+d x))}\right )}{d} \] Input:
Integrate[Sin[c + d*x]^4/(a + b*Tan[c + d*x])^3,x]
Output:
(b*(-((a^3*(a^2 - 5*b^2)*ArcTan[Tan[c + d*x]])/(b*(a^2 + b^2)^4)) + (3*a*( a^2 - 3*b^2)*ArcTan[Tan[c + d*x]])/(8*b*(a^2 + b^2)^3) - (3*a^2*(a - b)*(a + b)*Cos[c + d*x]^2)/(a^2 + b^2)^4 + ((3*a^2 - b^2)*Cos[c + d*x]^4)/(4*(a ^2 + b^2)^3) - (a^2*(3*a^4 - 15*a^2*b^2 + 6*b^4 - (a^5 - 13*a^3*b^2 + 10*a *b^4)/Sqrt[-b^2])*Log[Sqrt[-b^2] - b*Tan[c + d*x]])/(2*(a^2 + b^2)^5) + (3 *a^2*(a^4 - 5*a^2*b^2 + 2*b^4)*Log[a + b*Tan[c + d*x]])/(a^2 + b^2)^5 - (a ^2*(3*a^4 - 15*a^2*b^2 + 6*b^4 + (a^5 - 13*a^3*b^2 + 10*a*b^4)/Sqrt[-b^2]) *Log[Sqrt[-b^2] + b*Tan[c + d*x]])/(2*(a^2 + b^2)^5) - (a^3*(a^2 - 5*b^2)* Cos[c + d*x]*Sin[c + d*x])/(b*(a^2 + b^2)^4) + (3*a*(a^2 - 3*b^2)*Cos[c + d*x]*Sin[c + d*x])/(8*b*(a^2 + b^2)^3) + (a*(a^2 - 3*b^2)*Cos[c + d*x]^3*S in[c + d*x])/(4*b*(a^2 + b^2)^3) - a^4/(2*(a^2 + b^2)^3*(a + b*Tan[c + d*x ])^2) - (2*a^3*(a^2 - 2*b^2))/((a^2 + b^2)^4*(a + b*Tan[c + d*x]))))/d
Time = 1.40 (sec) , antiderivative size = 378, normalized size of antiderivative = 1.33, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3999, 601, 2178, 2160, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\sin ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^4}{(a+b \tan (c+d x))^3}dx\) |
| \(\Big \downarrow \) 3999 |
| \(\displaystyle \frac {b \int \frac {b^4 \tan ^4(c+d x)}{(a+b \tan (c+d x))^3 \left (\tan ^2(c+d x) b^2+b^2\right )^3}d(b \tan (c+d x))}{d}\) |
| \(\Big \downarrow \) 601 |
| \(\displaystyle \frac {b \left (\frac {b^2 \left (a b \left (a^2-3 b^2\right ) \tan (c+d x)+b^2 \left (3 a^2-b^2\right )\right )}{4 \left (a^2+b^2\right )^3 \left (b^2 \tan ^2(c+d x)+b^2\right )^2}-\frac {\int \frac {-\frac {3 a \left (a^2-3 b^2\right ) \tan ^3(c+d x) b^7}{\left (a^2+b^2\right )^3}-\frac {a^3 \left (9 a^2+5 b^2\right ) \tan (c+d x) b^5}{\left (a^2+b^2\right )^3}-\frac {a^2 \left (4 a^4+21 b^2 a^2-3 b^4\right ) \tan ^2(c+d x) b^4}{\left (a^2+b^2\right )^3}+\frac {a^4 \left (a^2-3 b^2\right ) b^4}{\left (a^2+b^2\right )^3}}{(a+b \tan (c+d x))^3 \left (\tan ^2(c+d x) b^2+b^2\right )^2}d(b \tan (c+d x))}{4 b^2}\right )}{d}\) |
| \(\Big \downarrow \) 2178 |
| \(\displaystyle \frac {b \left (\frac {b^2 \left (a b \left (a^2-3 b^2\right ) \tan (c+d x)+b^2 \left (3 a^2-b^2\right )\right )}{4 \left (a^2+b^2\right )^3 \left (b^2 \tan ^2(c+d x)+b^2\right )^2}-\frac {\frac {a b^2 \left (24 a b^2 \left (a^2-b^2\right )+b \left (5 a^4-34 a^2 b^2+9 b^4\right ) \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^4 \left (b^2 \tan ^2(c+d x)+b^2\right )}-\frac {\int \frac {-\frac {a \left (5 a^4-34 b^2 a^2+9 b^4\right ) \tan ^3(c+d x) b^7}{\left (a^2+b^2\right )^4}-\frac {3 a^2 \left (5 a^4-18 b^2 a^2-7 b^4\right ) \tan ^2(c+d x) b^6}{\left (a^2+b^2\right )^4}-\frac {a^3 \left (15 a^4+26 b^2 a^2-37 b^4\right ) \tan (c+d x) b^5}{\left (a^2+b^2\right )^4}+\frac {3 a^4 \left (a^4-10 b^2 a^2+5 b^4\right ) b^4}{\left (a^2+b^2\right )^4}}{(a+b \tan (c+d x))^3 \left (\tan ^2(c+d x) b^2+b^2\right )}d(b \tan (c+d x))}{2 b^2}}{4 b^2}\right )}{d}\) |
| \(\Big \downarrow \) 2160 |
