\(\int \frac {\sin ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx\) [69]

Optimal result

Integrand size = 21, antiderivative size = 206 \[ \int \frac {\sin ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {a \left (a^4-14 a^2 b^2+9 b^4\right ) x}{2 \left (a^2+b^2\right )^4}+\frac {b \left (3 a^4-8 a^2 b^2+b^4\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right )^4 d}-\frac {a^2 b}{2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))^2}-\frac {2 a b \left (a^2-b^2\right )}{\left (a^2+b^2\right )^3 d (a+b \tan (c+d x))}-\frac {\cos ^2(c+d x) \left (b \left (3 a^2-b^2\right )+a \left (a^2-3 b^2\right ) \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^3 d} \] Output:

1/2*a*(a^4-14*a^2*b^2+9*b^4)*x/(a^2+b^2)^4+b*(3*a^4-8*a^2*b^2+b^4)*ln(a*co 
s(d*x+c)+b*sin(d*x+c))/(a^2+b^2)^4/d-1/2*a^2*b/(a^2+b^2)^2/d/(a+b*tan(d*x+ 
c))^2-2*a*b*(a^2-b^2)/(a^2+b^2)^3/d/(a+b*tan(d*x+c))-1/2*cos(d*x+c)^2*(b*( 
3*a^2-b^2)+a*(a^2-3*b^2)*tan(d*x+c))/(a^2+b^2)^3/d
 

Mathematica [A] (verified)

Time = 2.61 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.53 \[ \int \frac {\sin ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx=-\frac {b \left (\frac {a \left (a^2-3 b^2\right ) \left (a^2+b^2\right ) \arctan (\tan (c+d x))}{b}+\left (3 a^2-b^2\right ) \left (a^2+b^2\right ) \cos ^2(c+d x)+\left (3 a^4-8 a^2 b^2+b^4-\frac {a^5-8 a^3 b^2+3 a b^4}{\sqrt {-b^2}}\right ) \log \left (\sqrt {-b^2}-b \tan (c+d x)\right )-2 \left (3 a^4-8 a^2 b^2+b^4\right ) \log (a+b \tan (c+d x))+\left (3 a^4-8 a^2 b^2+b^4+\frac {a^5-8 a^3 b^2+3 a b^4}{\sqrt {-b^2}}\right ) \log \left (\sqrt {-b^2}+b \tan (c+d x)\right )+\frac {a \left (a^2-3 b^2\right ) \left (a^2+b^2\right ) \sin (2 (c+d x))}{2 b}+\frac {a^2 \left (a^2+b^2\right )^2}{(a+b \tan (c+d x))^2}+\frac {4 \left (a^5-a b^4\right )}{a+b \tan (c+d x)}\right )}{2 \left (a^2+b^2\right )^4 d} \] Input:

Integrate[Sin[c + d*x]^2/(a + b*Tan[c + d*x])^3,x]
 

Output:

-1/2*(b*((a*(a^2 - 3*b^2)*(a^2 + b^2)*ArcTan[Tan[c + d*x]])/b + (3*a^2 - b 
^2)*(a^2 + b^2)*Cos[c + d*x]^2 + (3*a^4 - 8*a^2*b^2 + b^4 - (a^5 - 8*a^3*b 
^2 + 3*a*b^4)/Sqrt[-b^2])*Log[Sqrt[-b^2] - b*Tan[c + d*x]] - 2*(3*a^4 - 8* 
a^2*b^2 + b^4)*Log[a + b*Tan[c + d*x]] + (3*a^4 - 8*a^2*b^2 + b^4 + (a^5 - 
 8*a^3*b^2 + 3*a*b^4)/Sqrt[-b^2])*Log[Sqrt[-b^2] + b*Tan[c + d*x]] + (a*(a 
^2 - 3*b^2)*(a^2 + b^2)*Sin[2*(c + d*x)])/(2*b) + (a^2*(a^2 + b^2)^2)/(a + 
 b*Tan[c + d*x])^2 + (4*(a^5 - a*b^4))/(a + b*Tan[c + d*x])))/((a^2 + b^2) 
^4*d)
 

Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.31, 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.

