\(\int \frac {\tan ^5(c+d x)}{(a+b \sin (c+d x))^3} \, dx\) [192]

Optimal result

Integrand size = 21, antiderivative size = 373 \[ \int \frac {\tan ^5(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {\left (8 a^2-5 a b-b^2\right ) \log (1-\sin (c+d x))}{16 (a+b)^5 d}-\frac {\left (8 a^2+5 a b-b^2\right ) \log (1+\sin (c+d x))}{16 (a-b)^5 d}+\frac {a^3 \left (a^4+13 a^2 b^2+10 b^4\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^5 d}-\frac {8 a^4-19 a^3 b+21 a^2 b^2-a b^3-b^4}{16 (a-b)^3 (a+b)^4 d (1-\sin (c+d x))}-\frac {8 a^4+19 a^3 b+21 a^2 b^2+a b^3-b^4}{16 (a-b)^4 (a+b)^3 d (1+\sin (c+d x))}-\frac {a^5}{2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^2}-\frac {a^4 \left (a^2+5 b^2\right )}{\left (a^2-b^2\right )^4 d (a+b \sin (c+d x))}+\frac {\sec ^4(c+d x) \left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{4 \left (a^2-b^2\right )^3 d} \] Output:

-1/16*(8*a^2-5*a*b-b^2)*ln(1-sin(d*x+c))/(a+b)^5/d-1/16*(8*a^2+5*a*b-b^2)* 
ln(1+sin(d*x+c))/(a-b)^5/d+a^3*(a^4+13*a^2*b^2+10*b^4)*ln(a+b*sin(d*x+c))/ 
(a^2-b^2)^5/d-1/16*(8*a^4-19*a^3*b+21*a^2*b^2-a*b^3-b^4)/(a-b)^3/(a+b)^4/d 
/(1-sin(d*x+c))-1/16*(8*a^4+19*a^3*b+21*a^2*b^2+a*b^3-b^4)/(a-b)^4/(a+b)^3 
/d/(1+sin(d*x+c))-1/2*a^5/(a^2-b^2)^3/d/(a+b*sin(d*x+c))^2-a^4*(a^2+5*b^2) 
/(a^2-b^2)^4/d/(a+b*sin(d*x+c))+1/4*sec(d*x+c)^4*(a*(a^2+3*b^2)-b*(3*a^2+b 
^2)*sin(d*x+c))/(a^2-b^2)^3/d
 

Mathematica [A] (verified)

Time = 6.29 (sec) , antiderivative size = 304, normalized size of antiderivative = 0.82 \[ \int \frac {\tan ^5(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {\left (8 a^2-5 a b-b^2\right ) \log (1-\sin (c+d x))}{16 (a+b)^5 d}-\frac {\left (8 a^2+5 a b-b^2\right ) \log (1+\sin (c+d x))}{16 (a-b)^5 d}+\frac {a^3 \left (a^4+13 a^2 b^2+10 b^4\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^5 d}+\frac {1}{16 (a+b)^3 d (1-\sin (c+d x))^2}-\frac {7 a+b}{16 (a+b)^4 d (1-\sin (c+d x))}+\frac {1}{16 (a-b)^3 d (1+\sin (c+d x))^2}-\frac {7 a-b}{16 (a-b)^4 d (1+\sin (c+d x))}-\frac {a^5}{2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))^2}-\frac {a^4 \left (a^2+5 b^2\right )}{\left (a^2-b^2\right )^4 d (a+b \sin (c+d x))} \] Input:

Integrate[Tan[c + d*x]^5/(a + b*Sin[c + d*x])^3,x]
 

Output:

-1/16*((8*a^2 - 5*a*b - b^2)*Log[1 - Sin[c + d*x]])/((a + b)^5*d) - ((8*a^ 
2 + 5*a*b - b^2)*Log[1 + Sin[c + d*x]])/(16*(a - b)^5*d) + (a^3*(a^4 + 13* 
a^2*b^2 + 10*b^4)*Log[a + b*Sin[c + d*x]])/((a^2 - b^2)^5*d) + 1/(16*(a + 
b)^3*d*(1 - Sin[c + d*x])^2) - (7*a + b)/(16*(a + b)^4*d*(1 - Sin[c + d*x] 
)) + 1/(16*(a - b)^3*d*(1 + Sin[c + d*x])^2) - (7*a - b)/(16*(a - b)^4*d*( 
1 + Sin[c + d*x])) - a^5/(2*(a^2 - b^2)^3*d*(a + b*Sin[c + d*x])^2) - (a^4 
*(a^2 + 5*b^2))/((a^2 - b^2)^4*d*(a + b*Sin[c + d*x]))
 

Rubi [A] (verified)

Time = 1.50 (sec) , antiderivative size = 365, normalized size of antiderivative = 0.98, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3200, 601, 25, 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.

\(\displaystyle \int \frac {\tan ^5(c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x)^5}{(a+b \sin (c+d x))^3}dx\)

\(\Big \downarrow \) 3200

\(\displaystyle \frac {\int \frac {b^5 \sin ^5(c+d x)}{(a+b \sin (c+d x))^3 \left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\)

\(\Big \downarrow \) 601

\(\displaystyle \frac {\frac {b^4 \left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{4 \left (a^2-b^2\right )^3 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\int -\frac {-\frac {a \left (23 a^2-3 b^2\right ) \sin ^2(c+d x) b^8}{\left (a^2-b^2\right )^3}+\frac {a^3 \left (3 a^2+b^2\right ) b^6}{\left (a^2-b^2\right )^3}-\frac {\left (4 a^6-12 b^2 a^4+21 b^4 a^2-b^6\right ) \sin ^3(c+d x) b^5}{\left (a^2-b^2\right )^3}-\frac {a^2 \left (4 a^4+3 b^2 a^2-3 b^4\right ) \sin (c+d x) b^5}{\left (a^2-b^2\right )^3}}{(a+b \sin (c+d x))^3 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {-\frac {a \left (23 a^2-3 b^2\right ) \sin ^2(c+d x) b^8}{\left (a^2-b^2\right )^3}+\frac {a^3 \left (3 a^2+b^2\right ) b^6}{\left (a^2-b^2\right )^3}-\frac {\left (4 a^6-12 b^2 a^4+21 b^4 a^2-b^6\right ) \sin ^3(c+d x) b^5}{\left (a^2-b^2\right )^3}-\frac {a^2 \left (4 a^4+3 b^2 a^2-3 b^4\right ) \sin (c+d x) b^5}{\left (a^2-b^2\right )^3}}{(a+b \sin (c+d x))^3 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}+\frac {b^4 \left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{4 \left (a^2-b^2\right )^3 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}}{d}\)

\(\Big \downarrow \) 2178

\(\displaystyle \frac {\frac {\frac {\int -\frac {-\frac {\left (27 a^4+22 b^2 a^2-b^4\right ) \sin ^3(c+d x) b^9}{\left (a^2-b^2\right )^4}-\frac {a \left (65 a^4-14 b^2 a^2-3 b^4\right ) \sin ^2(c+d x) b^8}{\left (a^2-b^2\right )^4}+\frac {a^3 \left (21 a^4+26 b^2 a^2+b^4\right ) b^6}{\left (a^2-b^2\right )^4}-\frac {a^2 \left (8 a^6+b^2 a^4-54 b^4 a^2-3 b^6\right ) \sin (c+d x) b^5}{\left (a^2-b^2\right )^4}}{(a+b \sin (c+d x))^3 \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2}-\frac {b^4 \left (8 a^3 \left (a^2+5 b^2\right )-b \left (27 a^4+22 a^2 b^2-b^4\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^4 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}+\frac {b^4 \left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{4 \left (a^2-b^2\right )^3 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {-\frac {\int \frac {-\frac {\left (27 a^4+22 b^2 a^2-b^4\right ) \sin ^3(c+d x) b^9}{\left (a^2-b^2\right )^4}-\frac {a \left (65 a^4-14 b^2 a^2-3 b^4\right ) \sin ^2(c+d x) b^8}{\left (a^2-b^2\right )^4}+\frac {a^3 \left (21 a^4+26 b^2 a^2+b^4\right ) b^6}{\left (a^2-b^2\right )^4}-\frac {a^2 \left (8 a^6+b^2 a^4-54 b^4 a^2-3 b^6\right ) \sin (c+d x) b^5}{\left (a^2-b^2\right )^4}}{(a+b \sin (c+d x))^3 \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2}-\frac {b^4 \left (8 a^3 \left (a^2+5 b^2\right )-b \left (27 a^4+22 a^2 b^2-b^4\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^4 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}+\frac {b^4 \left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{4 \left (a^2-b^2\right )^3 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}}{d}\)

