Integrand size = 29, antiderivative size = 269 \[ \int \frac {\sin ^3(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\left (35 a^2+57 a b+24 b^2\right ) \log (1-\sin (c+d x))}{16 (a+b)^3 d}+\frac {\left (35 a^2-57 a b+24 b^2\right ) \log (1+\sin (c+d x))}{16 (a-b)^3 d}-\frac {a^8 \log (a+b \sin (c+d x))}{b^3 \left (a^2-b^2\right )^3 d}-\frac {13 a^2-3 a b-12 b^2}{16 (a-b) (a+b)^2 d (1-\sin (c+d x))}+\frac {a \sin (c+d x)}{b^2 d}-\frac {\sin ^2(c+d x)}{2 b d}+\frac {13 a^2+3 a b-12 b^2}{16 (a-b)^2 (a+b) d (1+\sin (c+d x))}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x))}{4 \left (a^2-b^2\right ) d} \] Output:
-1/16*(35*a^2+57*a*b+24*b^2)*ln(1-sin(d*x+c))/(a+b)^3/d+1/16*(35*a^2-57*a* b+24*b^2)*ln(1+sin(d*x+c))/(a-b)^3/d-a^8*ln(a+b*sin(d*x+c))/b^3/(a^2-b^2)^ 3/d-1/16*(13*a^2-3*a*b-12*b^2)/(a-b)/(a+b)^2/d/(1-sin(d*x+c))+a*sin(d*x+c) /b^2/d-1/2*sin(d*x+c)^2/b/d+1/16*(13*a^2+3*a*b-12*b^2)/(a-b)^2/(a+b)/d/(1+ sin(d*x+c))-1/4*sec(d*x+c)^4*(b-a*sin(d*x+c))/(a^2-b^2)/d
Time = 3.77 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.79 \[ \int \frac {\sin ^3(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-\frac {\left (35 a^2+57 a b+24 b^2\right ) \log (1-\sin (c+d x))}{(a+b)^3}+\frac {\left (35 a^2-57 a b+24 b^2\right ) \log (1+\sin (c+d x))}{(a-b)^3}-\frac {16 a^8 \log (a+b \sin (c+d x))}{(a-b)^3 b^3 (a+b)^3}+\frac {1}{(a+b) (-1+\sin (c+d x))^2}+\frac {13 a+11 b}{(a+b)^2 (-1+\sin (c+d x))}+\frac {16 a \sin (c+d x)}{b^2}-\frac {8 \sin ^2(c+d x)}{b}-\frac {1}{(a-b) (1+\sin (c+d x))^2}+\frac {13 a-11 b}{(a-b)^2 (1+\sin (c+d x))}}{16 d} \] Input:
Integrate[(Sin[c + d*x]^3*Tan[c + d*x]^5)/(a + b*Sin[c + d*x]),x]
Output:
(-(((35*a^2 + 57*a*b + 24*b^2)*Log[1 - Sin[c + d*x]])/(a + b)^3) + ((35*a^ 2 - 57*a*b + 24*b^2)*Log[1 + Sin[c + d*x]])/(a - b)^3 - (16*a^8*Log[a + b* Sin[c + d*x]])/((a - b)^3*b^3*(a + b)^3) + 1/((a + b)*(-1 + Sin[c + d*x])^ 2) + (13*a + 11*b)/((a + b)^2*(-1 + Sin[c + d*x])) + (16*a*Sin[c + d*x])/b ^2 - (8*Sin[c + d*x]^2)/b - 1/((a - b)*(1 + Sin[c + d*x])^2) + (13*a - 11* b)/((a - b)^2*(1 + Sin[c + d*x])))/(16*d)
Time = 1.00 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.10, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {3042, 3316, 27, 601, 2178, 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.
