Integrand size = 21, antiderivative size = 264 \[ \int \frac {\sin ^2(c+d x)}{(a+b \tan (c+d x))^4} \, dx=\frac {\left (a^6-25 a^4 b^2+35 a^2 b^4-3 b^6\right ) x}{2 \left (a^2+b^2\right )^5}+\frac {4 a b \left (a^4-5 a^2 b^2+2 b^4\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right )^5 d}-\frac {a^2 b}{3 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))^3}-\frac {a b \left (a^2-b^2\right )}{\left (a^2+b^2\right )^3 d (a+b \tan (c+d x))^2}-\frac {b \left (3 a^4-8 a^2 b^2+b^4\right )}{\left (a^2+b^2\right )^4 d (a+b \tan (c+d x))}-\frac {\cos ^2(c+d x) \left (4 a b \left (a^2-b^2\right )+\left (a^4-6 a^2 b^2+b^4\right ) \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^4 d} \] Output:
1/2*(a^6-25*a^4*b^2+35*a^2*b^4-3*b^6)*x/(a^2+b^2)^5+4*a*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-1/3*a^2*b/(a^2+b^2)^2/d/ (a+b*tan(d*x+c))^3-a*b*(a^2-b^2)/(a^2+b^2)^3/d/(a+b*tan(d*x+c))^2-b*(3*a^4 -8*a^2*b^2+b^4)/(a^2+b^2)^4/d/(a+b*tan(d*x+c))-1/2*cos(d*x+c)^2*(4*a*b*(a^ 2-b^2)+(a^4-6*a^2*b^2+b^4)*tan(d*x+c))/(a^2+b^2)^4/d
Time = 2.40 (sec) , antiderivative size = 395, normalized size of antiderivative = 1.50 \[ \int \frac {\sin ^2(c+d x)}{(a+b \tan (c+d x))^4} \, dx=-\frac {b \left (\frac {3 \left (a^2+b^2\right ) \left (a^4-6 a^2 b^2+b^4\right ) \arctan (\tan (c+d x))}{b}+12 a (a-b) (a+b) \left (a^2+b^2\right ) \cos ^2(c+d x)+3 \left (4 a^5-20 a^3 b^2+8 a b^4+\frac {-a^6+15 a^4 b^2-15 a^2 b^4+b^6}{\sqrt {-b^2}}\right ) \log \left (\sqrt {-b^2}-b \tan (c+d x)\right )-24 a \left (a^4-5 a^2 b^2+2 b^4\right ) \log (a+b \tan (c+d x))+3 \left (4 a^5-20 a^3 b^2+8 a b^4+\frac {a^6-15 a^4 b^2+15 a^2 b^4-b^6}{\sqrt {-b^2}}\right ) \log \left (\sqrt {-b^2}+b \tan (c+d x)\right )+\frac {3 \left (a^2+b^2\right ) \left (a^4-6 a^2 b^2+b^4\right ) \sin (2 (c+d x))}{2 b}+\frac {2 a^2 \left (a^2+b^2\right )^3}{(a+b \tan (c+d x))^3}+\frac {6 a (a-b) (a+b) \left (a^2+b^2\right )^2}{(a+b \tan (c+d x))^2}+\frac {6 \left (a^2+b^2\right ) \left (3 a^4-8 a^2 b^2+b^4\right )}{a+b \tan (c+d x)}\right )}{6 \left (a^2+b^2\right )^5 d} \] Input:
Integrate[Sin[c + d*x]^2/(a + b*Tan[c + d*x])^4,x]
Output:
-1/6*(b*((3*(a^2 + b^2)*(a^4 - 6*a^2*b^2 + b^4)*ArcTan[Tan[c + d*x]])/b + 12*a*(a - b)*(a + b)*(a^2 + b^2)*Cos[c + d*x]^2 + 3*(4*a^5 - 20*a^3*b^2 + 8*a*b^4 + (-a^6 + 15*a^4*b^2 - 15*a^2*b^4 + b^6)/Sqrt[-b^2])*Log[Sqrt[-b^2 ] - b*Tan[c + d*x]] - 24*a*(a^4 - 5*a^2*b^2 + 2*b^4)*Log[a + b*Tan[c + d*x ]] + 3*(4*a^5 - 20*a^3*b^2 + 8*a*b^4 + (a^6 - 15*a^4*b^2 + 15*a^2*b^4 - b^ 6)/Sqrt[-b^2])*Log[Sqrt[-b^2] + b*Tan[c + d*x]] + (3*(a^2 + b^2)*(a^4 - 6* a^2*b^2 + b^4)*Sin[2*(c + d*x)])/(2*b) + (2*a^2*(a^2 + b^2)^3)/(a + b*Tan[ c + d*x])^3 + (6*a*(a - b)*(a + b)*(a^2 + b^2)^2)/(a + b*Tan[c + d*x])^2 + (6*(a^2 + b^2)*(3*a^4 - 8*a^2*b^2 + b^4))/(a + b*Tan[c + d*x])))/((a^2 + b^2)^5*d)
Time = 1.02 (sec) , antiderivative size = 329, normalized size of antiderivative = 1.25, 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, 25, 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.