\(\Big \downarrow \) 2160

\(\displaystyle \frac {\frac {-\frac {\int \left (-\frac {8 b^4 a^5}{\left (a^2-b^2\right )^3 (a+b \sin (c+d x))^3}-\frac {8 b^4 \left (a^2+5 b^2\right ) a^4}{\left (a^2-b^2\right )^4 (a+b \sin (c+d x))^2}-\frac {8 b^4 \left (a^4+13 b^2 a^2+10 b^4\right ) a^3}{\left (a^2-b^2\right )^5 (a+b \sin (c+d x))}+\frac {b^4 \left (-8 a^2+5 b a+b^2\right )}{2 (a+b)^5 (b-b \sin (c+d x))}+\frac {b^4 \left (-8 a^2-5 b a+b^2\right )}{2 (b-a)^5 (\sin (c+d x) b+b)}\right )d(b \sin (c+d x))}{2 b^2}-\frac {b^4 \left (8 a^3 \left (a^2+5 b^2\right )-b \left (27 a^4+22 a^2 b^2-b^4\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^4 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}+\frac {b^4 \left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{4 \left (a^2-b^2\right )^3 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {b^4 \left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{4 \left (a^2-b^2\right )^3 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}+\frac {-\frac {b^4 \left (8 a^3 \left (a^2+5 b^2\right )-b \left (27 a^4+22 a^2 b^2-b^4\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^4 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {\frac {b^4 \left (8 a^2-5 a b-b^2\right ) \log (b-b \sin (c+d x))}{2 (a+b)^5}+\frac {b^4 \left (8 a^2+5 a b-b^2\right ) \log (b \sin (c+d x)+b)}{2 (a-b)^5}+\frac {4 a^5 b^4}{\left (a^2-b^2\right )^3 (a+b \sin (c+d x))^2}+\frac {8 a^4 b^4 \left (a^2+5 b^2\right )}{\left (a^2-b^2\right )^4 (a+b \sin (c+d x))}-\frac {8 a^3 b^4 \left (a^4+13 a^2 b^2+10 b^4\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^5}}{2 b^2}}{4 b^2}}{d}\)

Input:

Int[Tan[c + d*x]^5/(a + b*Sin[c + d*x])^3,x]
 

Output:

((b^4*(a*(a^2 + 3*b^2) - b*(3*a^2 + b^2)*Sin[c + d*x]))/(4*(a^2 - b^2)^3*( 
b^2 - b^2*Sin[c + d*x]^2)^2) + (-1/2*(b^4*(8*a^3*(a^2 + 5*b^2) - b*(27*a^4 
 + 22*a^2*b^2 - b^4)*Sin[c + d*x]))/((a^2 - b^2)^4*(b^2 - b^2*Sin[c + d*x] 
^2)) - ((b^4*(8*a^2 - 5*a*b - b^2)*Log[b - b*Sin[c + d*x]])/(2*(a + b)^5) 
- (8*a^3*b^4*(a^4 + 13*a^2*b^2 + 10*b^4)*Log[a + b*Sin[c + d*x]])/(a^2 - b 
^2)^5 + (b^4*(8*a^2 + 5*a*b - b^2)*Log[b + b*Sin[c + d*x]])/(2*(a - b)^5) 
+ (4*a^5*b^4)/((a^2 - b^2)^3*(a + b*Sin[c + d*x])^2) + (8*a^4*b^4*(a^2 + 5 
*b^2))/((a^2 - b^2)^4*(a + b*Sin[c + d*x])))/(2*b^2))/(4*b^2))/d
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 601