| \(\displaystyle \frac {b \left (\frac {b^2 \left (a b \left (a^2-3 b^2\right ) \tan (c+d x)+b^2 \left (3 a^2-b^2\right )\right )}{4 \left (a^2+b^2\right )^3 \left (b^2 \tan ^2(c+d x)+b^2\right )^2}-\frac {\frac {a b^2 \left (24 a b^2 \left (a^2-b^2\right )+b \left (5 a^4-34 a^2 b^2+9 b^4\right ) \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^4 \left (b^2 \tan ^2(c+d x)+b^2\right )}-\frac {\int \left (\frac {24 a^2 \left (a^4-5 b^2 a^2+2 b^4\right ) b^4}{\left (a^2+b^2\right )^5 (a+b \tan (c+d x))}+\frac {3 a \left (a^6-25 b^2 a^4+35 b^4 a^2-8 b \left (a^4-5 b^2 a^2+2 b^4\right ) \tan (c+d x) a-3 b^6\right ) b^4}{\left (a^2+b^2\right )^5 \left (\tan ^2(c+d x) b^2+b^2\right )}+\frac {16 a^3 \left (a^2-2 b^2\right ) b^4}{\left (a^2+b^2\right )^4 (a+b \tan (c+d x))^2}+\frac {8 a^4 b^4}{\left (a^2+b^2\right )^3 (a+b \tan (c+d x))^3}\right )d(b \tan (c+d x))}{2 b^2}}{4 b^2}\right )}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b \left (\frac {b^2 \left (a b \left (a^2-3 b^2\right ) \tan (c+d x)+b^2 \left (3 a^2-b^2\right )\right )}{4 \left (a^2+b^2\right )^3 \left (b^2 \tan ^2(c+d x)+b^2\right )^2}-\frac {\frac {a b^2 \left (24 a b^2 \left (a^2-b^2\right )+b \left (5 a^4-34 a^2 b^2+9 b^4\right ) \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^4 \left (b^2 \tan ^2(c+d x)+b^2\right )}-\frac {-\frac {4 a^4 b^4}{\left (a^2+b^2\right )^3 (a+b \tan (c+d x))^2}-\frac {12 a^2 b^4 \left (a^4-5 a^2 b^2+2 b^4\right ) \log \left (b^2 \tan ^2(c+d x)+b^2\right )}{\left (a^2+b^2\right )^5}+\frac {24 a^2 b^4 \left (a^4-5 a^2 b^2+2 b^4\right ) \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^5}-\frac {16 a^3 b^4 \left (a^2-2 b^2\right )}{\left (a^2+b^2\right )^4 (a+b \tan (c+d x))}+\frac {3 a b^3 \left (a^6-25 a^4 b^2+35 a^2 b^4-3 b^6\right ) \arctan (\tan (c+d x))}{\left (a^2+b^2\right )^5}}{2 b^2}}{4 b^2}\right )}{d}\) |
Input:
Int[Sin[c + d*x]^4/(a + b*Tan[c + d*x])^3,x]
Output:
(b*((b^2*(b^2*(3*a^2 - b^2) + a*b*(a^2 - 3*b^2)*Tan[c + d*x]))/(4*(a^2 + b ^2)^3*(b^2 + b^2*Tan[c + d*x]^2)^2) - ((a*b^2*(24*a*b^2*(a^2 - b^2) + b*(5 *a^4 - 34*a^2*b^2 + 9*b^4)*Tan[c + d*x]))/(2*(a^2 + b^2)^4*(b^2 + b^2*Tan[ c + d*x]^2)) - ((3*a*b^3*(a^6 - 25*a^4*b^2 + 35*a^2*b^4 - 3*b^6)*ArcTan[Ta n[c + d*x]])/(a^2 + b^2)^5 + (24*a^2*b^4*(a^4 - 5*a^2*b^2 + 2*b^4)*Log[a + b*Tan[c + d*x]])/(a^2 + b^2)^5 - (12*a^2*b^4*(a^4 - 5*a^2*b^2 + 2*b^4)*Lo g[b^2 + b^2*Tan[c + d*x]^2])/(a^2 + b^2)^5 - (4*a^4*b^4)/((a^2 + b^2)^3*(a + b*Tan[c + d*x])^2) - (16*a^3*b^4*(a^2 - 2*b^2))/((a^2 + b^2)^4*(a + b*T an[c + d*x])))/(2*b^2))/(4*b^2)))/d
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) *((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[(c + d*x)^n*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Qx)/(c + d*x)^n + (e* (2*p + 3))/(c + d*x)^n, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 1] && LtQ[p, -1] && ILtQ[n, 0] && NeQ[b*c^2 + a*d^2, 0]