| \(\displaystyle \int \frac {\sin ^2(c+d x)}{(a+b \tan (c+d x))^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^2}{(a+b \tan (c+d x))^4}dx\) |
| \(\Big \downarrow \) 3999 |
| \(\displaystyle \frac {b \int \frac {b^2 \tan ^2(c+d x)}{(a+b \tan (c+d x))^4 \left (\tan ^2(c+d x) b^2+b^2\right )^2}d(b \tan (c+d x))}{d}\) |
| \(\Big \downarrow \) 601 |
| \(\displaystyle \frac {b \left (-\frac {\int -\frac {-\frac {\left (a^4-6 b^2 a^2+b^4\right ) \tan ^4(c+d x) b^6}{\left (a^2+b^2\right )^4}-\frac {4 a \left (a^4-4 b^2 a^2-b^4\right ) \tan ^3(c+d x) b^5}{\left (a^2+b^2\right )^4}-\frac {2 \left (3 a^4-6 b^2 a^2-b^4\right ) \tan ^2(c+d x) b^4}{\left (a^2+b^2\right )^3}-\frac {4 a^3 \left (a^4+4 b^2 a^2-b^4\right ) \tan (c+d x) b^3}{\left (a^2+b^2\right )^4}+\frac {a^4 \left (a^4-6 b^2 a^2+b^4\right ) b^2}{\left (a^2+b^2\right )^4}}{(a+b \tan (c+d x))^4 \left (\tan ^2(c+d x) b^2+b^2\right )}d(b \tan (c+d x))}{2 b^2}-\frac {4 a b^2 \left (a^2-b^2\right )+b \left (a^4-6 a^2 b^2+b^4\right ) \tan (c+d x)}{2 \left (a^2+b^2\right )^4 \left (b^2 \tan ^2(c+d x)+b^2\right )}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b \left (\frac {\int \frac {-\frac {\left (a^4-6 b^2 a^2+b^4\right ) \tan ^4(c+d x) b^6}{\left (a^2+b^2\right )^4}-\frac {4 a \left (a^4-4 b^2 a^2-b^4\right ) \tan ^3(c+d x) b^5}{\left (a^2+b^2\right )^4}-\frac {2 \left (3 a^4-6 b^2 a^2-b^4\right ) \tan ^2(c+d x) b^4}{\left (a^2+b^2\right )^3}-\frac {4 a^3 \left (a^4+4 b^2 a^2-b^4\right ) \tan (c+d x) b^3}{\left (a^2+b^2\right )^4}+\frac {a^4 \left (a^4-6 b^2 a^2+b^4\right ) b^2}{\left (a^2+b^2\right )^4}}{(a+b \tan (c+d x))^4 \left (\tan ^2(c+d x) b^2+b^2\right )}d(b \tan (c+d x))}{2 b^2}-\frac {4 a b^2 \left (a^2-b^2\right )+b \left (a^4-6 a^2 b^2+b^4\right ) \tan (c+d x)}{2 \left (a^2+b^2\right )^4 \left (b^2 \tan ^2(c+d x)+b^2\right )}\right )}{d}\) |
| \(\Big \downarrow \) 2160 |
| \(\displaystyle \frac {b \left (\frac {\int \left (\frac {8 a \left (a^4-5 b^2 a^2+2 b^4\right ) b^2}{\left (a^2+b^2\right )^5 (a+b \tan (c+d x))}+\frac {\left (a^6-25 b^2 a^4+35 b^4 a^2-8 b \left (a^4-5 b^2 a^2+2 b^4\right ) \tan (c+d x) a-3 b^6\right ) b^2}{\left (a^2+b^2\right )^5 \left (\tan ^2(c+d x) b^2+b^2\right )}+\frac {4 a \left (a^2-b^2\right ) b^2}{\left (a^2+b^2\right )^3 (a+b \tan (c+d x))^3}+\frac {2 a^2 b^2}{\left (a^2+b^2\right )^2 (a+b \tan (c+d x))^4}+\frac {2 \left (b^6-8 a^2 b^4+3 a^4 b^2\right )}{\left (a^2+b^2\right )^4 (a+b \tan (c+d x))^2}\right )d(b \tan (c+d x))}{2 b^2}-\frac {4 a b^2 \left (a^2-b^2\right )+b \left (a^4-6 a^2 b^2+b^4\right ) \tan (c+d x)}{2 \left (a^2+b^2\right )^4 \left (b^2 \tan ^2(c+d x)+b^2\right )}\right )}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b \left (\frac {-\frac {2 a b^2 \left (a^2-b^2\right )}{\left (a^2+b^2\right )^3 (a+b \tan (c+d x))^2}-\frac {2 a^2 b^2}{3 \left (a^2+b^2\right )^2 (a+b \tan (c+d x))^3}-\frac {2 b^2 \left (3 a^4-8 a^2 b^2+b^4\right )}{\left (a^2+b^2\right )^4 (a+b \tan (c+d x))}-\frac {4 a b^2 \left (a^4-5 a^2 b^2+2 b^4\right ) \log \left (b^2 \tan ^2(c+d x)+b^2\right )}{\left (a^2+b^2\right )^5}+\frac {8 a b^2 \left (a^4-5 a^2 b^2+2 b^4\right ) \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^5}+\frac {b \left (a^6-25 a^4 b^2+35 a^2 b^4-3 b^6\right ) \arctan (\tan (c+d x))}{\left (a^2+b^2\right )^5}}{2 b^2}-\frac {4 a b^2 \left (a^2-b^2\right )+b \left (a^4-6 a^2 b^2+b^4\right ) \tan (c+d x)}{2 \left (a^2+b^2\right )^4 \left (b^2 \tan ^2(c+d x)+b^2\right )}\right )}{d}\) |
Input:
Int[Sin[c + d*x]^2/(a + b*Tan[c + d*x])^4,x]
Output:
(b*(-1/2*(4*a*b^2*(a^2 - b^2) + b*(a^4 - 6*a^2*b^2 + b^4)*Tan[c + d*x])/(( a^2 + b^2)^4*(b^2 + b^2*Tan[c + d*x]^2)) + ((b*(a^6 - 25*a^4*b^2 + 35*a^2* b^4 - 3*b^6)*ArcTan[Tan[c + d*x]])/(a^2 + b^2)^5 + (8*a*b^2*(a^4 - 5*a^2*b ^2 + 2*b^4)*Log[a + b*Tan[c + d*x]])/(a^2 + b^2)^5 - (4*a*b^2*(a^4 - 5*a^2 *b^2 + 2*b^4)*Log[b^2 + b^2*Tan[c + d*x]^2])/(a^2 + b^2)^5 - (2*a^2*b^2)/( 3*(a^2 + b^2)^2*(a + b*Tan[c + d*x])^3) - (2*a*b^2*(a^2 - b^2))/((a^2 + b^ 2)^3*(a + b*Tan[c + d*x])^2) - (2*b^2*(3*a^4 - 8*a^2*b^2 + b^4))/((a^2 + b ^2)^4*(a + b*Tan[c + d*x])))/(2*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[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 = 29.54 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.09
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {\left (-\frac {1}{2} a^{6}+\frac {5}{2} a^{4} b^{2}+\frac {5}{2} a^{2} b^{4}-\frac {1}{2} b^{6}\right ) \tan \left (d x +c \right )-2 a^{5} b +2 a \,b^{5}}{1+\tan \left (d x +c \right )^{2}}+\frac {\left (-8 a^{5} b +40 a^{3} b^{3}-16 a \,b^{5}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{4}+\frac {\left (a^{6}-25 a^{4} b^{2}+35 a^{2} b^{4}-3 b^{6}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{2}}{\left (a^{2}+b^{2}\right )^{5}}-\frac {a^{2} b}{3 \left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )^{3}}-\frac {b \left (3 a^{4}-8 b^{2} a^{2}+b^{4}\right )}{\left (a^{2}+b^{2}\right )^{4} \left (a +b \tan \left (d x +c \right )\right )}-\frac {a b \left (a^{2}-b^{2}\right )}{\left (a^{2}+b^{2}\right )^{3} \left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {4 a b \left (a^{4}-5 b^{2} a^{2}+2 b^{4}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{5}}}{d}\) | \(288\) |
| default | \(\frac {\frac {\frac {\left (-\frac {1}{2} a^{6}+\frac {5}{2} a^{4} b^{2}+\frac {5}{2} a^{2} b^{4}-\frac {1}{2} b^{6}\right ) \tan \left (d x +c \right )-2 a^{5} b +2 a \,b^{5}}{1+\tan \left (d x +c \right )^{2}}+\frac {\left (-8 a^{5} b +40 a^{3} b^{3}-16 a \,b^{5}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{4}+\frac {\left (a^{6}-25 a^{4} b^{2}+35 a^{2} b^{4}-3 b^{6}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{2}}{\left (a^{2}+b^{2}\right )^{5}}-\frac {a^{2} b}{3 \left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )^{3}}-\frac {b \left (3 a^{4}-8 b^{2} a^{2}+b^{4}\right )}{\left (a^{2}+b^{2}\right )^{4} \left (a +b \tan \left (d x +c \right )\right )}-\frac {a b \left (a^{2}-b^{2}\right )}{\left (a^{2}+b^{2}\right )^{3} \left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {4 a b \left (a^{4}-5 b^{2} a^{2}+2 b^{4}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{5}}}{d}\) | \(288\) |