\(\displaystyle \int \frac {\sin ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (c+d x)^2}{(a+b \tan (c+d x))^3}dx\)

\(\Big \downarrow \) 3999

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

\(\Big \downarrow \) 601

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2160

\(\displaystyle \frac {b \left (\frac {\int \left (\frac {\left (a \left (a^4-14 b^2 a^2+9 b^4\right )-2 b \left (3 a^4-8 b^2 a^2+b^4\right ) \tan (c+d x)\right ) b^2}{\left (a^2+b^2\right )^4 \left (\tan ^2(c+d x) b^2+b^2\right )}+\frac {4 a \left (a^2-b^2\right ) b^2}{\left (a^2+b^2\right )^3 (a+b \tan (c+d x))^2}+\frac {2 a^2 b^2}{\left (a^2+b^2\right )^2 (a+b \tan (c+d x))^3}+\frac {2 \left (b^6-8 a^2 b^4+3 a^4 b^2\right )}{\left (a^2+b^2\right )^4 (a+b \tan (c+d x))}\right )d(b \tan (c+d x))}{2 b^2}-\frac {a b \left (a^2-3 b^2\right ) \tan (c+d x)+b^2 \left (3 a^2-b^2\right )}{2 \left (a^2+b^2\right )^3 \left (b^2 \tan ^2(c+d x)+b^2\right )}\right )}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b \left (\frac {-\frac {4 a b^2 \left (a^2-b^2\right )}{\left (a^2+b^2\right )^3 (a+b \tan (c+d x))}-\frac {a^2 b^2}{\left (a^2+b^2\right )^2 (a+b \tan (c+d x))^2}+\frac {a b \left (a^4-14 a^2 b^2+9 b^4\right ) \arctan (\tan (c+d x))}{\left (a^2+b^2\right )^4}-\frac {b^2 \left (3 a^4-8 a^2 b^2+b^4\right ) \log \left (b^2 \tan ^2(c+d x)+b^2\right )}{\left (a^2+b^2\right )^4}+\frac {2 b^2 \left (3 a^4-8 a^2 b^2+b^4\right ) \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^4}}{2 b^2}-\frac {a b \left (a^2-3 b^2\right ) \tan (c+d x)+b^2 \left (3 a^2-b^2\right )}{2 \left (a^2+b^2\right )^3 \left (b^2 \tan ^2(c+d x)+b^2\right )}\right )}{d}\)

Input:

Int[Sin[c + d*x]^2/(a + b*Tan[c + d*x])^3,x]
 

Output:

(b*(-1/2*(b^2*(3*a^2 - b^2) + a*b*(a^2 - 3*b^2)*Tan[c + d*x])/((a^2 + b^2) 
^3*(b^2 + b^2*Tan[c + d*x]^2)) + ((a*b*(a^4 - 14*a^2*b^2 + 9*b^4)*ArcTan[T 
an[c + d*x]])/(a^2 + b^2)^4 + (2*b^2*(3*a^4 - 8*a^2*b^2 + b^4)*Log[a + b*T 
an[c + d*x]])/(a^2 + b^2)^4 - (b^2*(3*a^4 - 8*a^2*b^2 + b^4)*Log[b^2 + b^2 
*Tan[c + d*x]^2])/(a^2 + b^2)^4 - (a^2*b^2)/((a^2 + b^2)^2*(a + b*Tan[c + 
d*x])^2) - (4*a*b^2*(a^2 - b^2))/((a^2 + b^2)^3*(a + b*Tan[c + d*x])))/(2* 
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 3042

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

rule 3999

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]
 

Maple [A] (verified)