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : > With[{Qx = PolynomialQuotient[(d + e*x)^m*Pq, a + b*x^2, x], R = Coeff[Po lynomialRemainder[(d + e*x)^m*Pq, a + b*x^2, x], x, 0], S = Coeff[Polynomia lRemainder[(d + e*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*S - b*R*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*b*(p + 1)) Int[(d + e*x )^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*b*(p + 1)*Qx)/(d + e*x)^m + (b*R*( 2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, b, d, e}, x] && PolyQ[Pq, x ] && NeQ[b*d^2 + a*e^2, 0] && LtQ[p, -1] && ILtQ[m, 0]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[b/f Subst[Int[x^m*((a + x)^n/(b^2 + x^2)^(m/2 + 1)), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[m/2]
Time = 36.31 (sec) , antiderivative size = 331, normalized size of antiderivative = 1.16
| method | result | size |
| derivativedivides | \(\frac {-\frac {a^{4} b}{2 \left (a^{2}+b^{2}\right )^{3} \left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {3 a^{2} b \left (a^{4}-5 b^{2} a^{2}+2 b^{4}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{5}}-\frac {2 a^{3} b \left (a^{2}-2 b^{2}\right )}{\left (a^{2}+b^{2}\right )^{4} \left (a +b \tan \left (d x +c \right )\right )}+\frac {\frac {\left (-\frac {5}{8} a^{7}+\frac {29}{8} a^{5} b^{2}+\frac {25}{8} a^{3} b^{4}-\frac {9}{8} a \,b^{6}\right ) \tan \left (d x +c \right )^{3}+\left (-3 a^{6} b +3 b^{5} a^{2}\right ) \tan \left (d x +c \right )^{2}+\left (-\frac {3}{8} a^{7}+\frac {27}{8} a^{5} b^{2}+\frac {15}{8} a^{3} b^{4}-\frac {15}{8} a \,b^{6}\right ) \tan \left (d x +c \right )-\frac {9 a^{6} b}{4}+\frac {5 a^{4} b^{3}}{4}+\frac {13 b^{5} a^{2}}{4}-\frac {b^{7}}{4}}{\left (1+\tan \left (d x +c \right )^{2}\right )^{2}}+\frac {3 a \left (\frac {\left (-8 a^{5} b +40 a^{3} b^{3}-16 a \,b^{5}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (a^{6}-25 a^{4} b^{2}+35 a^{2} b^{4}-3 b^{6}\right ) \arctan \left (\tan \left (d x +c \right )\right )\right )}{8}}{\left (a^{2}+b^{2}\right )^{5}}}{d}\) | \(331\) |
| default | \(\frac {-\frac {a^{4} b}{2 \left (a^{2}+b^{2}\right )^{3} \left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {3 a^{2} b \left (a^{4}-5 b^{2} a^{2}+2 b^{4}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{5}}-\frac {2 a^{3} b \left (a^{2}-2 b^{2}\right )}{\left (a^{2}+b^{2}\right )^{4} \left (a +b \tan \left (d x +c \right )\right )}+\frac {\frac {\left (-\frac {5}{8} a^{7}+\frac {29}{8} a^{5} b^{2}+\frac {25}{8} a^{3} b^{4}-\frac {9}{8} a \,b^{6}\right ) \tan \left (d x +c \right )^{3}+\left (-3 a^{6} b +3 b^{5} a^{2}\right ) \tan \left (d x +c \right )^{2}+\left (-\frac {3}{8} a^{7}+\frac {27}{8} a^{5} b^{2}+\frac {15}{8} a^{3} b^{4}-\frac {15}{8} a \,b^{6}\right ) \tan \left (d x +c \right )-\frac {9 a^{6} b}{4}+\frac {5 a^{4} b^{3}}{4}+\frac {13 b^{5} a^{2}}{4}-\frac {b^{7}}{4}}{\left (1+\tan \left (d x +c \right )^{2}\right )^{2}}+\frac {3 a \left (\frac {\left (-8 a^{5} b +40 a^{3} b^{3}-16 a \,b^{5}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (a^{6}-25 a^{4} b^{2}+35 a^{2} b^{4}-3 b^{6}\right ) \arctan \left (\tan \left (d x +c \right )\right )\right )}{8}}{\left (a^{2}+b^{2}\right )^{5}}}{d}\) | \(331\) |