| \(\displaystyle \int \frac {\sin ^3(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^8}{\cos (c+d x)^5 (a+b \sin (c+d x))}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {b^5 \int \frac {\sin ^8(c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {b^8 \sin ^8(c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{b^3 d}\) |
| \(\Big \downarrow \) 601 |
| \(\displaystyle \frac {-\frac {\int \frac {-\frac {3 a \sin (c+d x) b^9}{a^2-b^2}+4 \sin ^6(c+d x) b^8+4 \sin ^4(c+d x) b^8+4 \sin ^2(c+d x) b^8+\frac {a^2 b^8}{a^2-b^2}}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}-\frac {b^6 \left (\frac {b^2}{a^2-b^2}-\frac {a b \sin (c+d x)}{a^2-b^2}\right )}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}}{b^3 d}\) |
| \(\Big \downarrow \) 2178 |
| \(\displaystyle \frac {-\frac {\frac {\int -\frac {-\frac {a \left (13 a^2-9 b^2\right ) \sin (c+d x) b^9}{\left (a^2-b^2\right )^2}+8 \sin ^4(c+d x) b^8+16 \sin ^2(c+d x) b^8+\frac {a^2 \left (11 a^2-7 b^2\right ) b^8}{\left (a^2-b^2\right )^2}}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2}-\frac {b^6 \left (4 b^2 \left (4 a^2-3 b^2\right )-a b \left (13 a^2-9 b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}-\frac {b^6 \left (\frac {b^2}{a^2-b^2}-\frac {a b \sin (c+d x)}{a^2-b^2}\right )}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}}{b^3 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {-\frac {\int \frac {-\frac {a \left (13 a^2-9 b^2\right ) \sin (c+d x) b^9}{\left (a^2-b^2\right )^2}+8 \sin ^4(c+d x) b^8+16 \sin ^2(c+d x) b^8+\frac {a^2 \left (11 a^2-7 b^2\right ) b^8}{\left (a^2-b^2\right )^2}}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2}-\frac {b^6 \left (4 b^2 \left (4 a^2-3 b^2\right )-a b \left (13 a^2-9 b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}-\frac {b^6 \left (\frac {b^2}{a^2-b^2}-\frac {a b \sin (c+d x)}{a^2-b^2}\right )}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}}{b^3 d}\) |
| \(\Big \downarrow \) 2160 |
| \(\displaystyle \frac {-\frac {-\frac {\int \left (-\frac {8 b^4 a^8}{(a-b)^3 (a+b)^3 (a+b \sin (c+d x))}+8 b^4 a-8 b^5 \sin (c+d x)+\frac {b^7 \left (35 a^2+57 b a+24 b^2\right )}{2 (a+b)^3 (b-b \sin (c+d x))}+\frac {b^7 \left (35 a^2-57 b a+24 b^2\right )}{2 (a-b)^3 (\sin (c+d x) b+b)}\right )d(b \sin (c+d x))}{2 b^2}-\frac {b^6 \left (4 b^2 \left (4 a^2-3 b^2\right )-a b \left (13 a^2-9 b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}-\frac {b^6 \left (\frac {b^2}{a^2-b^2}-\frac {a b \sin (c+d x)}{a^2-b^2}\right )}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}}{b^3 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {b^6 \left (\frac {b^2}{a^2-b^2}-\frac {a b \sin (c+d x)}{a^2-b^2}\right )}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {-\frac {b^6 \left (4 b^2 \left (4 a^2-3 b^2\right )-a b \left (13 a^2-9 b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {-\frac {b^7 \left (35 a^2+57 a b+24 b^2\right ) \log (b-b \sin (c+d x))}{2 (a+b)^3}+\frac {b^7 \left (35 a^2-57 a b+24 b^2\right ) \log (b \sin (c+d x)+b)}{2 (a-b)^3}-\frac {8 a^8 b^4 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3}+8 a b^5 \sin (c+d x)-4 b^6 \sin ^2(c+d x)}{2 b^2}}{4 b^2}}{b^3 d}\) |