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]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2160

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]
 

rule 2178

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]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3200

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p 
_.), x_Symbol] :> Simp[1/f   Subst[Int[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1) 
/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b 
^2, 0] && IntegerQ[(p + 1)/2]
 

Maple [A] (verified)

Time = 4.01 (sec) , antiderivative size = 263, normalized size of antiderivative = 0.71

method	result	size
derivativedivides	\(\frac {\frac {1}{16 \left (a +b \right )^{3} \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {-b -7 a}{16 \left (a +b \right )^{4} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-8 a^{2}+5 a b +b^{2}\right ) \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a +b \right )^{5}}-\frac {a^{5}}{2 \left (a +b \right )^{3} \left (a -b \right )^{3} \left (a +b \sin \left (d x +c \right )\right )^{2}}+\frac {a^{3} \left (a^{4}+13 b^{2} a^{2}+10 b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{5} \left (a -b \right )^{5}}-\frac {a^{4} \left (a^{2}+5 b^{2}\right )}{\left (a +b \right )^{4} \left (a -b \right )^{4} \left (a +b \sin \left (d x +c \right )\right )}+\frac {1}{16 \left (a -b \right )^{3} \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {-b +7 a}{16 \left (a -b \right )^{4} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (-8 a^{2}-5 a b +b^{2}\right ) \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{5}}}{d}\)	\(263\)
default	\(\frac {\frac {1}{16 \left (a +b \right )^{3} \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {-b -7 a}{16 \left (a +b \right )^{4} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-8 a^{2}+5 a b +b^{2}\right ) \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a +b \right )^{5}}-\frac {a^{5}}{2 \left (a +b \right )^{3} \left (a -b \right )^{3} \left (a +b \sin \left (d x +c \right )\right )^{2}}+\frac {a^{3} \left (a^{4}+13 b^{2} a^{2}+10 b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{5} \left (a -b \right )^{5}}-\frac {a^{4} \left (a^{2}+5 b^{2}\right )}{\left (a +b \right )^{4} \left (a -b \right )^{4} \left (a +b \sin \left (d x +c \right )\right )}+\frac {1}{16 \left (a -b \right )^{3} \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {-b +7 a}{16 \left (a -b \right )^{4} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (-8 a^{2}-5 a b +b^{2}\right ) \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{5}}}{d}\)	\(263\)
risch	\(\text {Expression too large to display}\)	\(2267\)

Input:

int(tan(d*x+c)^5/(a+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

1/d*(1/16/(a+b)^3/(sin(d*x+c)-1)^2-1/16*(-b-7*a)/(a+b)^4/(sin(d*x+c)-1)+1/ 
16/(a+b)^5*(-8*a^2+5*a*b+b^2)*ln(sin(d*x+c)-1)-1/2*a^5/(a+b)^3/(a-b)^3/(a+ 
b*sin(d*x+c))^2+a^3*(a^4+13*a^2*b^2+10*b^4)/(a+b)^5/(a-b)^5*ln(a+b*sin(d*x 
+c))-a^4*(a^2+5*b^2)/(a+b)^4/(a-b)^4/(a+b*sin(d*x+c))+1/16/(a-b)^3/(1+sin( 
d*x+c))^2-1/16*(-b+7*a)/(a-b)^4/(1+sin(d*x+c))+1/16/(a-b)^5*(-8*a^2-5*a*b+ 
b^2)*ln(1+sin(d*x+c)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 981 vs. \(2 (359) = 718\).