| risch | \(-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} a}{8 \left (i b +a \right )^{2} \left (2 i a b +a^{2}-b^{2}\right ) d}-\frac {3 a^{2} x}{8 \left (5 i b \,a^{4}-10 i a^{2} b^{3}+i b^{5}-a^{5}+10 a^{3} b^{2}-5 a \,b^{4}\right )}-\frac {12 i b^{5} a^{2} c}{d \left (a^{10}+5 a^{8} b^{2}+10 a^{6} b^{4}+10 a^{4} b^{6}+5 a^{2} b^{8}+b^{10}\right )}-\frac {{\mathrm e}^{2 i \left (d x +c \right )} b}{16 \left (-4 i a^{3} b +4 i a \,b^{3}+a^{4}-6 b^{2} a^{2}+b^{4}\right ) d}-\frac {9 i a x b}{8 \left (5 i b \,a^{4}-10 i a^{2} b^{3}+i b^{5}-a^{5}+10 a^{3} b^{2}-5 a \,b^{4}\right )}-\frac {{\mathrm e}^{-2 i \left (d x +c \right )} b}{16 \left (i b +a \right )^{2} \left (2 i a b +a^{2}-b^{2}\right ) d}+\frac {30 i a^{4} b^{3} c}{d \left (a^{10}+5 a^{8} b^{2}+10 a^{6} b^{4}+10 a^{4} b^{6}+5 a^{2} b^{8}+b^{10}\right )}-\frac {i {\mathrm e}^{4 i \left (d x +c \right )}}{64 \left (-3 i a^{2} b +i b^{3}+a^{3}-3 a \,b^{2}\right ) d}+\frac {30 i a^{4} b^{3} x}{a^{10}+5 a^{8} b^{2}+10 a^{6} b^{4}+10 a^{4} b^{6}+5 a^{2} b^{8}+b^{10}}-\frac {6 i a^{6} b x}{a^{10}+5 a^{8} b^{2}+10 a^{6} b^{4}+10 a^{4} b^{6}+5 a^{2} b^{8}+b^{10}}+\frac {i {\mathrm e}^{-4 i \left (d x +c \right )}}{64 \left (2 i a b +a^{2}-b^{2}\right ) \left (i b +a \right ) d}-\frac {6 i a^{6} b c}{d \left (a^{10}+5 a^{8} b^{2}+10 a^{6} b^{4}+10 a^{4} b^{6}+5 a^{2} b^{8}+b^{10}\right )}-\frac {12 i b^{5} a^{2} x}{a^{10}+5 a^{8} b^{2}+10 a^{6} b^{4}+10 a^{4} b^{6}+5 a^{2} b^{8}+b^{10}}+\frac {i {\mathrm e}^{2 i \left (d x +c \right )} a}{8 \left (-4 i a^{3} b +4 i a \,b^{3}+a^{4}-6 b^{2} a^{2}+b^{4}\right ) d}+\frac {2 i a^{3} b^{2} \left (-3 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+3 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-2 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}+4 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-3 i a^{3}+4 i a \,b^{2}+3 a^{2} b -4 b^{3}\right )}{\left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+i a \,{\mathrm e}^{2 i \left (d x +c \right )}-b +i a \right )^{2} \left (-i a +b \right )^{4} d \left (i a +b \right )^{5}}+\frac {3 a^{6} b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{d \left (a^{10}+5 a^{8} b^{2}+10 a^{6} b^{4}+10 a^{4} b^{6}+5 a^{2} b^{8}+b^{10}\right )}-\frac {15 a^{4} b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{d \left (a^{10}+5 a^{8} b^{2}+10 a^{6} b^{4}+10 a^{4} b^{6}+5 a^{2} b^{8}+b^{10}\right )}+\frac {6 b^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right ) a^{2}}{d \left (a^{10}+5 a^{8} b^{2}+10 a^{6} b^{4}+10 a^{4} b^{6}+5 a^{2} b^{8}+b^{10}\right )}\) | \(1052\) |
Input:
int(sin(d*x+c)^4/(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d*(-1/2*a^4*b/(a^2+b^2)^3/(a+b*tan(d*x+c))^2+3*a^2*b*(a^4-5*a^2*b^2+2*b^ 4)/(a^2+b^2)^5*ln(a+b*tan(d*x+c))-2*a^3*b*(a^2-2*b^2)/(a^2+b^2)^4/(a+b*tan (d*x+c))+1/(a^2+b^2)^5*(((-5/8*a^7+29/8*a^5*b^2+25/8*a^3*b^4-9/8*a*b^6)*ta n(d*x+c)^3+(-3*a^6*b+3*a^2*b^5)*tan(d*x+c)^2+(-3/8*a^7+27/8*a^5*b^2+15/8*a ^3*b^4-15/8*a*b^6)*tan(d*x+c)-9/4*a^6*b+5/4*a^4*b^3+13/4*b^5*a^2-1/4*b^7)/ (1+tan(d*x+c)^2)^2+3/8*a*(1/2*(-8*a^5*b+40*a^3*b^3-16*a*b^5)*ln(1+tan(d*x+ c)^2)+(a^6-25*a^4*b^2+35*a^2*b^4-3*b^6)*arctan(tan(d*x+c)))))
Leaf count of result is larger than twice the leaf count of optimal. 705 vs. \(2 (277) = 554\).