Input:
Int[(Sin[c + d*x]^3*Tan[c + d*x]^5)/(a + b*Sin[c + d*x]),x]
Output:
(-1/4*(b^6*(b^2/(a^2 - b^2) - (a*b*Sin[c + d*x])/(a^2 - b^2)))/(b^2 - b^2* Sin[c + d*x]^2)^2 - (-1/2*(b^6*(4*b^2*(4*a^2 - 3*b^2) - a*b*(13*a^2 - 9*b^ 2)*Sin[c + d*x]))/((a^2 - b^2)^2*(b^2 - b^2*Sin[c + d*x]^2)) - (-1/2*(b^7* (35*a^2 + 57*a*b + 24*b^2)*Log[b - b*Sin[c + d*x]])/(a + b)^3 - (8*a^8*b^4 *Log[a + b*Sin[c + d*x]])/(a^2 - b^2)^3 + (b^7*(35*a^2 - 57*a*b + 24*b^2)* Log[b + b*Sin[c + d*x]])/(2*(a - b)^3) + 8*a*b^5*Sin[c + d*x] - 4*b^6*Sin[ c + d*x]^2)/(2*b^2))/(4*b^2))/(b^3*d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) /2] && NeQ[a^2 - b^2, 0]
Time = 2.33 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.81
| method | result | size |
| derivativedivides | \(\frac {\frac {-\frac {\sin \left (d x +c \right )^{2} b}{2}+a \sin \left (d x +c \right )}{b^{2}}-\frac {1}{2 \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {-13 a +11 b}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (35 a^{2}-57 a b +24 b^{2}\right ) \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}-\frac {a^{8} \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{3} \left (a +b \right )^{3} \left (a -b \right )^{3}}+\frac {1}{2 \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {-13 a -11 b}{16 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-35 a^{2}-57 a b -24 b^{2}\right ) \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a +b \right )^{3}}}{d}\) | \(217\) |
| default | \(\frac {\frac {-\frac {\sin \left (d x +c \right )^{2} b}{2}+a \sin \left (d x +c \right )}{b^{2}}-\frac {1}{2 \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {-13 a +11 b}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (35 a^{2}-57 a b +24 b^{2}\right ) \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}-\frac {a^{8} \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{3} \left (a +b \right )^{3} \left (a -b \right )^{3}}+\frac {1}{2 \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {-13 a -11 b}{16 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-35 a^{2}-57 a b -24 b^{2}\right ) \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a +b \right )^{3}}}{d}\) | \(217\) |
| risch | \(\text {Expression too large to display}\) | \(1099\) |
Input:
int(sin(d*x+c)^3*tan(d*x+c)^5/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(1/b^2*(-1/2*sin(d*x+c)^2*b+a*sin(d*x+c))-1/2/(8*a-8*b)/(1+sin(d*x+c)) ^2-1/16*(-13*a+11*b)/(a-b)^2/(1+sin(d*x+c))+1/16*(35*a^2-57*a*b+24*b^2)/(a -b)^3*ln(1+sin(d*x+c))-1/b^3*a^8/(a+b)^3/(a-b)^3*ln(a+b*sin(d*x+c))+1/2/(8 *a+8*b)/(sin(d*x+c)-1)^2-1/16*(-13*a-11*b)/(a+b)^2/(sin(d*x+c)-1)+1/16/(a+ b)^3*(-35*a^2-57*a*b-24*b^2)*ln(sin(d*x+c)-1))