Time = 0.56 (sec) , antiderivative size = 981, normalized size of antiderivative = 2.63 \[ \int \frac {\tan ^5(c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:

integrate(tan(d*x+c)^5/(a+b*sin(d*x+c))^3,x, algorithm="fricas")
 

Output:

-1/16*(4*a^9 - 16*a^7*b^2 + 24*a^5*b^4 - 16*a^3*b^6 + 4*a*b^8 - 4*(6*a^9 + 
 35*a^7*b^2 - 39*a^5*b^4 - 3*a^3*b^6 + a*b^8)*cos(d*x + c)^4 - 16*(a^9 - 3 
*a^7*b^2 + 3*a^5*b^4 - a^3*b^6)*cos(d*x + c)^2 - 16*((a^7*b^2 + 13*a^5*b^4 
 + 10*a^3*b^6)*cos(d*x + c)^6 - 2*(a^8*b + 13*a^6*b^3 + 10*a^4*b^5)*cos(d* 
x + c)^4*sin(d*x + c) - (a^9 + 14*a^7*b^2 + 23*a^5*b^4 + 10*a^3*b^6)*cos(d 
*x + c)^4)*log(b*sin(d*x + c) + a) + ((8*a^7*b^2 + 45*a^6*b^3 + 104*a^5*b^ 
4 + 125*a^4*b^5 + 80*a^3*b^6 + 23*a^2*b^7 - b^9)*cos(d*x + c)^6 - 2*(8*a^8 
*b + 45*a^7*b^2 + 104*a^6*b^3 + 125*a^5*b^4 + 80*a^4*b^5 + 23*a^3*b^6 - a* 
b^8)*cos(d*x + c)^4*sin(d*x + c) - (8*a^9 + 45*a^8*b + 112*a^7*b^2 + 170*a 
^6*b^3 + 184*a^5*b^4 + 148*a^4*b^5 + 80*a^3*b^6 + 22*a^2*b^7 - b^9)*cos(d* 
x + c)^4)*log(sin(d*x + c) + 1) + ((8*a^7*b^2 - 45*a^6*b^3 + 104*a^5*b^4 - 
 125*a^4*b^5 + 80*a^3*b^6 - 23*a^2*b^7 + b^9)*cos(d*x + c)^6 - 2*(8*a^8*b 
- 45*a^7*b^2 + 104*a^6*b^3 - 125*a^5*b^4 + 80*a^4*b^5 - 23*a^3*b^6 + a*b^8 
)*cos(d*x + c)^4*sin(d*x + c) - (8*a^9 - 45*a^8*b + 112*a^7*b^2 - 170*a^6* 
b^3 + 184*a^5*b^4 - 148*a^4*b^5 + 80*a^3*b^6 - 22*a^2*b^7 + b^9)*cos(d*x + 
 c)^4)*log(-sin(d*x + c) + 1) - 2*(2*a^8*b - 8*a^6*b^3 + 12*a^4*b^5 - 8*a^ 
2*b^7 + 2*b^9 + (8*a^8*b + 59*a^6*b^3 - 45*a^4*b^5 - 23*a^2*b^7 + b^9)*cos 
(d*x + c)^4 - (11*a^8*b - 36*a^6*b^3 + 42*a^4*b^5 - 20*a^2*b^7 + 3*b^9)*co 
s(d*x + c)^2)*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*cos(d*x + c)^6 - 2*(a^11*b - 5*a^9*b^3 + 10*a...
 

Sympy [F]

\[ \int \frac {\tan ^5(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\int \frac {\tan ^{5}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{3}}\, dx \] Input:

integrate(tan(d*x+c)**5/(a+b*sin(d*x+c))**3,x)
 

Output:

Integral(tan(c + d*x)**5/(a + b*sin(c + d*x))**3, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 730 vs. \(2 (359) = 718\).