Time = 0.16 (sec) , antiderivative size = 705, normalized size of antiderivative = 2.47 \[ \int \frac {\sin ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx =\text {Too large to display} \] Input:
integrate(sin(d*x+c)^4/(a+b*tan(d*x+c))^3,x, algorithm="fricas")
Output:
1/32*(119*a^6*b^3 - 159*a^4*b^5 - 51*a^2*b^7 + 3*b^9 + 8*(a^8*b + 4*a^6*b^ 3 + 6*a^4*b^5 + 4*a^2*b^7 + b^9)*cos(d*x + c)^6 - 8*(5*a^8*b + 16*a^6*b^3 + 18*a^4*b^5 + 8*a^2*b^7 + b^9)*cos(d*x + c)^4 + 12*(a^7*b^2 - 25*a^5*b^4 + 35*a^3*b^6 - 3*a*b^8)*d*x - (a^8*b + 110*a^6*b^3 - 420*a^4*b^5 - 78*a^2* b^7 + 3*b^9 - 12*(a^9 - 26*a^7*b^2 + 60*a^5*b^4 - 38*a^3*b^6 + 3*a*b^8)*d* x)*cos(d*x + c)^2 + 48*(a^6*b^3 - 5*a^4*b^5 + 2*a^2*b^7 + (a^8*b - 6*a^6*b ^3 + 7*a^4*b^5 - 2*a^2*b^7)*cos(d*x + c)^2 + 2*(a^7*b^2 - 5*a^5*b^4 + 2*a^ 3*b^6)*cos(d*x + c)*sin(d*x + c))*log(2*a*b*cos(d*x + c)*sin(d*x + c) + (a ^2 - b^2)*cos(d*x + c)^2 + b^2) + 2*(4*(a^9 + 4*a^7*b^2 + 6*a^5*b^4 + 4*a^ 3*b^6 + a*b^8)*cos(d*x + c)^5 - 2*(5*a^9 + 12*a^7*b^2 + 6*a^5*b^4 - 4*a^3* b^6 - 3*a*b^8)*cos(d*x + c)^3 + (77*a^7*b^2 - 69*a^5*b^4 + 63*a^3*b^6 - 15 *a*b^8 + 12*(a^8*b - 25*a^6*b^3 + 35*a^4*b^5 - 3*a^2*b^7)*d*x)*cos(d*x + c ))*sin(d*x + c))/((a^12 + 4*a^10*b^2 + 5*a^8*b^4 - 5*a^4*b^8 - 4*a^2*b^10 - b^12)*d*cos(d*x + c)^2 + 2*(a^11*b + 5*a^9*b^3 + 10*a^7*b^5 + 10*a^5*b^7 + 5*a^3*b^9 + a*b^11)*d*cos(d*x + c)*sin(d*x + c) + (a^10*b^2 + 5*a^8*b^4 + 10*a^6*b^6 + 10*a^4*b^8 + 5*a^2*b^10 + b^12)*d)
Exception generated. \[ \int \frac {\sin ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\text {Exception raised: AttributeError} \] Input:
integrate(sin(d*x+c)**4/(a+b*tan(d*x+c))**3,x)
Output:
Exception raised: AttributeError >> 'NoneType' object has no attribute 'pr imitive'
Leaf count of result is larger than twice the leaf count of optimal. 744 vs. \(2 (277) = 554\).
Time = 0.13 (sec) , antiderivative size = 744, normalized size of antiderivative = 2.61 \[ \int \frac {\sin ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx =\text {Too large to display} \] Input:
integrate(sin(d*x+c)^4/(a+b*tan(d*x+c))^3,x, algorithm="maxima")
Output:
1/8*(3*(a^7 - 25*a^5*b^2 + 35*a^3*b^4 - 3*a*b^6)*(d*x + c)/(a^10 + 5*a^8*b ^2 + 10*a^6*b^4 + 10*a^4*b^6 + 5*a^2*b^8 + b^10) + 24*(a^6*b - 5*a^4*b^3 + 2*a^2*b^5)*log(b*tan(d*x + c) + a)/(a^10 + 5*a^8*b^2 + 10*a^6*b^4 + 10*a^ 4*b^6 + 5*a^2*b^8 + b^10) - 12*(a^6*b - 5*a^4*b^3 + 2*a^2*b^5)*log(tan(d*x + c)^2 + 1)/(a^10 + 5*a^8*b^2 + 10*a^6*b^4 + 10*a^4*b^6 + 5*a^2*b^8 + b^1 0) - (38*a^6*b - 56*a^4*b^3 + 2*a^2*b^5 + 3*(7*a^5*b^2 - 22*a^3*b^4 + 3*a* b^6)*tan(d*x + c)^5 + 6*(5*a^6*b - 12*a^4*b^3 - a^2*b^5)*tan(d*x + c)^4 + (5*a^7 + 49*a^5*b^2 - 133*a^3*b^4 + 15*a*b^6)*tan(d*x + c)^3 + 2*(35*a^6*b - 61*a^4*b^3 + a^2*b^5 + b^7)*tan(d*x + c)^2 + (3*a^7 + 22*a^5*b^2 - 73*a ^3*b^4 + 4*a*b^6)*tan(d*x + c))/(a^10 + 4*a^8*b^2 + 6*a^6*b^4 + 4*a^4*b^6 + a^2*b^8 + (a^8*b^2 + 4*a^6*b^4 + 6*a^4*b^6 + 4*a^2*b^8 + b^10)*tan(d*x + c)^6 + 2*(a^9*b + 4*a^7*b^3 + 6*a^5*b^5 + 4*a^3*b^7 + a*b^9)*tan(d*x + c) ^5 + (a^10 + 6*a^8*b^2 + 14*a^6*b^4 + 16*a^4*b^6 + 9*a^2*b^8 + 2*b^10)*tan (d*x + c)^4 + 4*(a^9*b + 4*a^7*b^3 + 6*a^5*b^5 + 4*a^3*b^7 + a*b^9)*tan(d* x + c)^3 + (2*a^10 + 9*a^8*b^2 + 16*a^6*b^4 + 14*a^4*b^6 + 6*a^2*b^8 + b^1 0)*tan(d*x + c)^2 + 2*(a^9*b + 4*a^7*b^3 + 6*a^5*b^5 + 4*a^3*b^7 + a*b^9)* tan(d*x + c)))/d
Leaf count of result is larger than twice the leaf count of optimal. 613 vs. \(2 (277) = 554\).