Time = 9.99 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.15

method	result	size
derivativedivides	\(\frac {\frac {\frac {\left (-\frac {1}{2} a^{5}+a^{3} b^{2}+\frac {3}{2} a \,b^{4}\right ) \tan \left (d x +c \right )-\frac {3 a^{4} b}{2}-a^{2} b^{3}+\frac {b^{5}}{2}}{1+\tan \left (d x +c \right )^{2}}+\frac {\left (-6 a^{4} b +16 a^{2} b^{3}-2 b^{5}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{4}+\frac {\left (a^{5}-14 a^{3} b^{2}+9 a \,b^{4}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{2}}{\left (a^{2}+b^{2}\right )^{4}}-\frac {a^{2} b}{2 \left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {b \left (3 a^{4}-8 b^{2} a^{2}+b^{4}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{4}}-\frac {2 a b \left (a^{2}-b^{2}\right )}{\left (a^{2}+b^{2}\right )^{3} \left (a +b \tan \left (d x +c \right )\right )}}{d}\)	\(236\)
default	\(\frac {\frac {\frac {\left (-\frac {1}{2} a^{5}+a^{3} b^{2}+\frac {3}{2} a \,b^{4}\right ) \tan \left (d x +c \right )-\frac {3 a^{4} b}{2}-a^{2} b^{3}+\frac {b^{5}}{2}}{1+\tan \left (d x +c \right )^{2}}+\frac {\left (-6 a^{4} b +16 a^{2} b^{3}-2 b^{5}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{4}+\frac {\left (a^{5}-14 a^{3} b^{2}+9 a \,b^{4}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{2}}{\left (a^{2}+b^{2}\right )^{4}}-\frac {a^{2} b}{2 \left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {b \left (3 a^{4}-8 b^{2} a^{2}+b^{4}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{4}}-\frac {2 a b \left (a^{2}-b^{2}\right )}{\left (a^{2}+b^{2}\right )^{3} \left (a +b \tan \left (d x +c \right )\right )}}{d}\)	\(236\)
risch	\(-\frac {i x b}{4 i a^{3} b -4 i a \,b^{3}-a^{4}+6 b^{2} a^{2}-b^{4}}-\frac {x a}{2 \left (4 i a^{3} b -4 i a \,b^{3}-a^{4}+6 b^{2} a^{2}-b^{4}\right )}+\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{8 \left (-3 i a^{2} b +i b^{3}+a^{3}-3 a \,b^{2}\right ) d}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 \left (3 i a^{2} b -i b^{3}+a^{3}-3 a \,b^{2}\right ) d}-\frac {6 i b \,a^{4} x}{a^{8}+4 a^{6} b^{2}+6 a^{4} b^{4}+4 a^{2} b^{6}+b^{8}}+\frac {16 i b^{3} a^{2} x}{a^{8}+4 a^{6} b^{2}+6 a^{4} b^{4}+4 a^{2} b^{6}+b^{8}}-\frac {2 i b^{5} x}{a^{8}+4 a^{6} b^{2}+6 a^{4} b^{4}+4 a^{2} b^{6}+b^{8}}-\frac {6 i b \,a^{4} c}{d \left (a^{8}+4 a^{6} b^{2}+6 a^{4} b^{4}+4 a^{2} b^{6}+b^{8}\right )}+\frac {16 i b^{3} a^{2} c}{d \left (a^{8}+4 a^{6} b^{2}+6 a^{4} b^{4}+4 a^{2} b^{6}+b^{8}\right )}-\frac {2 i b^{5} c}{d \left (a^{8}+4 a^{6} b^{2}+6 a^{4} b^{4}+4 a^{2} b^{6}+b^{8}\right )}+\frac {2 i a \,b^{2} \left (-3 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-2 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-3 i a^{3}+2 i a \,b^{2}+3 a^{2} b -2 b^{3}\right )}{\left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+i a \,{\mathrm e}^{2 i \left (d x +c \right )}-b +i a \right )^{2} \left (-i a +b \right )^{3} d \left (i a +b \right )^{4}}+\frac {3 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right ) a^{4}}{d \left (a^{8}+4 a^{6} b^{2}+6 a^{4} b^{4}+4 a^{2} b^{6}+b^{8}\right )}-\frac {8 b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right ) a^{2}}{d \left (a^{8}+4 a^{6} b^{2}+6 a^{4} b^{4}+4 a^{2} b^{6}+b^{8}\right )}+\frac {b^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{d \left (a^{8}+4 a^{6} b^{2}+6 a^{4} b^{4}+4 a^{2} b^{6}+b^{8}\right )}\)	\(776\)