Time = 0.34 (sec) , antiderivative size = 429, normalized size of antiderivative = 1.59 \[ \int \frac {\sin ^3(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {16 \, a^{8} \cos \left (d x + c\right )^{4} \log \left (b \sin \left (d x + c\right ) + a\right ) + 4 \, a^{4} b^{4} - 8 \, a^{2} b^{6} + 4 \, b^{8} - 8 \, {\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} \cos \left (d x + c\right )^{6} - {\left (35 \, a^{5} b^{3} + 48 \, a^{4} b^{4} - 42 \, a^{3} b^{5} - 64 \, a^{2} b^{6} + 15 \, a b^{7} + 24 \, b^{8}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (35 \, a^{5} b^{3} - 48 \, a^{4} b^{4} - 42 \, a^{3} b^{5} + 64 \, a^{2} b^{6} + 15 \, a b^{7} - 24 \, b^{8}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, {\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} \cos \left (d x + c\right )^{4} - 8 \, {\left (4 \, a^{4} b^{4} - 7 \, a^{2} b^{6} + 3 \, b^{8}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, a^{5} b^{3} - 4 \, a^{3} b^{5} + 2 \, a b^{7} + 8 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} \cos \left (d x + c\right )^{4} - {\left (13 \, a^{5} b^{3} - 22 \, a^{3} b^{5} + 9 \, a b^{7}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left (a^{6} b^{3} - 3 \, a^{4} b^{5} + 3 \, a^{2} b^{7} - b^{9}\right )} d \cos \left (d x + c\right )^{4}} \] Input:
integrate(sin(d*x+c)^3*tan(d*x+c)^5/(a+b*sin(d*x+c)),x, algorithm="fricas" )
Output:
-1/16*(16*a^8*cos(d*x + c)^4*log(b*sin(d*x + c) + a) + 4*a^4*b^4 - 8*a^2*b ^6 + 4*b^8 - 8*(a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*cos(d*x + c)^6 - (3 5*a^5*b^3 + 48*a^4*b^4 - 42*a^3*b^5 - 64*a^2*b^6 + 15*a*b^7 + 24*b^8)*cos( d*x + c)^4*log(sin(d*x + c) + 1) + (35*a^5*b^3 - 48*a^4*b^4 - 42*a^3*b^5 + 64*a^2*b^6 + 15*a*b^7 - 24*b^8)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 4 *(a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*cos(d*x + c)^4 - 8*(4*a^4*b^4 - 7 *a^2*b^6 + 3*b^8)*cos(d*x + c)^2 - 2*(2*a^5*b^3 - 4*a^3*b^5 + 2*a*b^7 + 8* (a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*cos(d*x + c)^4 - (13*a^5*b^3 - 22* a^3*b^5 + 9*a*b^7)*cos(d*x + c)^2)*sin(d*x + c))/((a^6*b^3 - 3*a^4*b^5 + 3 *a^2*b^7 - b^9)*d*cos(d*x + c)^4)
Timed out. \[ \int \frac {\sin ^3(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(sin(d*x+c)**3*tan(d*x+c)**5/(a+b*sin(d*x+c)),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.17 \[ \int \frac {\sin ^3(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {16 \, a^{8} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} b^{3} - 3 \, a^{4} b^{5} + 3 \, a^{2} b^{7} - b^{9}} - \frac {{\left (35 \, a^{2} - 57 \, a b + 24 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (35 \, a^{2} + 57 \, a b + 24 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {2 \, {\left ({\left (13 \, a^{3} - 9 \, a b^{2}\right )} \sin \left (d x + c\right )^{3} + 14 \, a^{2} b - 10 \, b^{3} - 4 \, {\left (4 \, a^{2} b - 3 \, b^{3}\right )} \sin \left (d x + c\right )^{2} - {\left (11 \, a^{3} - 7 \, a b^{2}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}} + \frac {8 \, {\left (b \sin \left (d x + c\right )^{2} - 2 \, a \sin \left (d x + c\right )\right )}}{b^{2}}}{16 \, d} \] Input:
integrate(sin(d*x+c)^3*tan(d*x+c)^5/(a+b*sin(d*x+c)),x, algorithm="maxima" )
Output:
-1/16*(16*a^8*log(b*sin(d*x + c) + a)/(a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b ^9) - (35*a^2 - 57*a*b + 24*b^2)*log(sin(d*x + c) + 1)/(a^3 - 3*a^2*b + 3* a*b^2 - b^3) + (35*a^2 + 57*a*b + 24*b^2)*log(sin(d*x + c) - 1)/(a^3 + 3*a ^2*b + 3*a*b^2 + b^3) - 2*((13*a^3 - 9*a*b^2)*sin(d*x + c)^3 + 14*a^2*b - 10*b^3 - 4*(4*a^2*b - 3*b^3)*sin(d*x + c)^2 - (11*a^3 - 7*a*b^2)*sin(d*x + c))/((a^4 - 2*a^2*b^2 + b^4)*sin(d*x + c)^4 + a^4 - 2*a^2*b^2 + b^4 - 2*( a^4 - 2*a^2*b^2 + b^4)*sin(d*x + c)^2) + 8*(b*sin(d*x + c)^2 - 2*a*sin(d*x + c))/b^2)/d
Time = 0.22 (sec) , antiderivative size = 340, normalized size of antiderivative = 1.26 \[ \int \frac {\sin ^3(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a^{8} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b^{3} d - 3 \, a^{4} b^{5} d + 3 \, a^{2} b^{7} d - b^{9} d} - \frac {{\left (35 \, a^{2} + 57 \, a b + 24 \, b^{2}\right )} \log \left ({\left | -\sin \left (d x + c\right ) + 1 \right |}\right )}{16 \, {\left (a^{3} d + 3 \, a^{2} b d + 3 \, a b^{2} d + b^{3} d\right )}} + \frac {{\left (35 \, a^{2} - 57 \, a b + 24 \, b^{2}\right )} \log \left ({\left | -\sin \left (d x + c\right ) - 1 \right |}\right )}{16 \, {\left (a^{3} d - 3 \, a^{2} b d + 3 \, a b^{2} d - b^{3} d\right )}} - \frac {b d \sin \left (d x + c\right )^{2} - 2 \, a d \sin \left (d x + c\right )}{2 \, b^{2} d^{2}} + \frac {14 \, a^{4} b - 24 \, a^{2} b^{3} + 10 \, b^{5} + {\left (13 \, a^{5} - 22 \, a^{3} b^{2} + 9 \, a b^{4}\right )} \sin \left (d x + c\right )^{3} - 4 \, {\left (4 \, a^{4} b - 7 \, a^{2} b^{3} + 3 \, b^{5}\right )} \sin \left (d x + c\right )^{2} - {\left (11 \, a^{5} - 18 \, a^{3} b^{2} + 7 \, a b^{4}\right )} \sin \left (d x + c\right )}{8 \, {\left (a + b\right )}^{3} {\left (a - b\right )}^{3} d {\left (\sin \left (d x + c\right ) + 1\right )}^{2} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} \] Input:
integrate(sin(d*x+c)^3*tan(d*x+c)^5/(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
-a^8*log(abs(b*sin(d*x + c) + a))/(a^6*b^3*d - 3*a^4*b^5*d + 3*a^2*b^7*d - b^9*d) - 1/16*(35*a^2 + 57*a*b + 24*b^2)*log(abs(-sin(d*x + c) + 1))/(a^3 *d + 3*a^2*b*d + 3*a*b^2*d + b^3*d) + 1/16*(35*a^2 - 57*a*b + 24*b^2)*log( abs(-sin(d*x + c) - 1))/(a^3*d - 3*a^2*b*d + 3*a*b^2*d - b^3*d) - 1/2*(b*d *sin(d*x + c)^2 - 2*a*d*sin(d*x + c))/(b^2*d^2) + 1/8*(14*a^4*b - 24*a^2*b ^3 + 10*b^5 + (13*a^5 - 22*a^3*b^2 + 9*a*b^4)*sin(d*x + c)^3 - 4*(4*a^4*b - 7*a^2*b^3 + 3*b^5)*sin(d*x + c)^2 - (11*a^5 - 18*a^3*b^2 + 7*a*b^4)*sin( d*x + c))/((a + b)^3*(a - b)^3*d*(sin(d*x + c) + 1)^2*(sin(d*x + c) - 1)^2 )
Time = 21.47 (sec) , antiderivative size = 806, normalized size of antiderivative = 3.00 \[ \int \frac {\sin ^3(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
int((sin(c + d*x)^3*tan(c + d*x)^5)/(a + b*sin(c + d*x)),x)
Output:
(log(tan(c/2 + (d*x)/2) + 1)*(b^2/(4*(a - b)^3) + (13*b)/(8*(a - b)^2) + 3 5/(8*(a - b))))/d - ((4*tan(c/2 + (d*x)/2)^6*(b^4 - 3*a^4 + a^2*b^2))/(b*( a^2 - b^2)^2) - (2*tan(c/2 + (d*x)/2)^2*(a^4 + 3*b^4 - 5*a^2*b^2))/(b*(a^2 - b^2)^2) + (tan(c/2 + (d*x)/2)^5*(8*a^5 - 11*a*b^4 + 7*a^3*b^2))/(2*b^2* (a^4 + b^4 - 2*a^2*b^2)) + (tan(c/2 + (d*x)/2)^7*(8*a^5 - 11*a*b^4 + 7*a^3 *b^2))/(2*b^2*(a^4 + b^4 - 2*a^2*b^2)) + (tan(c/2 + (d*x)/2)^11*(15*a*b^4 + 8*a^5 - 27*a^3*b^2))/(4*b^2*(a^4 + b^4 - 2*a^2*b^2)) - (tan(c/2 + (d*x)/ 2)^3*(25*a*b^4 + 24*a^5 - 45*a^3*b^2))/(4*b^2*(a^4 + b^4 - 2*a^2*b^2)) - ( tan(c/2 + (d*x)/2)^9*(25*a*b^4 + 24*a^5 - 45*a^3*b^2))/(4*b^2*(a^4 + b^4 - 2*a^2*b^2)) + (4*tan(c/2 + (d*x)/2)^4*(2*a^2 - 3*b^2))/(b*(a^2 - b^2)) + (4*tan(c/2 + (d*x)/2)^8*(2*a^2 - 3*b^2))/(b*(a^2 - b^2)) - (2*tan(c/2 + (d *x)/2)^10*(a^4 + 3*b^4 - 5*a^2*b^2))/(b*(a^4 + b^4 - 2*a^2*b^2)) + (a*tan( c/2 + (d*x)/2)*(8*a^4 + 15*b^4 - 27*a^2*b^2))/(4*b^2*(a^4 + b^4 - 2*a^2*b^ 2)))/(d*(2*tan(c/2 + (d*x)/2)^2 + tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x) /2)^6 + tan(c/2 + (d*x)/2)^8 + 2*tan(c/2 + (d*x)/2)^10 - tan(c/2 + (d*x)/2 )^12 - 1)) - (log(tan(c/2 + (d*x)/2) - 1)*(35/(8*(a + b)) - (13*b)/(8*(a + b)^2) + b^2/(4*(a + b)^3)))/d + (log(tan(c/2 + (d*x)/2)^2 + 1)*(a^2 + 3*b ^2))/(b^3*d) + (a^8*log(a + 2*b*tan(c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/2)^ 2))/(d*(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3))
Time = 0.22 (sec) , antiderivative size = 5684, normalized size of antiderivative = 21.13 \[ \int \frac {\sin ^3(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
int(sin(d*x+c)^3*tan(d*x+c)^5/(a+b*sin(d*x+c)),x)
Output:
(4*cos(c + d*x)**2*sin(c + d*x)**4*tan(c + d*x)**2*a**6*b**2 - 12*cos(c + d*x)**2*sin(c + d*x)**4*tan(c + d*x)**2*a**4*b**4 + 12*cos(c + d*x)**2*sin (c + d*x)**4*tan(c + d*x)**2*a**2*b**6 - 4*cos(c + d*x)**2*sin(c + d*x)**4 *tan(c + d*x)**2*b**8 - 4*cos(c + d*x)**2*sin(c + d*x)**4*a**6*b**2 + 12*c os(c + d*x)**2*sin(c + d*x)**4*a**4*b**4 - 12*cos(c + d*x)**2*sin(c + d*x) **4*a**2*b**6 + 4*cos(c + d*x)**2*sin(c + d*x)**4*b**8 - 8*cos(c + d*x)**2 *sin(c + d*x)**2*tan(c + d*x)**2*a**6*b**2 + 24*cos(c + d*x)**2*sin(c + d* x)**2*tan(c + d*x)**2*a**4*b**4 - 24*cos(c + d*x)**2*sin(c + d*x)**2*tan(c + d*x)**2*a**2*b**6 + 8*cos(c + d*x)**2*sin(c + d*x)**2*tan(c + d*x)**2*b **8 + 8*cos(c + d*x)**2*sin(c + d*x)**2*a**6*b**2 - 24*cos(c + d*x)**2*sin (c + d*x)**2*a**4*b**4 + 24*cos(c + d*x)**2*sin(c + d*x)**2*a**2*b**6 - 8* cos(c + d*x)**2*sin(c + d*x)**2*b**8 + 4*cos(c + d*x)**2*tan(c + d*x)**2*a **6*b**2 - 12*cos(c + d*x)**2*tan(c + d*x)**2*a**4*b**4 + 12*cos(c + d*x)* *2*tan(c + d*x)**2*a**2*b**6 - 4*cos(c + d*x)**2*tan(c + d*x)**2*b**8 - 4* cos(c + d*x)**2*a**6*b**2 + 12*cos(c + d*x)**2*a**4*b**4 - 12*cos(c + d*x) **2*a**2*b**6 + 4*cos(c + d*x)**2*b**8 - 8*cos(c + d*x)*sin(c + d*x)**5*ta n(c + d*x)**3*a**6*b**2 + 24*cos(c + d*x)*sin(c + d*x)**5*tan(c + d*x)**3* a**4*b**4 - 24*cos(c + d*x)*sin(c + d*x)**5*tan(c + d*x)**3*a**2*b**6 + 8* cos(c + d*x)*sin(c + d*x)**5*tan(c + d*x)**3*b**8 - 8*cos(c + d*x)*sin(c + d*x)**5*tan(c + d*x)*a**6*b**2 + 24*cos(c + d*x)*sin(c + d*x)**5*tan(c...