| risch | \(-\frac {3 i x b}{2 \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 {x a}{2 \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 {i {\mathrm e}^{2 i \left (d x +c \right )}}{8 \left (-4 i a^{3} b +4 i a \,b^{3}+a^{4}-6 b^{2} a^{2}+b^{4}\right ) d}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 \left (4 i a^{3} b -4 i a \,b^{3}+a^{4}-6 b^{2} a^{2}+b^{4}\right ) d}-\frac {8 i a^{5} 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 {40 i a^{3} 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 {16 i a \,b^{5} 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 {8 i a^{5} 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 {40 i a^{3} 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 {16 i a \,b^{5} 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 {2 i b^{2} \left (6 i a \,b^{5} {\mathrm e}^{2 i \left (d x +c \right )}-12 i a^{5} b \,{\mathrm e}^{2 i \left (d x +c \right )}-18 a^{6} {\mathrm e}^{4 i \left (d x +c \right )}+27 a^{4} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-24 a^{2} b^{4} {\mathrm e}^{4 i \left (d x +c \right )}+3 b^{6} {\mathrm e}^{4 i \left (d x +c \right )}-6 i a \,b^{5}-48 i a^{3} b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-6 i a^{3} b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-36 a^{6} {\mathrm e}^{2 i \left (d x +c \right )}+24 a^{4} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+54 a^{2} b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-6 b^{6} {\mathrm e}^{2 i \left (d x +c \right )}-36 i a^{5} b +62 i a^{3} b^{3}+24 i a^{5} b \,{\mathrm e}^{4 i \left (d x +c \right )}-18 a^{6}+49 a^{4} b^{2}-34 a^{2} b^{4}+3 b^{6}\right )}{3 \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 )^{3} \left (-i a +b \right )^{4} d \left (i a +b \right )^{5}}+\frac {4 a^{5} 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 {20 a^{3} 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 {8 a \,b^{5} \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 )}\) | \(1054\) |
Input:
int(sin(d*x+c)^2/(a+b*tan(d*x+c))^4,x,method=_RETURNVERBOSE)
Output:
1/d*(1/(a^2+b^2)^5*(((-1/2*a^6+5/2*a^4*b^2+5/2*a^2*b^4-1/2*b^6)*tan(d*x+c) -2*a^5*b+2*a*b^5)/(1+tan(d*x+c)^2)+1/4*(-8*a^5*b+40*a^3*b^3-16*a*b^5)*ln(1 +tan(d*x+c)^2)+1/2*(a^6-25*a^4*b^2+35*a^2*b^4-3*b^6)*arctan(tan(d*x+c)))-1 /3*a^2*b/(a^2+b^2)^2/(a+b*tan(d*x+c))^3-b*(3*a^4-8*a^2*b^2+b^4)/(a^2+b^2)^ 4/(a+b*tan(d*x+c))-a*b*(a^2-b^2)/(a^2+b^2)^3/(a+b*tan(d*x+c))^2+4*a*b*(a^4 -5*a^2*b^2+2*b^4)/(a^2+b^2)^5*ln(a+b*tan(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 802 vs. \(2 (258) = 516\).