Time = 0.05 (sec) , antiderivative size = 730, normalized size of antiderivative = 1.96 \[ \int \frac {\tan ^5(c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:

integrate(tan(d*x+c)^5/(a+b*sin(d*x+c))^3,x, algorithm="maxima")
 

Output:

1/16*(16*(a^7 + 13*a^5*b^2 + 10*a^3*b^4)*log(b*sin(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) - (8*a^2 + 5*a*b - 
b^2)*log(sin(d*x + c) + 1)/(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a* 
b^4 - b^5) - (8*a^2 - 5*a*b - b^2)*log(sin(d*x + c) - 1)/(a^5 + 5*a^4*b + 
10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5) - 2*(18*a^7 + 72*a^5*b^2 + 6*a^3* 
b^4 + (8*a^6*b + 67*a^4*b^3 + 22*a^2*b^5 - b^7)*sin(d*x + c)^5 + 2*(6*a^7 
+ 41*a^5*b^2 + 2*a^3*b^4 - a*b^6)*sin(d*x + c)^4 - (5*a^6*b + 159*a^4*b^3 
+ 27*a^2*b^5 + b^7)*sin(d*x + c)^3 - 4*(8*a^7 + 37*a^5*b^2 + 4*a^3*b^4 - a 
*b^6)*sin(d*x + c)^2 - (a^6*b - 86*a^4*b^3 - 11*a^2*b^5)*sin(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)*sin(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)*sin(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)*sin(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)*sin(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^10)*sin(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)*sin(d*x + c)))/d
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 607, normalized size of antiderivative = 1.63 \[ \int \frac {\tan ^5(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {{\left (a^{7} b + 13 \, a^{5} b^{3} + 10 \, a^{3} b^{5}\right )} \log \left ({\left | b \sin \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 {{\left (8 \, a^{2} + 5 \, a b - b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{16 \, {\left (a^{5} d - 5 \, a^{4} b d + 10 \, a^{3} b^{2} d - 10 \, a^{2} b^{3} d + 5 \, a b^{4} d - b^{5} d\right )}} - \frac {{\left (8 \, a^{2} - 5 \, a b - b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{16 \, {\left (a^{5} d + 5 \, a^{4} b d + 10 \, a^{3} b^{2} d + 10 \, a^{2} b^{3} d + 5 \, a b^{4} d + b^{5} d\right )}} - \frac {8 \, a^{6} b \sin \left (d x + c\right )^{5} + 67 \, a^{4} b^{3} \sin \left (d x + c\right )^{5} + 22 \, a^{2} b^{5} \sin \left (d x + c\right )^{5} - b^{7} \sin \left (d x + c\right )^{5} + 12 \, a^{7} \sin \left (d x + c\right )^{4} + 82 \, a^{5} b^{2} \sin \left (d x + c\right )^{4} + 4 \, a^{3} b^{4} \sin \left (d x + c\right )^{4} - 2 \, a b^{6} \sin \left (d x + c\right )^{4} - 5 \, a^{6} b \sin \left (d x + c\right )^{3} - 159 \, a^{4} b^{3} \sin \left (d x + c\right )^{3} - 27 \, a^{2} b^{5} \sin \left (d x + c\right )^{3} - b^{7} \sin \left (d x + c\right )^{3} - 32 \, a^{7} \sin \left (d x + c\right )^{2} - 148 \, a^{5} b^{2} \sin \left (d x + c\right )^{2} - 16 \, a^{3} b^{4} \sin \left (d x + c\right )^{2} + 4 \, a b^{6} \sin \left (d x + c\right )^{2} - a^{6} b \sin \left (d x + c\right ) + 86 \, a^{4} b^{3} \sin \left (d x + c\right ) + 11 \, a^{2} b^{5} \sin \left (d x + c\right ) + 18 \, a^{7} + 72 \, a^{5} b^{2} + 6 \, a^{3} b^{4}}{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 \sin \left (d x + c\right )^{3} + a \sin \left (d x + c\right )^{2} - b \sin \left (d x + c\right ) - a\right )}^{2}} \] Input:

integrate(tan(d*x+c)^5/(a+b*sin(d*x+c))^3,x, algorithm="giac")
 

Output:

(a^7*b + 13*a^5*b^3 + 10*a^3*b^5)*log(abs(b*sin(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) - 1/16* 
(8*a^2 + 5*a*b - b^2)*log(abs(sin(d*x + c) + 1))/(a^5*d - 5*a^4*b*d + 10*a 
^3*b^2*d - 10*a^2*b^3*d + 5*a*b^4*d - b^5*d) - 1/16*(8*a^2 - 5*a*b - b^2)* 
log(abs(sin(d*x + c) - 1))/(a^5*d + 5*a^4*b*d + 10*a^3*b^2*d + 10*a^2*b^3* 
d + 5*a*b^4*d + b^5*d) - 1/8*(8*a^6*b*sin(d*x + c)^5 + 67*a^4*b^3*sin(d*x 
+ c)^5 + 22*a^2*b^5*sin(d*x + c)^5 - b^7*sin(d*x + c)^5 + 12*a^7*sin(d*x + 
 c)^4 + 82*a^5*b^2*sin(d*x + c)^4 + 4*a^3*b^4*sin(d*x + c)^4 - 2*a*b^6*sin 
(d*x + c)^4 - 5*a^6*b*sin(d*x + c)^3 - 159*a^4*b^3*sin(d*x + c)^3 - 27*a^2 
*b^5*sin(d*x + c)^3 - b^7*sin(d*x + c)^3 - 32*a^7*sin(d*x + c)^2 - 148*a^5 
*b^2*sin(d*x + c)^2 - 16*a^3*b^4*sin(d*x + c)^2 + 4*a*b^6*sin(d*x + c)^2 - 
 a^6*b*sin(d*x + c) + 86*a^4*b^3*sin(d*x + c) + 11*a^2*b^5*sin(d*x + c) + 
18*a^7 + 72*a^5*b^2 + 6*a^3*b^4)/((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*sin(d*x + c)^3 + a*sin(d*x + c)^2 - b*sin(d*x + c) - 
a)^2)
 

Mupad [B] (verification not implemented)

Time = 20.63 (sec) , antiderivative size = 1229, normalized size of antiderivative = 3.29 \[ \int \frac {\tan ^5(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Too large to display} \] Input:

int(tan(c + d*x)^5/(a + b*sin(c + d*x))^3,x)
 

Output:

((tan(c/2 + (d*x)/2)^2*(a*b^6 - 2*a^7 + 38*a^3*b^4 + 11*a^5*b^2))/(a^8 + b 
^8 - 4*a^2*b^6 + 6*a^4*b^4 - 4*a^6*b^2) - (4*tan(c/2 + (d*x)/2)^4*(4*a*b^6 
 - a^7 + 33*a^3*b^4 + 12*a^5*b^2))/(a^8 + b^8 - 4*a^2*b^6 + 6*a^4*b^4 - 4* 
a^6*b^2) - (4*tan(c/2 + (d*x)/2)^8*(4*a*b^6 - a^7 + 33*a^3*b^4 + 12*a^5*b^ 
2))/(a^8 + b^8 - 4*a^2*b^6 + 6*a^4*b^4 - 4*a^6*b^2) + (tan(c/2 + (d*x)/2)^ 
10*(a*b^6 - 2*a^7 + 38*a^3*b^4 + 11*a^5*b^2))/(a^8 + b^8 - 4*a^2*b^6 + 6*a 
^4*b^4 - 4*a^6*b^2) + (2*tan(c/2 + (d*x)/2)^6*(7*a*b^6 + 6*a^7 + 118*a^3*b 
^4 + 13*a^5*b^2))/(a^8 + b^8 - 4*a^2*b^6 + 6*a^4*b^4 - 4*a^6*b^2) + (b*tan 
(c/2 + (d*x)/2)^11*(37*a^6 + a^2*b^4 + 58*a^4*b^2))/(4*(a^8 + b^8 - 4*a^2* 
b^6 + 6*a^4*b^4 - 4*a^6*b^2)) + (b*tan(c/2 + (d*x)/2)^5*(7*a^6 + 14*b^6 - 
57*a^2*b^4 + 132*a^4*b^2))/(2*(a^8 + b^8 - 4*a^2*b^6 + 6*a^4*b^4 - 4*a^6*b 
^2)) + (b*tan(c/2 + (d*x)/2)^7*(7*a^6 + 14*b^6 - 57*a^2*b^4 + 132*a^4*b^2) 
)/(2*(a^8 + b^8 - 4*a^2*b^6 + 6*a^4*b^4 - 4*a^6*b^2)) - (b*tan(c/2 + (d*x) 
/2)^3*(83*a^6 - 4*b^6 - 17*a^2*b^4 + 226*a^4*b^2))/(4*(a^8 + b^8 - 4*a^2*b 
^6 + 6*a^4*b^4 - 4*a^6*b^2)) - (b*tan(c/2 + (d*x)/2)^9*(83*a^6 - 4*b^6 - 1 
7*a^2*b^4 + 226*a^4*b^2))/(4*(a^8 + b^8 - 4*a^2*b^6 + 6*a^4*b^4 - 4*a^6*b^ 
2)) + (b*tan(c/2 + (d*x)/2)*(37*a^6 + a^2*b^4 + 58*a^4*b^2))/(4*(a^8 + b^8 
 - 4*a^2*b^6 + 6*a^4*b^4 - 4*a^6*b^2)))/(d*(tan(c/2 + (d*x)/2)^6*(4*a^2 + 
24*b^2) - tan(c/2 + (d*x)/2)^10*(2*a^2 - 4*b^2) - tan(c/2 + (d*x)/2)^2*(2* 
a^2 - 4*b^2) + a^2*tan(c/2 + (d*x)/2)^12 + a^2 - tan(c/2 + (d*x)/2)^4*(...
 

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 4732, normalized size of antiderivative = 12.69 \[ \int \frac {\tan ^5(c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:

int(tan(d*x+c)^5/(a+b*sin(d*x+c))^3,x)
 

Output:

( - 16*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**6*a**7*b**2 + 90*log(tan((c 
 + d*x)/2) - 1)*sin(c + d*x)**6*a**6*b**3 - 208*log(tan((c + d*x)/2) - 1)* 
sin(c + d*x)**6*a**5*b**4 + 250*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**6* 
a**4*b**5 - 160*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**6*a**3*b**6 + 46*l 
og(tan((c + d*x)/2) - 1)*sin(c + d*x)**6*a**2*b**7 - 2*log(tan((c + d*x)/2 
) - 1)*sin(c + d*x)**6*b**9 - 32*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5 
*a**8*b + 180*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5*a**7*b**2 - 416*lo 
g(tan((c + d*x)/2) - 1)*sin(c + d*x)**5*a**6*b**3 + 500*log(tan((c + d*x)/ 
2) - 1)*sin(c + d*x)**5*a**5*b**4 - 320*log(tan((c + d*x)/2) - 1)*sin(c + 
d*x)**5*a**4*b**5 + 92*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5*a**3*b**6 
 - 4*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5*a*b**8 - 16*log(tan((c + d* 
x)/2) - 1)*sin(c + d*x)**4*a**9 + 90*log(tan((c + d*x)/2) - 1)*sin(c + d*x 
)**4*a**8*b - 176*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**7*b**2 + 70 
*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**6*b**3 + 256*log(tan((c + d* 
x)/2) - 1)*sin(c + d*x)**4*a**5*b**4 - 454*log(tan((c + d*x)/2) - 1)*sin(c 
 + d*x)**4*a**4*b**5 + 320*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**3* 
b**6 - 94*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**2*b**7 + 4*log(tan( 
(c + d*x)/2) - 1)*sin(c + d*x)**4*b**9 + 64*log(tan((c + d*x)/2) - 1)*sin( 
c + d*x)**3*a**8*b - 360*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*a**7*b* 
*2 + 832*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*a**6*b**3 - 1000*log...