Input:

int(sin(d*x+c)^2/(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

1/d*(1/(a^2+b^2)^4*(((-1/2*a^5+a^3*b^2+3/2*a*b^4)*tan(d*x+c)-3/2*a^4*b-a^2 
*b^3+1/2*b^5)/(1+tan(d*x+c)^2)+1/4*(-6*a^4*b+16*a^2*b^3-2*b^5)*ln(1+tan(d* 
x+c)^2)+1/2*(a^5-14*a^3*b^2+9*a*b^4)*arctan(tan(d*x+c)))-1/2*a^2*b/(a^2+b^ 
2)^2/(a+b*tan(d*x+c))^2+b*(3*a^4-8*a^2*b^2+b^4)/(a^2+b^2)^4*ln(a+b*tan(d*x 
+c))-2*a*b*(a^2-b^2)/(a^2+b^2)^3/(a+b*tan(d*x+c)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 526 vs. \(2 (200) = 400\).

Time = 0.13 (sec) , antiderivative size = 526, normalized size of antiderivative = 2.55 \[ \int \frac {\sin ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {13 \, a^{4} b^{3} - 8 \, a^{2} b^{5} - b^{7} - 2 \, {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (a^{5} b^{2} - 14 \, a^{3} b^{4} + 9 \, a b^{6}\right )} d x - {\left (a^{6} b + 23 \, a^{4} b^{3} - 21 \, a^{2} b^{5} - 3 \, b^{7} - 2 \, {\left (a^{7} - 15 \, a^{5} b^{2} + 23 \, a^{3} b^{4} - 9 \, a b^{6}\right )} d x\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, a^{4} b^{3} - 8 \, a^{2} b^{5} + b^{7} + {\left (3 \, a^{6} b - 11 \, a^{4} b^{3} + 9 \, a^{2} b^{5} - b^{7}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, a^{5} b^{2} - 8 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) - 2 \, {\left ({\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (4 \, a^{5} b^{2} - 3 \, a^{3} b^{4} + 3 \, a b^{6} + {\left (a^{6} b - 14 \, a^{4} b^{3} + 9 \, a^{2} b^{5}\right )} d x\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{10} + 3 \, a^{8} b^{2} + 2 \, a^{6} b^{4} - 2 \, a^{4} b^{6} - 3 \, a^{2} b^{8} - b^{10}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{9} b + 4 \, a^{7} b^{3} + 6 \, a^{5} b^{5} + 4 \, a^{3} b^{7} + a b^{9}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{8} b^{2} + 4 \, a^{6} b^{4} + 6 \, a^{4} b^{6} + 4 \, a^{2} b^{8} + b^{10}\right )} d\right )}} \] Input:

integrate(sin(d*x+c)^2/(a+b*tan(d*x+c))^3,x, algorithm="fricas")
 

Output:

1/4*(13*a^4*b^3 - 8*a^2*b^5 - b^7 - 2*(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7 
)*cos(d*x + c)^4 + 2*(a^5*b^2 - 14*a^3*b^4 + 9*a*b^6)*d*x - (a^6*b + 23*a^ 
4*b^3 - 21*a^2*b^5 - 3*b^7 - 2*(a^7 - 15*a^5*b^2 + 23*a^3*b^4 - 9*a*b^6)*d 
*x)*cos(d*x + c)^2 + 2*(3*a^4*b^3 - 8*a^2*b^5 + b^7 + (3*a^6*b - 11*a^4*b^ 
3 + 9*a^2*b^5 - b^7)*cos(d*x + c)^2 + 2*(3*a^5*b^2 - 8*a^3*b^4 + a*b^6)*co 
s(d*x + c)*sin(d*x + c))*log(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2) 
*cos(d*x + c)^2 + b^2) - 2*((a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cos(d*x 
+ c)^3 - 2*(4*a^5*b^2 - 3*a^3*b^4 + 3*a*b^6 + (a^6*b - 14*a^4*b^3 + 9*a^2* 
b^5)*d*x)*cos(d*x + c))*sin(d*x + c))/((a^10 + 3*a^8*b^2 + 2*a^6*b^4 - 2*a 
^4*b^6 - 3*a^2*b^8 - b^10)*d*cos(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)*d*cos(d*x + c)*sin(d*x + c) + (a^8*b^2 + 4*a^6*b 
^4 + 6*a^4*b^6 + 4*a^2*b^8 + b^10)*d)
 

Sympy [F(-2)]

Exception generated. \[ \int \frac {\sin ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\text {Exception raised: AttributeError} \] Input:

integrate(sin(d*x+c)**2/(a+b*tan(d*x+c))**3,x)
 

Output:

Exception raised: AttributeError >> 'NoneType' object has no attribute 'pr 
imitive'
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 463 vs. \(2 (200) = 400\).

Time = 0.15 (sec) , antiderivative size = 463, normalized size of antiderivative = 2.25 \[ \int \frac {\sin ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\frac {{\left (a^{5} - 14 \, a^{3} b^{2} + 9 \, a b^{4}\right )} {\left (d x + c\right )}}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} + \frac {2 \, {\left (3 \, a^{4} b - 8 \, a^{2} b^{3} + b^{5}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} - \frac {{\left (3 \, a^{4} b - 8 \, a^{2} b^{3} + b^{5}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} - \frac {8 \, a^{4} b - 4 \, a^{2} b^{3} + {\left (5 \, a^{3} b^{2} - 7 \, a b^{4}\right )} \tan \left (d x + c\right )^{3} + {\left (7 \, a^{4} b - 6 \, a^{2} b^{3} - b^{5}\right )} \tan \left (d x + c\right )^{2} + {\left (a^{5} + 7 \, a^{3} b^{2} - 6 \, a b^{4}\right )} \tan \left (d x + c\right )}{a^{8} + 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} + a^{2} b^{6} + {\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\right )} \tan \left (d x + c\right )^{4} + 2 \, {\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} \tan \left (d x + c\right )^{3} + {\left (a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} \tan \left (d x + c\right )}}{2 \, d} \] Input:

integrate(sin(d*x+c)^2/(a+b*tan(d*x+c))^3,x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

1/2*((a^5 - 14*a^3*b^2 + 9*a*b^4)*(d*x + c)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 
 4*a^2*b^6 + b^8) + 2*(3*a^4*b - 8*a^2*b^3 + b^5)*log(b*tan(d*x + c) + a)/ 
(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) - (3*a^4*b - 8*a^2*b^3 + b 
^5)*log(tan(d*x + c)^2 + 1)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8 
) - (8*a^4*b - 4*a^2*b^3 + (5*a^3*b^2 - 7*a*b^4)*tan(d*x + c)^3 + (7*a^4*b 
 - 6*a^2*b^3 - b^5)*tan(d*x + c)^2 + (a^5 + 7*a^3*b^2 - 6*a*b^4)*tan(d*x + 
 c))/(a^8 + 3*a^6*b^2 + 3*a^4*b^4 + a^2*b^6 + (a^6*b^2 + 3*a^4*b^4 + 3*a^2 
*b^6 + b^8)*tan(d*x + c)^4 + 2*(a^7*b + 3*a^5*b^3 + 3*a^3*b^5 + a*b^7)*tan 
(d*x + c)^3 + (a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*tan(d*x + c) 
^2 + 2*(a^7*b + 3*a^5*b^3 + 3*a^3*b^5 + a*b^7)*tan(d*x + c)))/d
 