Time = 0.16 (sec) , antiderivative size = 802, normalized size of antiderivative = 3.04 \[ \int \frac {\sin ^2(c+d x)}{(a+b \tan (c+d x))^4} \, dx =\text {Too large to display} \] Input:
integrate(sin(d*x+c)^2/(a+b*tan(d*x+c))^4,x, algorithm="fricas")
Output:
-1/6*(3*(a^8*b + 4*a^6*b^3 + 6*a^4*b^5 + 4*a^2*b^7 + b^9)*cos(d*x + c)^5 + (3*a^8*b + 111*a^6*b^3 - 231*a^4*b^5 + 65*a^2*b^7 - 12*b^9 - 3*(a^9 - 28* a^7*b^2 + 110*a^5*b^4 - 108*a^3*b^6 + 9*a*b^8)*d*x)*cos(d*x + c)^3 - 3*(25 *a^6*b^3 - 51*a^4*b^5 + 25*a^2*b^7 - 3*b^9 + 3*(a^7*b^2 - 25*a^5*b^4 + 35* a^3*b^6 - 3*a*b^8)*d*x)*cos(d*x + c) - 12*((a^8*b - 8*a^6*b^3 + 17*a^4*b^5 - 6*a^2*b^7)*cos(d*x + c)^3 + 3*(a^6*b^3 - 5*a^4*b^5 + 2*a^2*b^7)*cos(d*x + c) + (a^5*b^4 - 5*a^3*b^6 + 2*a*b^8 + (3*a^7*b^2 - 16*a^5*b^4 + 11*a^3* b^6 - 2*a*b^8)*cos(d*x + c)^2)*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) - (32*a^5*b^4 - 66*a^3*b^6 + 6* a*b^8 - 3*(a^9 + 4*a^7*b^2 + 6*a^5*b^4 + 4*a^3*b^6 + a*b^8)*cos(d*x + c)^4 + 3*(a^6*b^3 - 25*a^4*b^5 + 35*a^2*b^7 - 3*b^9)*d*x + (45*a^7*b^2 - 143*a ^5*b^4 + 219*a^3*b^6 - 9*a*b^8 + 3*(3*a^8*b - 76*a^6*b^3 + 130*a^4*b^5 - 4 4*a^2*b^7 + 3*b^9)*d*x)*cos(d*x + c)^2)*sin(d*x + c))/((a^13 + 2*a^11*b^2 - 5*a^9*b^4 - 20*a^7*b^6 - 25*a^5*b^8 - 14*a^3*b^10 - 3*a*b^12)*d*cos(d*x + c)^3 + 3*(a^11*b^2 + 5*a^9*b^4 + 10*a^7*b^6 + 10*a^5*b^8 + 5*a^3*b^10 + a*b^12)*d*cos(d*x + c) + ((3*a^12*b + 14*a^10*b^3 + 25*a^8*b^5 + 20*a^6*b^ 7 + 5*a^4*b^9 - 2*a^2*b^11 - b^13)*d*cos(d*x + c)^2 + (a^10*b^3 + 5*a^8*b^ 5 + 10*a^6*b^7 + 10*a^4*b^9 + 5*a^2*b^11 + b^13)*d)*sin(d*x + c))
Exception generated. \[ \int \frac {\sin ^2(c+d x)}{(a+b \tan (c+d x))^4} \, dx=\text {Exception raised: AttributeError} \] Input:
integrate(sin(d*x+c)**2/(a+b*tan(d*x+c))**4,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. 662 vs. \(2 (258) = 516\).
Time = 0.13 (sec) , antiderivative size = 662, normalized size of antiderivative = 2.51 \[ \int \frac {\sin ^2(c+d x)}{(a+b \tan (c+d x))^4} \, dx=\frac {\frac {3 \, {\left (a^{6} - 25 \, a^{4} b^{2} + 35 \, a^{2} b^{4} - 3 \, b^{6}\right )} {\left (d x + c\right )}}{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 {24 \, {\left (a^{5} b - 5 \, a^{3} b^{3} + 2 \, a b^{5}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{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 {12 \, {\left (a^{5} b - 5 \, a^{3} b^{3} + 2 \, a b^{5}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{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 {38 \, a^{6} b - 56 \, a^{4} b^{3} + 2 \, a^{2} b^{5} + 3 \, {\left (7 \, a^{4} b^{3} - 22 \, a^{2} b^{5} + 3 \, b^{7}\right )} \tan \left (d x + c\right )^{4} + 3 \, {\left (17 \, a^{5} b^{2} - 46 \, a^{3} b^{4} + a b^{6}\right )} \tan \left (d x + c\right )^{3} + {\left (35 \, a^{6} b - 44 \, a^{4} b^{3} - 73 \, a^{2} b^{5} + 6 \, b^{7}\right )} \tan \left (d x + c\right )^{2} + 3 \, {\left (a^{7} + 20 \, a^{5} b^{2} - 43 \, a^{3} b^{4} + 2 \, a b^{6}\right )} \tan \left (d x + c\right )}{a^{11} + 4 \, a^{9} b^{2} + 6 \, a^{7} b^{4} + 4 \, a^{5} b^{6} + a^{3} b^{8} + {\left (a^{8} b^{3} + 4 \, a^{6} b^{5} + 6 \, a^{4} b^{7} + 4 \, a^{2} b^{9} + b^{11}\right )} \tan \left (d x + c\right )^{5} + 3 \, {\left (a^{9} b^{2} + 4 \, a^{7} b^{4} + 6 \, a^{5} b^{6} + 4 \, a^{3} b^{8} + a b^{10}\right )} \tan \left (d x + c\right )^{4} + {\left (3 \, a^{10} b + 13 \, a^{8} b^{3} + 22 \, a^{6} b^{5} + 18 \, a^{4} b^{7} + 7 \, a^{2} b^{9} + b^{11}\right )} \tan \left (d x + c\right )^{3} + {\left (a^{11} + 7 \, a^{9} b^{2} + 18 \, a^{7} b^{4} + 22 \, a^{5} b^{6} + 13 \, a^{3} b^{8} + 3 \, a b^{10}\right )} \tan \left (d x + c\right )^{2} + 3 \, {\left (a^{10} b + 4 \, a^{8} b^{3} + 6 \, a^{6} b^{5} + 4 \, a^{4} b^{7} + a^{2} b^{9}\right )} \tan \left (d x + c\right )}}{6 \, d} \] Input:
integrate(sin(d*x+c)^2/(a+b*tan(d*x+c))^4,x, algorithm="maxima")
Output:
1/6*(3*(a^6 - 25*a^4*b^2 + 35*a^2*b^4 - 3*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^5*b - 5*a^3*b^3 + 2 *a*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^5*b - 5*a^3*b^3 + 2*a*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^10) - ( 38*a^6*b - 56*a^4*b^3 + 2*a^2*b^5 + 3*(7*a^4*b^3 - 22*a^2*b^5 + 3*b^7)*tan (d*x + c)^4 + 3*(17*a^5*b^2 - 46*a^3*b^4 + a*b^6)*tan(d*x + c)^3 + (35*a^6 *b - 44*a^4*b^3 - 73*a^2*b^5 + 6*b^7)*tan(d*x + c)^2 + 3*(a^7 + 20*a^5*b^2 - 43*a^3*b^4 + 2*a*b^6)*tan(d*x + c))/(a^11 + 4*a^9*b^2 + 6*a^7*b^4 + 4*a ^5*b^6 + a^3*b^8 + (a^8*b^3 + 4*a^6*b^5 + 6*a^4*b^7 + 4*a^2*b^9 + b^11)*ta n(d*x + c)^5 + 3*(a^9*b^2 + 4*a^7*b^4 + 6*a^5*b^6 + 4*a^3*b^8 + a*b^10)*ta n(d*x + c)^4 + (3*a^10*b + 13*a^8*b^3 + 22*a^6*b^5 + 18*a^4*b^7 + 7*a^2*b^ 9 + b^11)*tan(d*x + c)^3 + (a^11 + 7*a^9*b^2 + 18*a^7*b^4 + 22*a^5*b^6 + 1 3*a^3*b^8 + 3*a*b^10)*tan(d*x + c)^2 + 3*(a^10*b + 4*a^8*b^3 + 6*a^6*b^5 + 4*a^4*b^7 + a^2*b^9)*tan(d*x + c)))/d
Time = 0.28 (sec) , antiderivative size = 486, normalized size of antiderivative = 1.84 \[ \int \frac {\sin ^2(c+d x)}{(a+b \tan (c+d x))^4} \, dx=\frac {{\left (a^{6} - 25 \, a^{4} b^{2} + 35 \, a^{2} b^{4} - 3 \, b^{6}\right )} {\left (d x + c\right )}}{2 \, {\left (a^{10} d + 5 \, a^{8} b^{2} d + 10 \, a^{6} b^{4} d + 10 \, a^{4} b^{6} d + 5 \, a^{2} b^{8} d + b^{10} d\right )}} - \frac {2 \, {\left (a^{5} b - 5 \, a^{3} b^{3} + 2 \, a b^{5}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{10} d + 5 \, a^{8} b^{2} d + 10 \, a^{6} b^{4} d + 10 \, a^{4} b^{6} d + 5 \, a^{2} b^{8} d + b^{10} d} + \frac {4 \, {\left (a^{5} b^{2} - 5 \, a^{3} b^{4} + 2 \, a b^{6}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{10} b d + 5 \, a^{8} b^{3} d + 10 \, a^{6} b^{5} d + 10 \, a^{4} b^{7} d + 5 \, a^{2} b^{9} d + b^{11} d} - \frac {38 \, a^{8} b - 18 \, a^{6} b^{3} - 54 \, a^{4} b^{5} + 2 \, a^{2} b^{7} + 3 \, {\left (7 \, a^{6} b^{3} - 15 \, a^{4} b^{5} - 19 \, a^{2} b^{7} + 3 \, b^{9}\right )} \tan \left (d x + c\right )^{4} + 3 \, {\left (17 \, a^{7} b^{2} - 