\(\int \frac {\sin (c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx\) [1357]

Optimal result

Integrand size = 27, antiderivative size = 237 \[ \int \frac {\sin (c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\left (15 a^2+21 a b+8 b^2\right ) \log (1-\sin (c+d x))}{16 (a+b)^3 d}+\frac {\left (15 a^2-21 a b+8 b^2\right ) \log (1+\sin (c+d x))}{16 (a-b)^3 d}-\frac {a^6 \log (a+b \sin (c+d x))}{b \left (a^2-b^2\right )^3 d}-\frac {9 a^2-3 a b-8 b^2}{16 (a-b) (a+b)^2 d (1-\sin (c+d x))}+\frac {9 a^2+3 a b-8 b^2}{16 (a-b)^2 (a+b) d (1+\sin (c+d x))}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x))}{4 \left (a^2-b^2\right ) d} \] Output:

-1/16*(15*a^2+21*a*b+8*b^2)*ln(1-sin(d*x+c))/(a+b)^3/d+1/16*(15*a^2-21*a*b 
+8*b^2)*ln(1+sin(d*x+c))/(a-b)^3/d-a^6*ln(a+b*sin(d*x+c))/b/(a^2-b^2)^3/d- 
1/16*(9*a^2-3*a*b-8*b^2)/(a-b)/(a+b)^2/d/(1-sin(d*x+c))+1/16*(9*a^2+3*a*b- 
8*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
 

Mathematica [A] (verified)

Time = 1.95 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.79 \[ \int \frac {\sin (c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-\frac {\left (15 a^2+21 a b+8 b^2\right ) \log (1-\sin (c+d x))}{(a+b)^3}+\frac {\left (15 a^2-21 a b+8 b^2\right ) \log (1+\sin (c+d x))}{(a-b)^3}-\frac {16 a^6 \log (a+b \sin (c+d x))}{(a-b)^3 b (a+b)^3}+\frac {1}{(a+b) (-1+\sin (c+d x))^2}+\frac {9 a+7 b}{(a+b)^2 (-1+\sin (c+d x))}-\frac {1}{(a-b) (1+\sin (c+d x))^2}+\frac {9 a-7 b}{(a-b)^2 (1+\sin (c+d x))}}{16 d} \] Input:

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

Output:

(-(((15*a^2 + 21*a*b + 8*b^2)*Log[1 - Sin[c + d*x]])/(a + b)^3) + ((15*a^2 
 - 21*a*b + 8*b^2)*Log[1 + Sin[c + d*x]])/(a - b)^3 - (16*a^6*Log[a + b*Si 
n[c + d*x]])/((a - b)^3*b*(a + b)^3) + 1/((a + b)*(-1 + Sin[c + d*x])^2) + 
 (9*a + 7*b)/((a + b)^2*(-1 + Sin[c + d*x])) - 1/((a - b)*(1 + Sin[c + d*x 
])^2) + (9*a - 7*b)/((a - b)^2*(1 + Sin[c + d*x])))/(16*d)
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.14, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (c+d x)^6}{\cos (c+d x)^5 (a+b \sin (c+d x))}dx\)

\(\Big \downarrow \) 3316

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {b^6 \sin ^6(c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{b d}\)

\(\Big \downarrow \) 601

\(\displaystyle \frac {-\frac {\int \frac {-\frac {3 a \sin (c+d x) b^7}{a^2-b^2}+4 \sin ^4(c+d x) b^6+4 \sin ^2(c+d x) b^6+\frac {a^2 b^6}{a^2-b^2}}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}-\frac {b^4 \left (\frac {b^2}{a^2-b^2}-\frac {a b \sin (c+d x)}{a^2-b^2}\right )}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}}{b d}\)