Time = 0.25 (sec) , antiderivative size = 613, normalized size of antiderivative = 2.15 \[ \int \frac {\sin ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {3 \, {\left (a^{7} - 25 \, a^{5} b^{2} + 35 \, a^{3} b^{4} - 3 \, a b^{6}\right )} {\left (d x + c\right )}}{8 \, {\left (a^{10} d + 5 \, a^{8} b^{2} d + 10 \, a^{6} b^{4} d + 10 \, a^{4} b^{6} d + 5 \, a^{2} b^{8} d + b^{10} d\right )}} - \frac {3 \, {\left (a^{6} b - 5 \, a^{4} b^{3} + 2 \, a^{2} b^{5}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, {\left (a^{10} d + 5 \, a^{8} b^{2} d + 10 \, a^{6} b^{4} d + 10 \, a^{4} b^{6} d + 5 \, a^{2} b^{8} d + b^{10} d\right )}} + \frac {3 \, {\left (a^{6} b^{2} - 5 \, a^{4} b^{4} + 2 \, a^{2} b^{6}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{10} b d + 5 \, a^{8} b^{3} d + 10 \, a^{6} b^{5} d + 10 \, a^{4} b^{7} d + 5 \, a^{2} b^{9} d + b^{11} d} - \frac {21 \, a^{5} b^{2} \tan \left (d x + c\right )^{5} - 66 \, a^{3} b^{4} \tan \left (d x + c\right )^{5} + 9 \, a b^{6} \tan \left (d x + c\right )^{5} + 30 \, a^{6} b \tan \left (d x + c\right )^{4} - 72 \, a^{4} b^{3} \tan \left (d x + c\right )^{4} - 6 \, a^{2} b^{5} \tan \left (d x + c\right )^{4} + 5 \, a^{7} \tan \left (d x + c\right )^{3} + 49 \, a^{5} b^{2} \tan \left (d x + c\right )^{3} - 133 \, a^{3} b^{4} \tan \left (d x + c\right )^{3} + 15 \, a b^{6} \tan \left (d x + c\right )^{3} + 70 \, a^{6} b \tan \left (d x + c\right )^{2} - 122 \, a^{4} b^{3} \tan \left (d x + c\right )^{2} + 2 \, a^{2} b^{5} \tan \left (d x + c\right )^{2} + 2 \, b^{7} \tan \left (d x + c\right )^{2} + 3 \, a^{7} \tan \left (d x + c\right ) + 22 \, a^{5} b^{2} \tan \left (d x + c\right ) - 73 \, a^{3} b^{4} \tan \left (d x + c\right ) + 4 \, a b^{6} \tan \left (d x + c\right ) + 38 \, a^{6} b - 56 \, a^{4} b^{3} + 2 \, a^{2} b^{5}}{8 \, {\left (a^{8} d + 4 \, a^{6} b^{2} d + 6 \, a^{4} b^{4} d + 4 \, a^{2} b^{6} d + b^{8} d\right )} {\left (b \tan \left (d x + c\right )^{3} + a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right ) + a\right )}^{2}} \] Input:
integrate(sin(d*x+c)^4/(a+b*tan(d*x+c))^3,x, algorithm="giac")
Output:
3/8*(a^7 - 25*a^5*b^2 + 35*a^3*b^4 - 3*a*b^6)*(d*x + c)/(a^10*d + 5*a^8*b^ 2*d + 10*a^6*b^4*d + 10*a^4*b^6*d + 5*a^2*b^8*d + b^10*d) - 3/2*(a^6*b - 5 *a^4*b^3 + 2*a^2*b^5)*log(tan(d*x + c)^2 + 1)/(a^10*d + 5*a^8*b^2*d + 10*a ^6*b^4*d + 10*a^4*b^6*d + 5*a^2*b^8*d + b^10*d) + 3*(a^6*b^2 - 5*a^4*b^4 + 2*a^2*b^6)*log(abs(b*tan(d*x + c) + a))/(a^10*b*d + 5*a^8*b^3*d + 10*a^6* b^5*d + 10*a^4*b^7*d + 5*a^2*b^9*d + b^11*d) - 1/8*(21*a^5*b^2*tan(d*x + c )^5 - 66*a^3*b^4*tan(d*x + c)^5 + 9*a*b^6*tan(d*x + c)^5 + 30*a^6*b*tan(d* x + c)^4 - 72*a^4*b^3*tan(d*x + c)^4 - 6*a^2*b^5*tan(d*x + c)^4 + 5*a^7*ta n(d*x + c)^3 + 49*a^5*b^2*tan(d*x + c)^3 - 133*a^3*b^4*tan(d*x + c)^3 + 15 *a*b^6*tan(d*x + c)^3 + 70*a^6*b*tan(d*x + c)^2 - 122*a^4*b^3*tan(d*x + c) ^2 + 2*a^2*b^5*tan(d*x + c)^2 + 2*b^7*tan(d*x + c)^2 + 3*a^7*tan(d*x + c) + 22*a^5*b^2*tan(d*x + c) - 73*a^3*b^4*tan(d*x + c) + 4*a*b^6*tan(d*x + c) + 38*a^6*b - 56*a^4*b^3 + 2*a^2*b^5)/((a^8*d + 4*a^6*b^2*d + 6*a^4*b^4*d + 4*a^2*b^6*d + b^8*d)*(b*tan(d*x + c)^3 + a*tan(d*x + c)^2 + b*tan(d*x + c) + a)^2)