29 \, a^{5} b^{4} - 45 \, a^{3} b^{6} + a b^{8}\right )} \tan \left (d x + c\right )^{3} + {\left (35 \, a^{8} b - 9 \, a^{6} b^{3} - 117 \, a^{4} b^{5} - 67 \, a^{2} b^{7} + 6 \, b^{9}\right )} \tan \left (d x + c\right )^{2} + 3 \, {\left (a^{9} + 21 \, a^{7} b^{2} - 23 \, a^{5} b^{4} - 41 \, a^{3} b^{6} + 2 \, a b^{8}\right )} \tan \left (d x + c\right )}{6 \, {\left (a^{2} + b^{2}\right )}^{5} {\left (b \tan \left (d x + c\right ) + a\right )}^{3} {\left (\tan \left (d x + c\right )^{2} + 1\right )} d} \] Input:
integrate(sin(d*x+c)^2/(a+b*tan(d*x+c))^4,x, algorithm="giac")
Output:
1/2*(a^6 - 25*a^4*b^2 + 35*a^2*b^4 - 3*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) - 2*(a^5*b - 5*a^3 *b^3 + 2*a*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) + 4*(a^5*b^2 - 5*a^3*b^4 + 2*a*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) - 1/6*(38*a^8*b - 18*a^6*b^3 - 54*a^4 *b^5 + 2*a^2*b^7 + 3*(7*a^6*b^3 - 15*a^4*b^5 - 19*a^2*b^7 + 3*b^9)*tan(d*x + c)^4 + 3*(17*a^7*b^2 - 29*a^5*b^4 - 45*a^3*b^6 + a*b^8)*tan(d*x + c)^3 + (35*a^8*b - 9*a^6*b^3 - 117*a^4*b^5 - 67*a^2*b^7 + 6*b^9)*tan(d*x + c)^2 + 3*(a^9 + 21*a^7*b^2 - 23*a^5*b^4 - 41*a^3*b^6 + 2*a*b^8)*tan(d*x + c))/ ((a^2 + b^2)^5*(b*tan(d*x + c) + a)^3*(tan(d*x + c)^2 + 1)*d)
Time = 2.28 (sec) , antiderivative size = 597, normalized size of antiderivative = 2.26 \[ \int \frac {\sin ^2(c+d x)}{(a+b \tan (c+d x))^4} \, dx=\frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (\frac {4\,a\,b}{{\left (a^2+b^2\right )}^3}-\frac {28\,a\,b^3}{{\left (a^2+b^2\right )}^4}+\frac {32\,a\,b^5}{{\left (a^2+b^2\right )}^5}\right )}{d}-\frac {\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (35\,a^4\,b-79\,a^2\,b^3+6\,b^5\right )}{6\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (7\,a^4\,b^3-22\,a^2\,b^5+3\,b^7\right )}{2\,\left (a^8+4\,a^6\,b^2+6\,a^4\,b^4+4\,a^2\,b^6+b^8\right )}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (17\,a^5\,b^2-46\,a^3\,b^4+a\,b^6\right )}{2\,\left (a^8+4\,a^6\,b^2+6\,a^4\,b^4+4\,a^2\,b^6+b^8\right )}+\frac {a^2\,\left (19\,a^4\,b-28\,a^2\,b^3+b^5\right )}{3\,\left (a^2+b^2\right )\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}+\frac {a\,\mathrm {tan}\left (c+d\,x\right )\,\left (a^6+20\,a^4\,b^2-43\,a^2\,b^4+2\,b^6\right )}{2\,\left (a^2+b^2\right )\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}}{d\,\left (a^3+{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (a^3+3\,a\,b^2\right )+{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (3\,a^2\,b+b^3\right )+b^3\,{\mathrm {tan}\left (c+d\,x\right )}^5+3\,a\,b^2\,{\mathrm {tan}\left (c+d\,x\right )}^4+3\,a^2\,b\,\mathrm {tan}\left (c+d\,x\right )\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (3\,b+a\,1{}\mathrm {i}\right )}{4\,d\,\left (a^5+a^4\,b\,5{}\mathrm {i}-10\,a^3\,b^2-a^2\,b^3\,10{}\mathrm {i}+5\,a\,b^4+b^5\,1{}\mathrm {i}\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (-3\,b+a\,1{}\mathrm {i}\right )}{4\,d\,\left (a^5-a^4\,b\,5{}\mathrm {i}-10\,a^3\,b^2+a^2\,b^3\,10{}\mathrm {i}+5\,a\,b^4-b^5\,1{}\mathrm {i}\right )} \] Input:
int(sin(c + d*x)^2/(a + b*tan(c + d*x))^4,x)
Output:
(log(a + b*tan(c + d*x))*((4*a*b)/(a^2 + b^2)^3 - (28*a*b^3)/(a^2 + b^2)^4 + (32*a*b^5)/(a^2 + b^2)^5))/d - ((tan(c + d*x)^2*(35*a^4*b + 6*b^5 - 79* a^2*b^3))/(6*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)) + (tan(c + d*x)^4*(3*b^7 - 22*a^2*b^5 + 7*a^4*b^3))/(2*(a^8 + b^8 + 4*a^2*b^6 + 6*a^4*b^4 + 4*a^6* b^2)) + (tan(c + d*x)^3*(a*b^6 - 46*a^3*b^4 + 17*a^5*b^2))/(2*(a^8 + b^8 + 4*a^2*b^6 + 6*a^4*b^4 + 4*a^6*b^2)) + (a^2*(19*a^4*b + b^5 - 28*a^2*b^3)) /(3*(a^2 + b^2)*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)) + (a*tan(c + d*x)*(a^ 6 + 2*b^6 - 43*a^2*b^4 + 20*a^4*b^2))/(2*(a^2 + b^2)*(a^6 + b^6 + 3*a^2*b^ 4 + 3*a^4*b^2)))/(d*(a^3 + tan(c + d*x)^2*(3*a*b^2 + a^3) + tan(c + d*x)^3 *(3*a^2*b + b^3) + b^3*tan(c + d*x)^5 + 3*a*b^2*tan(c + d*x)^4 + 3*a^2*b*t an(c + d*x))) - (log(tan(c + d*x) - 1i)*(a*1i + 3*b))/(4*d*(5*a*b^4 + a^4* b*5i + a^5 + b^5*1i - a^2*b^3*10i - 10*a^3*b^2)) + (log(tan(c + d*x) + 1i) *(a*1i - 3*b))/(4*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 = 0.24 (sec) , antiderivative size = 2289, normalized size of antiderivative = 8.67 \[ \int \frac {\sin ^2(c+d x)}{(a+b \tan (c+d x))^4} \, dx =\text {Too large to display} \] Input:
int(sin(d*x+c)^2/(a+b*tan(d*x+c))^4,x)
Output:
( - 24*cos(c + d*x)*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*a**9*b**2 + 192*cos(c + d*x)*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*a**7*b**4 - 408*cos(c + d*x)*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*a**5*b**6 + 144*cos(c + d*x)*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*a**3*b**8 + 24*cos(c + d*x)*log(tan((c + d*x)/2)**2 + 1)*a**9*b**2 - 120*cos(c + d* x)*log(tan((c + d*x)/2)**2 + 1)*a**7*b**4 + 48*cos(c + d*x)*log(tan((c + d *x)/2)**2 + 1)*a**5*b**6 + 24*cos(c + d*x)*log(tan((c + d*x)/2)**2*a - 2*t an((c + d*x)/2)*b - a)*sin(c + d*x)**2*a**9*b**2 - 192*cos(c + d*x)*log(ta n((c + d*x)/2)**2*a - 2*tan((c + d*x)/2)*b - a)*sin(c + d*x)**2*a**7*b**4 + 408*cos(c + d*x)*log(tan((c + d*x)/2)**2*a - 2*tan((c + d*x)/2)*b - a)*s in(c + d*x)**2*a**5*b**6 - 144*cos(c + d*x)*log(tan((c + d*x)/2)**2*a - 2* tan((c + d*x)/2)*b - a)*sin(c + d*x)**2*a**3*b**8 - 24*cos(c + d*x)*log(ta n((c + d*x)/2)**2*a - 2*tan((c + d*x)/2)*b - a)*a**9*b**2 + 120*cos(c + d* x)*log(tan((c + d*x)/2)**2*a - 2*tan((c + d*x)/2)*b - a)*a**7*b**4 - 48*co s(c + d*x)*log(tan((c + d*x)/2)**2*a - 2*tan((c + d*x)/2)*b - a)*a**5*b**6 + 3*cos(c + d*x)*sin(c + d*x)**4*a**9*b**2 + 12*cos(c + d*x)*sin(c + d*x) **4*a**7*b**4 + 18*cos(c + d*x)*sin(c + d*x)**4*a**5*b**6 + 12*cos(c + d*x )*sin(c + d*x)**4*a**3*b**8 + 3*cos(c + d*x)*sin(c + d*x)**4*a*b**10 + cos (c + d*x)*sin(c + d*x)**2*a**11 + 3*cos(c + d*x)*sin(c + d*x)**2*a**10*b*d *x - 23*cos(c + d*x)*sin(c + d*x)**2*a**9*b**2 - 84*cos(c + d*x)*sin(c ...