\(\Big \downarrow \) 2178

\(\displaystyle \frac {-\frac {\frac {\int -\frac {-\frac {a \left (9 a^2-5 b^2\right ) \sin (c+d x) b^7}{\left (a^2-b^2\right )^2}+8 \sin ^2(c+d x) b^6+\frac {a^2 \left (7 a^2-3 b^2\right ) b^6}{\left (a^2-b^2\right )^2}}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2}-\frac {b^4 \left (4 b^2 \left (3 a^2-2 b^2\right )-a b \left (9 a^2-5 b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}-\frac {b^4 \left (\frac {b^2}{a^2-b^2}-\frac {a b \sin (c+d x)}{a^2-b^2}\right )}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}}{b d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {-\frac {-\frac {\int \frac {-\frac {a \left (9 a^2-5 b^2\right ) \sin (c+d x) b^7}{\left (a^2-b^2\right )^2}+8 \sin ^2(c+d x) b^6+\frac {a^2 \left (7 a^2-3 b^2\right ) b^6}{\left (a^2-b^2\right )^2}}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2}-\frac {b^4 \left (4 b^2 \left (3 a^2-2 b^2\right )-a b \left (9 a^2-5 b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}-\frac {b^4 \left (\frac {b^2}{a^2-b^2}-\frac {a b \sin (c+d x)}{a^2-b^2}\right )}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}}{b d}\)

\(\Big \downarrow \) 2160

\(\displaystyle \frac {-\frac {-\frac {\int \left (-\frac {8 b^4 a^6}{(a-b)^3 (a+b)^3 (a+b \sin (c+d x))}+\frac {b^5 \left (15 a^2+21 b a+8 b^2\right )}{2 (a+b)^3 (b-b \sin (c+d x))}+\frac {b^5 \left (15 a^2-21 b a+8 b^2\right )}{2 (a-b)^3 (\sin (c+d x) b+b)}\right )d(b \sin (c+d x))}{2 b^2}-\frac {b^4 \left (4 b^2 \left (3 a^2-2 b^2\right )-a b \left (9 a^2-5 b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}-\frac {b^4 \left (\frac {b^2}{a^2-b^2}-\frac {a b \sin (c+d x)}{a^2-b^2}\right )}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}}{b d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {b^4 \left (\frac {b^2}{a^2-b^2}-\frac {a b \sin (c+d x)}{a^2-b^2}\right )}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {-\frac {b^4 \left (4 b^2 \left (3 a^2-2 b^2\right )-a b \left (9 a^2-5 b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {-\frac {b^5 \left (15 a^2+21 a b+8 b^2\right ) \log (b-b \sin (c+d x))}{2 (a+b)^3}+\frac {b^5 \left (15 a^2-21 a b+8 b^2\right ) \log (b \sin (c+d x)+b)}{2 (a-b)^3}-\frac {8 a^6 b^4 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3}}{2 b^2}}{4 b^2}}{b d}\)

Input:

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

Output:

(-1/4*(b^4*(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*(-1/2*(b^5*(15*a^2 + 21*a*b + 8*b^2)*Log[b - b*S 
in[c + d*x]])/(a + b)^3 - (8*a^6*b^4*Log[a + b*Sin[c + d*x]])/(a^2 - b^2)^ 
3 + (b^5*(15*a^2 - 21*a*b + 8*b^2)*Log[b + b*Sin[c + d*x]])/(2*(a - b)^3)) 
/b^2 - (b^4*(4*b^2*(3*a^2 - 2*b^2) - a*b*(9*a^2 - 5*b^2)*Sin[c + d*x]))/(2 
*(a^2 - b^2)^2*(b^2 - b^2*Sin[c + d*x]^2)))/(4*b^2))/(b*d)
 

Defintions of rubi rules used

rule 25

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 3316

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]
 

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.81

method	result	size
derivativedivides	\(\frac {-\frac {1}{2 \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {-9 a +7 b}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (15 a^{2}-21 a b +8 b^{2}\right ) \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}+\frac {1}{2 \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {-9 a -7 b}{16 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-15 a^{2}-21 a b -8 b^{2}\right ) \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a +b \right )^{3}}-\frac {a^{6} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3} b}}{d}\)	\(193\)
default	\(\frac {-\frac {1}{2 \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {-9 a +7 b}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (15 a^{2}-21 a b +8 b^{2}\right ) \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}+\frac {1}{2 \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {-9 a -7 b}{16 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-15 a^{2}-21 a b -8 b^{2}\right ) \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a +b \right )^{3}}-\frac {a^{6} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3} b}}{d}\)	\(193\)
risch	\(\frac {2 i a^{6} x}{b \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}+\frac {21 i a b c}{8 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {i b^{2} c}{d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {15 i a^{2} c}{8 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {15 i a^{2} x}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {21 i a b x}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {21 i a b c}{8 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {i x}{b}+\frac {21 i a b x}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {i b^{2} x}{a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}}+\frac {i b^{2} x}{a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}}+\frac {i b^{2} c}{d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {15 i a^{2} c}{8 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {2 i a^{6} c}{b d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}+\frac {15 i a^{2} x}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {-9 i a^{3} {\mathrm e}^{7 i \left (d x +c \right )}+5 i a \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}-i a^{3} {\mathrm e}^{5 i \left (d x +c \right )}-3 i a \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-24 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}+16 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+i a^{3} {\mathrm e}^{3 i \left (d x +c \right )}+3 i a \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-32 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+16 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+9 i a^{3} {\mathrm e}^{i \left (d x +c \right )}-5 i a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}-24 \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2} b +16 \,{\mathrm e}^{2 i \left (d x +c \right )} b^{3}}{4 d \left (-a^{2}+b^{2}\right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2}}{8 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {21 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a b}{8 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2}}{8 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {21 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a b}{8 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {a^{6} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}\)	\(1010\)