Time = 2.57 (sec) , antiderivative size = 717, normalized size of antiderivative = 2.52 \[ \int \frac {\sin ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (\frac {3\,b}{{\left (a^2+b^2\right )}^2}-\frac {24\,b^3}{{\left (a^2+b^2\right )}^3}+\frac {45\,b^5}{{\left (a^2+b^2\right )}^4}-\frac {24\,b^7}{{\left (a^2+b^2\right )}^5}\right )}{d}-\frac {\frac {19\,a^6\,b-28\,a^4\,b^3+a^2\,b^5}{4\,\left (a^8+4\,a^6\,b^2+6\,a^4\,b^4+4\,a^2\,b^6+b^8\right )}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (35\,a^6\,b-61\,a^4\,b^3+a^2\,b^5+b^7\right )}{4\,\left (a^8+4\,a^6\,b^2+6\,a^4\,b^4+4\,a^2\,b^6+b^8\right )}-\frac {3\,{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (-5\,a^6\,b+12\,a^4\,b^3+a^2\,b^5\right )}{4\,\left (a^8+4\,a^6\,b^2+6\,a^4\,b^4+4\,a^2\,b^6+b^8\right )}+\frac {3\,{\mathrm {tan}\left (c+d\,x\right )}^5\,\left (7\,a^5\,b^2-22\,a^3\,b^4+3\,a\,b^6\right )}{8\,\left (a^8+4\,a^6\,b^2+6\,a^4\,b^4+4\,a^2\,b^6+b^8\right )}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (5\,a^7+49\,a^5\,b^2-133\,a^3\,b^4+15\,a\,b^6\right )}{8\,\left (a^8+4\,a^6\,b^2+6\,a^4\,b^4+4\,a^2\,b^6+b^8\right )}+\frac {a\,\mathrm {tan}\left (c+d\,x\right )\,\left (3\,a^6+22\,a^4\,b^2-73\,a^2\,b^4+4\,b^6\right )}{8\,\left (a^8+4\,a^6\,b^2+6\,a^4\,b^4+4\,a^2\,b^6+b^8\right )}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,\left (2\,a^2+b^2\right )+{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (a^2+2\,b^2\right )+a^2+b^2\,{\mathrm {tan}\left (c+d\,x\right )}^6+2\,a\,b\,\mathrm {tan}\left (c+d\,x\right )+4\,a\,b\,{\mathrm {tan}\left (c+d\,x\right )}^3+2\,a\,b\,{\mathrm {tan}\left (c+d\,x\right )}^5\right )}-\frac {3\,\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (a^2\,1{}\mathrm {i}+3\,b\,a\right )}{16\,d\,\left (a^5+a^4\,b\,5{}\mathrm {i}-10\,a^3\,b^2-a^2\,b^3\,10{}\mathrm {i}+5\,a\,b^4+b^5\,1{}\mathrm {i}\right )}-\frac {3\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (3\,a\,b-a^2\,1{}\mathrm {i}\right )}{16\,d\,\left (a^5-a^4\,b\,5{}\mathrm {i}-10\,a^3\,b^2+a^2\,b^3\,10{}\mathrm {i}+5\,a\,b^4-b^5\,1{}\mathrm {i}\right )} \] Input:
int(sin(c + d*x)^4/(a + b*tan(c + d*x))^3,x)
Output:
(log(a + b*tan(c + d*x))*((3*b)/(a^2 + b^2)^2 - (24*b^3)/(a^2 + b^2)^3 + ( 45*b^5)/(a^2 + b^2)^4 - (24*b^7)/(a^2 + b^2)^5))/d - ((19*a^6*b + a^2*b^5 - 28*a^4*b^3)/(4*(a^8 + b^8 + 4*a^2*b^6 + 6*a^4*b^4 + 4*a^6*b^2)) + (tan(c + d*x)^2*(35*a^6*b + b^7 + a^2*b^5 - 61*a^4*b^3))/(4*(a^8 + b^8 + 4*a^2*b ^6 + 6*a^4*b^4 + 4*a^6*b^2)) - (3*tan(c + d*x)^4*(a^2*b^5 - 5*a^6*b + 12*a ^4*b^3))/(4*(a^8 + b^8 + 4*a^2*b^6 + 6*a^4*b^4 + 4*a^6*b^2)) + (3*tan(c + d*x)^5*(3*a*b^6 - 22*a^3*b^4 + 7*a^5*b^2))/(8*(a^8 + b^8 + 4*a^2*b^6 + 6*a ^4*b^4 + 4*a^6*b^2)) + (tan(c + d*x)^3*(15*a*b^6 + 5*a^7 - 133*a^3*b^4 + 4 9*a^5*b^2))/(8*(a^8 + b^8 + 4*a^2*b^6 + 6*a^4*b^4 + 4*a^6*b^2)) + (a*tan(c + d*x)*(3*a^6 + 4*b^6 - 73*a^2*b^4 + 22*a^4*b^2))/(8*(a^8 + b^8 + 4*a^2*b ^6 + 6*a^4*b^4 + 4*a^6*b^2)))/(d*(tan(c + d*x)^2*(2*a^2 + b^2) + tan(c + d *x)^4*(a^2 + 2*b^2) + a^2 + b^2*tan(c + d*x)^6 + 2*a*b*tan(c + d*x) + 4*a* b*tan(c + d*x)^3 + 2*a*b*tan(c + d*x)^5)) - (3*log(tan(c + d*x) - 1i)*(3*a *b + a^2*1i))/(16*d*(5*a*b^4 + a^4*b*5i + a^5 + b^5*1i - a^2*b^3*10i - 10* a^3*b^2)) - (3*log(tan(c + d*x) + 1i)*(3*a*b - a^2*1i))/(16*d*(5*a*b^4 - a ^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2))
Time = 0.39 (sec) , antiderivative size = 1663, normalized size of antiderivative = 5.84 \[ \int \frac {\sin ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx =\text {Too large to display} \] Input:
int(sin(d*x+c)^4/(a+b*tan(d*x+c))^3,x)
Output:
( - 96*cos(c + d*x)*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)*a**7*b**3 + 480*cos(c + d*x)*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)*a**5*b**5 - 192 *cos(c + d*x)*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)*a**3*b**7 + 96*cos (c + d*x)*log(tan((c + d*x)/2)**2*a - 2*tan((c + d*x)/2)*b - a)*sin(c + d* x)*a**7*b**3 - 480*cos(c + d*x)*log(tan((c + d*x)/2)**2*a - 2*tan((c + d*x )/2)*b - a)*sin(c + d*x)*a**5*b**5 + 192*cos(c + d*x)*log(tan((c + d*x)/2) **2*a - 2*tan((c + d*x)/2)*b - a)*sin(c + d*x)*a**3*b**7 + 4*cos(c + d*x)* sin(c + d*x)**5*a**9*b + 16*cos(c + d*x)*sin(c + d*x)**5*a**7*b**3 + 24*co s(c + d*x)*sin(c + d*x)**5*a**5*b**5 + 16*cos(c + d*x)*sin(c + d*x)**5*a** 3*b**7 + 4*cos(c + d*x)*sin(c + d*x)**5*a*b**9 + 2*cos(c + d*x)*sin(c + d* x)**3*a**9*b - 8*cos(c + d*x)*sin(c + d*x)**3*a**7*b**3 - 36*cos(c + d*x)* sin(c + d*x)**3*a**5*b**5 - 40*cos(c + d*x)*sin(c + d*x)**3*a**3*b**7 - 14 *cos(c + d*x)*sin(c + d*x)**3*a*b**9 + 12*cos(c + d*x)*sin(c + d*x)*a**8*b **2*d*x - 300*cos(c + d*x)*sin(c + d*x)*a**6*b**4*d*x + 420*cos(c + d*x)*s in(c + d*x)*a**4*b**6*d*x - 36*cos(c + d*x)*sin(c + d*x)*a**2*b**8*d*x + 4 8*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*a**8*b**2 - 288*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*a**6*b**4 + 336*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*a**4*b**6 - 96*log(tan((c + d*x)/2)**2 + 1)*sin(c + d *x)**2*a**2*b**8 - 48*log(tan((c + d*x)/2)**2 + 1)*a**8*b**2 + 240*log(tan ((c + d*x)/2)**2 + 1)*a**6*b**4 - 96*log(tan((c + d*x)/2)**2 + 1)*a**4*...