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 372, normalized size of antiderivative = 1.81 \[ \int \frac {\sin ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {{\left (a^{5} - 14 \, a^{3} b^{2} + 9 \, a b^{4}\right )} {\left (d x + c\right )}}{2 \, {\left (a^{8} d + 4 \, a^{6} b^{2} d + 6 \, a^{4} b^{4} d + 4 \, a^{2} b^{6} d + b^{8} d\right )}} - \frac {{\left (3 \, a^{4} b - 8 \, a^{2} b^{3} + b^{5}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, {\left (a^{8} d + 4 \, a^{6} b^{2} d + 6 \, a^{4} b^{4} d + 4 \, a^{2} b^{6} d + b^{8} d\right )}} + \frac {{\left (3 \, a^{4} b^{2} - 8 \, a^{2} b^{4} + b^{6}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{8} b d + 4 \, a^{6} b^{3} d + 6 \, a^{4} b^{5} d + 4 \, a^{2} b^{7} d + b^{9} d} - \frac {8 \, a^{6} b + 4 \, a^{4} b^{3} - 4 \, a^{2} b^{5} + {\left (5 \, a^{5} b^{2} - 2 \, a^{3} b^{4} - 7 \, a b^{6}\right )} \tan \left (d x + c\right )^{3} + {\left (7 \, a^{6} b + a^{4} b^{3} - 7 \, a^{2} b^{5} - b^{7}\right )} \tan \left (d x + c\right )^{2} + {\left (a^{7} + 8 \, a^{5} b^{2} + a^{3} b^{4} - 6 \, a b^{6}\right )} \tan \left (d x + c\right )}{2 \, {\left (a^{2} + b^{2}\right )}^{4} {\left (b \tan \left (d x + c\right ) + a\right )}^{2} {\left (\tan \left (d x + c\right )^{2} + 1\right )} d} \] Input:

integrate(sin(d*x+c)^2/(a+b*tan(d*x+c))^3,x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 1.81 (sec) , antiderivative size = 433, normalized size of antiderivative = 2.10 \[ \int \frac {\sin ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (-7\,a^4\,b+6\,a^2\,b^3+b^5\right )}{2\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (7\,a\,b^4-5\,a^3\,b^2\right )}{2\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}-\frac {2\,a^2\,\left (2\,a^2\,b-b^3\right )}{\left (a^2+b^2\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}-\frac {a\,\mathrm {tan}\left (c+d\,x\right )\,\left (a^4+7\,a^2\,b^2-6\,b^4\right )}{2\,\left (a^2+b^2\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,\left (a^2+b^2\right )+a^2+b^2\,{\mathrm {tan}\left (c+d\,x\right )}^4+2\,a\,b\,\mathrm {tan}\left (c+d\,x\right )+2\,a\,b\,{\mathrm {tan}\left (c+d\,x\right )}^3\right )}+\frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (\frac {3\,b}{{\left (a^2+b^2\right )}^2}-\frac {14\,b^3}{{\left (a^2+b^2\right )}^3}+\frac {12\,b^5}{{\left (a^2+b^2\right )}^4}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (-2\,b+a\,1{}\mathrm {i}\right )}{4\,d\,\left (a^4-a^3\,b\,4{}\mathrm {i}-6\,a^2\,b^2+a\,b^3\,4{}\mathrm {i}+b^4\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (a-b\,2{}\mathrm {i}\right )}{4\,d\,\left (a^4\,1{}\mathrm {i}-4\,a^3\,b-a^2\,b^2\,6{}\mathrm {i}+4\,a\,b^3+b^4\,1{}\mathrm {i}\right )} \] Input:

int(sin(c + d*x)^2/(a + b*tan(c + d*x))^3,x)
 

Output:

((tan(c + d*x)^2*(b^5 - 7*a^4*b + 6*a^2*b^3))/(2*(a^6 + b^6 + 3*a^2*b^4 + 
3*a^4*b^2)) + (tan(c + d*x)^3*(7*a*b^4 - 5*a^3*b^2))/(2*(a^6 + b^6 + 3*a^2 
*b^4 + 3*a^4*b^2)) - (2*a^2*(2*a^2*b - b^3))/((a^2 + b^2)*(a^4 + b^4 + 2*a 
^2*b^2)) - (a*tan(c + d*x)*(a^4 - 6*b^4 + 7*a^2*b^2))/(2*(a^2 + b^2)*(a^4 
+ b^4 + 2*a^2*b^2)))/(d*(tan(c + d*x)^2*(a^2 + b^2) + a^2 + b^2*tan(c + d* 
x)^4 + 2*a*b*tan(c + d*x) + 2*a*b*tan(c + d*x)^3)) + (log(a + b*tan(c + d* 
x))*((3*b)/(a^2 + b^2)^2 - (14*b^3)/(a^2 + b^2)^3 + (12*b^5)/(a^2 + b^2)^4 
))/d + (log(tan(c + d*x) + 1i)*(a*1i - 2*b))/(4*d*(a*b^3*4i - a^3*b*4i + a 
^4 + b^4 - 6*a^2*b^2)) + (log(tan(c + d*x) - 1i)*(a - b*2i))/(4*d*(4*a*b^3 
 - 4*a^3*b + a^4*1i + b^4*1i - a^2*b^2*6i))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 1517, normalized size of antiderivative = 7.36 \[ \int \frac {\sin ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx =\text {Too large to display} \] Input:

int(sin(d*x+c)^2/(a+b*tan(d*x+c))^3,x)
 

Output:

( - 24*cos(c + d*x)*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)*a**5*b**3 + 
64*cos(c + d*x)*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)*a**3*b**5 - 8*co 
s(c + d*x)*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)*a*b**7 + 24*cos(c + d 
*x)*log(tan((c + d*x)/2)**2*a - 2*tan((c + d*x)/2)*b - a)*sin(c + d*x)*a** 
5*b**3 - 64*cos(c + d*x)*log(tan((c + d*x)/2)**2*a - 2*tan((c + d*x)/2)*b 
- a)*sin(c + d*x)*a**3*b**5 + 8*cos(c + d*x)*log(tan((c + d*x)/2)**2*a - 2 
*tan((c + d*x)/2)*b - a)*sin(c + d*x)*a*b**7 + 2*cos(c + d*x)*sin(c + d*x) 
**3*a**7*b + 6*cos(c + d*x)*sin(c + d*x)**3*a**5*b**3 + 6*cos(c + d*x)*sin 
(c + d*x)**3*a**3*b**5 + 2*cos(c + d*x)*sin(c + d*x)**3*a*b**7 + 4*cos(c + 
 d*x)*sin(c + d*x)*a**6*b**2*c + 4*cos(c + d*x)*sin(c + d*x)*a**6*b**2*d*x 
 - 56*cos(c + d*x)*sin(c + d*x)*a**4*b**4*c - 56*cos(c + d*x)*sin(c + d*x) 
*a**4*b**4*d*x + 36*cos(c + d*x)*sin(c + d*x)*a**2*b**6*c + 36*cos(c + d*x 
)*sin(c + d*x)*a**2*b**6*d*x + 12*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x 
)**2*a**6*b**2 - 44*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*a**4*b**4 
 + 36*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*a**2*b**6 - 4*log(tan(( 
c + d*x)/2)**2 + 1)*sin(c + d*x)**2*b**8 - 12*log(tan((c + d*x)/2)**2 + 1) 
*a**6*b**2 + 32*log(tan((c + d*x)/2)**2 + 1)*a**4*b**4 - 4*log(tan((c + d* 
x)/2)**2 + 1)*a**2*b**6 - 12*log(tan((c + d*x)/2)**2*a - 2*tan((c + d*x)/2 
)*b - a)*sin(c + d*x)**2*a**6*b**2 + 44*log(tan((c + d*x)/2)**2*a - 2*tan( 
(c + d*x)/2)*b - a)*sin(c + d*x)**2*a**4*b**4 - 36*log(tan((c + d*x)/2)...