Input:

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

Output:

1/d*(-1/2/(8*a-8*b)/(1+sin(d*x+c))^2-1/16*(-9*a+7*b)/(a-b)^2/(1+sin(d*x+c) 
)+1/16*(15*a^2-21*a*b+8*b^2)/(a-b)^3*ln(1+sin(d*x+c))+1/2/(8*a+8*b)/(sin(d 
*x+c)-1)^2-1/16*(-9*a-7*b)/(a+b)^2/(sin(d*x+c)-1)+1/16/(a+b)^3*(-15*a^2-21 
*a*b-8*b^2)*ln(sin(d*x+c)-1)-a^6/(a+b)^3/(a-b)^3/b*ln(a+b*sin(d*x+c)))
 

Fricas [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.28 \[ \int \frac {\sin (c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {16 \, a^{6} \cos \left (d x + c\right )^{4} \log \left (b \sin \left (d x + c\right ) + a\right ) + 4 \, a^{4} b^{2} - 8 \, a^{2} b^{4} + 4 \, b^{6} - {\left (15 \, a^{5} b + 24 \, a^{4} b^{2} - 10 \, a^{3} b^{3} - 24 \, a^{2} b^{4} + 3 \, a b^{5} + 8 \, b^{6}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (15 \, a^{5} b - 24 \, a^{4} b^{2} - 10 \, a^{3} b^{3} + 24 \, a^{2} b^{4} + 3 \, a b^{5} - 8 \, b^{6}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 8 \, {\left (3 \, a^{4} b^{2} - 5 \, a^{2} b^{4} + 2 \, b^{6}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, a^{5} b - 4 \, a^{3} b^{3} + 2 \, a b^{5} - {\left (9 \, a^{5} b - 14 \, a^{3} b^{3} + 5 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d \cos \left (d x + c\right )^{4}} \] Input:

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

Output:

-1/16*(16*a^6*cos(d*x + c)^4*log(b*sin(d*x + c) + a) + 4*a^4*b^2 - 8*a^2*b 
^4 + 4*b^6 - (15*a^5*b + 24*a^4*b^2 - 10*a^3*b^3 - 24*a^2*b^4 + 3*a*b^5 + 
8*b^6)*cos(d*x + c)^4*log(sin(d*x + c) + 1) + (15*a^5*b - 24*a^4*b^2 - 10* 
a^3*b^3 + 24*a^2*b^4 + 3*a*b^5 - 8*b^6)*cos(d*x + c)^4*log(-sin(d*x + c) + 
 1) - 8*(3*a^4*b^2 - 5*a^2*b^4 + 2*b^6)*cos(d*x + c)^2 - 2*(2*a^5*b - 4*a^ 
3*b^3 + 2*a*b^5 - (9*a^5*b - 14*a^3*b^3 + 5*a*b^5)*cos(d*x + c)^2)*sin(d*x 
 + c))/((a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*d*cos(d*x + c)^4)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.22 \[ \int \frac {\sin (c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {16 \, a^{6} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}} - \frac {{\left (15 \, a^{2} - 21 \, a b + 8 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (15 \, a^{2} + 21 \, a b + 8 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {2 \, {\left ({\left (9 \, a^{3} - 5 \, a b^{2}\right )} \sin \left (d x + c\right )^{3} + 10 \, a^{2} b - 6 \, b^{3} - 4 \, {\left (3 \, a^{2} b - 2 \, b^{3}\right )} \sin \left (d x + c\right )^{2} - {\left (7 \, a^{3} - 3 \, a b^{2}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}}}{16 \, d} \] Input:

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

Output:

-1/16*(16*a^6*log(b*sin(d*x + c) + a)/(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7 
) - (15*a^2 - 21*a*b + 8*b^2)*log(sin(d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a*b 
^2 - b^3) + (15*a^2 + 21*a*b + 8*b^2)*log(sin(d*x + c) - 1)/(a^3 + 3*a^2*b 
 + 3*a*b^2 + b^3) - 2*((9*a^3 - 5*a*b^2)*sin(d*x + c)^3 + 10*a^2*b - 6*b^3 
 - 4*(3*a^2*b - 2*b^3)*sin(d*x + c)^2 - (7*a^3 - 3*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))/d
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.30 \[ \int \frac {\sin (c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a^{6} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b d - 3 \, a^{4} b^{3} d + 3 \, a^{2} b^{5} d - b^{7} d} - \frac {{\left (15 \, a^{2} + 21 \, a b + 8 \, b^{2}\right )} \log \left ({\left | -\sin \left (d x + c\right ) + 1 \right |}\right )}{16 \, {\left (a^{3} d + 3 \, a^{2} b d + 3 \, a b^{2} d + b^{3} d\right )}} + \frac {{\left (15 \, a^{2} - 21 \, a b + 8 \, b^{2}\right )} \log \left ({\left | -\sin \left (d x + c\right ) - 1 \right |}\right )}{16 \, {\left (a^{3} d - 3 \, a^{2} b d + 3 \, a b^{2} d - b^{3} d\right )}} + \frac {10 \, a^{4} b - 16 \, a^{2} b^{3} + 6 \, b^{5} + {\left (9 \, a^{5} - 14 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \sin \left (d x + c\right )^{3} - 4 \, {\left (3 \, a^{4} b - 5 \, a^{2} b^{3} + 2 \, b^{5}\right )} \sin \left (d x + c\right )^{2} - {\left (7 \, a^{5} - 10 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \sin \left (d x + c\right )}{8 \, {\left (a + b\right )}^{3} {\left (a - b\right )}^{3} d {\left (\sin \left (d x + c\right ) + 1\right )}^{2} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} \] Input:

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

Output:

-a^6*log(abs(b*sin(d*x + c) + a))/(a^6*b*d - 3*a^4*b^3*d + 3*a^2*b^5*d - b 
^7*d) - 1/16*(15*a^2 + 21*a*b + 8*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*(15*a^2 - 21*a*b + 8*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) + 1/8*(10*a^4* 
b - 16*a^2*b^3 + 6*b^5 + (9*a^5 - 14*a^3*b^2 + 5*a*b^4)*sin(d*x + c)^3 - 4 
*(3*a^4*b - 5*a^2*b^3 + 2*b^5)*sin(d*x + c)^2 - (7*a^5 - 10*a^3*b^2 + 3*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)
 

Mupad [B] (verification not implemented)

Time = 20.73 (sec) , antiderivative size = 549, normalized size of antiderivative = 2.32 \[ \int \frac {\sin (c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {b^2}{4\,{\left (a-b\right )}^3}+\frac {9\,b}{8\,{\left (a-b\right )}^2}+\frac {15}{8\,\left (a-b\right )}\right )}{d}-\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (11\,a\,b^2-15\,a^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (3\,a\,b^2-7\,a^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (11\,a\,b^2-15\,a^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2\,b-b^3\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (3\,a^2\,b-2\,b^3\right )}{a^4-2\,a^2\,b^2+b^4}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (2\,a^2\,b-b^3\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (7\,a^2-3\,b^2\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{b\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (\frac {15}{8\,\left (a+b\right )}-\frac {9\,b}{8\,{\left (a+b\right )}^2}+\frac {b^2}{4\,{\left (a+b\right )}^3}\right )}{d}-\frac {a^6\,\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}{d\,\left (a^6\,b-3\,a^4\,b^3+3\,a^2\,b^5-b^7\right )} \] Input:

int((sin(c + d*x)*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) + (9*b)/(8*(a - b)^2) + 15 
/(8*(a - b))))/d - ((tan(c/2 + (d*x)/2)^3*(11*a*b^2 - 15*a^3))/(4*(a^4 + b 
^4 - 2*a^2*b^2)) - (tan(c/2 + (d*x)/2)^7*(3*a*b^2 - 7*a^3))/(4*(a^4 + b^4 
- 2*a^2*b^2)) + (tan(c/2 + (d*x)/2)^5*(11*a*b^2 - 15*a^3))/(4*(a^4 + b^4 - 
 2*a^2*b^2)) - (2*tan(c/2 + (d*x)/2)^2*(2*a^2*b - b^3))/(a^4 + b^4 - 2*a^2 
*b^2) + (4*tan(c/2 + (d*x)/2)^4*(3*a^2*b - 2*b^3))/(a^4 + b^4 - 2*a^2*b^2) 
 - (2*tan(c/2 + (d*x)/2)^6*(2*a^2*b - b^3))/(a^4 + b^4 - 2*a^2*b^2) + (a*t 
an(c/2 + (d*x)/2)*(7*a^2 - 3*b^2))/(4*(a^4 + b^4 - 2*a^2*b^2)))/(d*(6*tan( 
c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c 
/2 + (d*x)/2)^8 + 1)) + log(tan(c/2 + (d*x)/2)^2 + 1)/(b*d) - (log(tan(c/2 
 + (d*x)/2) - 1)*(15/(8*(a + b)) - (9*b)/(8*(a + b)^2) + b^2/(4*(a + b)^3) 
))/d - (a^6*log(a + 2*b*tan(c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/2)^2))/(d*( 
a^6*b - b^7 + 3*a^2*b^5 - 3*a^4*b^3))
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 1751, normalized size of antiderivative = 7.39 \[ \int \frac {\sin (c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

(4*log(tan(c + d*x)**2 + 1)*sin(c + d*x)**4*a**6 - 12*log(tan(c + d*x)**2 
+ 1)*sin(c + d*x)**4*a**4*b**2 + 12*log(tan(c + d*x)**2 + 1)*sin(c + d*x)* 
*4*a**2*b**4 - 4*log(tan(c + d*x)**2 + 1)*sin(c + d*x)**4*b**6 - 8*log(tan 
(c + d*x)**2 + 1)*sin(c + d*x)**2*a**6 + 24*log(tan(c + d*x)**2 + 1)*sin(c 
 + d*x)**2*a**4*b**2 - 24*log(tan(c + d*x)**2 + 1)*sin(c + d*x)**2*a**2*b* 
*4 + 8*log(tan(c + d*x)**2 + 1)*sin(c + d*x)**2*b**6 + 4*log(tan(c + d*x)* 
*2 + 1)*a**6 - 12*log(tan(c + d*x)**2 + 1)*a**4*b**2 + 12*log(tan(c + d*x) 
**2 + 1)*a**2*b**4 - 4*log(tan(c + d*x)**2 + 1)*b**6 + 8*log(tan((c + d*x) 
/2) - 1)*sin(c + d*x)**4*a**6 - 15*log(tan((c + d*x)/2) - 1)*sin(c + d*x)* 
*4*a**5*b + 10*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**3*b**3 - 3*log 
(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a*b**5 - 16*log(tan((c + d*x)/2) - 
1)*sin(c + d*x)**2*a**6 + 30*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a** 
5*b - 20*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**3*b**3 + 6*log(tan(( 
c + d*x)/2) - 1)*sin(c + d*x)**2*a*b**5 + 8*log(tan((c + d*x)/2) - 1)*a**6 
 - 15*log(tan((c + d*x)/2) - 1)*a**5*b + 10*log(tan((c + d*x)/2) - 1)*a**3 
*b**3 - 3*log(tan((c + d*x)/2) - 1)*a*b**5 + 8*log(tan((c + d*x)/2) + 1)*s 
in(c + d*x)**4*a**6 + 15*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a**5*b 
- 10*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a**3*b**3 + 3*log(tan((c + 
d*x)/2) + 1)*sin(c + d*x)**4*a*b**5 - 16*log(tan((c + d*x)/2) + 1)*sin(c + 
 d*x)**2*a**6 - 30*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**5*b + 2...