\(\int \frac {\tan ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx\) [1469]

Optimal result

Integrand size = 21, antiderivative size = 350 \[ \int \frac {\tan ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {4 a^2 b^2 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2} d}-\frac {a^2 \left (2 a^2+b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2} d}-\frac {2 b^2 \left (3 a^2+b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2} d}+\frac {\cos (c+d x)}{2 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 (a-b)^3 d (1+\sin (c+d x))}-\frac {a^2 b \cos (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}-\frac {3 a^3 b \cos (c+d x)}{2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}-\frac {2 a b^3 \cos (c+d x)}{\left (a^2-b^2\right )^3 d (a+b \sin (c+d x))} \] Output:

-4*a^2*b^2*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/(a^2-b^2)^(7/2 
)/d-a^2*(2*a^2+b^2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/(a^2- 
b^2)^(7/2)/d-2*b^2*(3*a^2+b^2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^( 
1/2))/(a^2-b^2)^(7/2)/d+1/2*cos(d*x+c)/(a+b)^3/d/(1-sin(d*x+c))-1/2*cos(d* 
x+c)/(a-b)^3/d/(1+sin(d*x+c))-1/2*a^2*b*cos(d*x+c)/(a^2-b^2)^2/d/(a+b*sin( 
d*x+c))^2-3/2*a^3*b*cos(d*x+c)/(a^2-b^2)^3/d/(a+b*sin(d*x+c))-2*a*b^3*cos( 
d*x+c)/(a^2-b^2)^3/d/(a+b*sin(d*x+c))
 

Mathematica [A] (verified)

Time = 3.53 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.61 \[ \int \frac {\tan ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {-\frac {2 \left (2 a^4+11 a^2 b^2+2 b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2}}+\sin \left (\frac {1}{2} (c+d x)\right ) \left (\frac {2}{(a+b)^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {2}{(a-b)^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )-\frac {a b \cos (c+d x) \left (4 a^3+3 a b^2+b \left (3 a^2+4 b^2\right ) \sin (c+d x)\right )}{(a-b)^3 (a+b)^3 (a+b \sin (c+d x))^2}}{2 d} \] Input:

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

Output:

((-2*(2*a^4 + 11*a^2*b^2 + 2*b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 
 - b^2]])/(a^2 - b^2)^(7/2) + Sin[(c + d*x)/2]*(2/((a + b)^3*(Cos[(c + d*x 
)/2] - Sin[(c + d*x)/2])) + 2/((a - b)^3*(Cos[(c + d*x)/2] + Sin[(c + d*x) 
/2]))) - (a*b*Cos[c + d*x]*(4*a^3 + 3*a*b^2 + b*(3*a^2 + 4*b^2)*Sin[c + d* 
x]))/((a - b)^3*(a + b)^3*(a + b*Sin[c + d*x])^2))/(2*d)
 

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 3210, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3210

\(\displaystyle \int \left (-\frac {a^2}{\left (a^2-b^2\right ) (a+b \sin (c+d x))^3}-\frac {2 a b^2}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}+\frac {b^2 \left (3 a^2+b^2\right )}{\left (b^2-a^2\right )^3 (a+b \sin (c+d x))}-\frac {1}{2 (a+b)^3 (\sin (c+d x)-1)}+\frac {1}{2 (a-b)^3 (\sin (c+d x)+1)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {a^2 \left (2 a^2+b^2\right ) \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{7/2}}-\frac {4 a^2 b^2 \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{7/2}}-\frac {2 b^2 \left (3 a^2+b^2\right ) \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{7/2}}-\frac {a^2 b \cos (c+d x)}{2 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}-\frac {2 a b^3 \cos (c+d x)}{d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}-\frac {3 a^3 b \cos (c+d x)}{2 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}+\frac {\cos (c+d x)}{2 d (a+b)^3 (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 d (a-b)^3 (\sin (c+d x)+1)}\)

Input:

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

Output:

(-4*a^2*b^2*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2) 
^(7/2)*d) - (a^2*(2*a^2 + b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - 
b^2]])/((a^2 - b^2)^(7/2)*d) - (2*b^2*(3*a^2 + b^2)*ArcTan[(b + a*Tan[(c + 
 d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(7/2)*d) + Cos[c + d*x]/(2*(a + b 
)^3*d*(1 - Sin[c + d*x])) - Cos[c + d*x]/(2*(a - b)^3*d*(1 + Sin[c + d*x]) 
) - (a^2*b*Cos[c + d*x])/(2*(a^2 - b^2)^2*d*(a + b*Sin[c + d*x])^2) - (3*a 
^3*b*Cos[c + d*x])/(2*(a^2 - b^2)^3*d*(a + b*Sin[c + d*x])) - (2*a*b^3*Cos 
[c + d*x])/((a^2 - b^2)^3*d*(a + b*Sin[c + d*x]))
 

Defintions of rubi rules used

rule 2009

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

rule 3042

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

rule 3210

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

Maple [A] (verified)

Time = 2.20 (sec) , antiderivative size = 258, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(\frac {-\frac {1}{\left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{\left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2 \left (\frac {\left (\frac {5}{2} a^{3} b^{2}+a \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\frac {b \left (4 a^{4}+11 a^{2} b^{2}+6 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}+\frac {a \,b^{2} \left (11 a^{2}+10 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+\frac {a^{2} b \left (4 a^{2}+3 b^{2}\right )}{2}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )^{2}}+\frac {\left (2 a^{4}+11 a^{2} b^{2}+2 b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3}}}{d}\)	\(258\)
default	\(\frac {-\frac {1}{\left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{\left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2 \left (\frac {\left (\frac {5}{2} a^{3} b^{2}+a \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\frac {b \left (4 a^{4}+11 a^{2} b^{2}+6 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}+\frac {a \,b^{2} \left (11 a^{2}+10 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+\frac {a^{2} b \left (4 a^{2}+3 b^{2}\right )}{2}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )^{2}}+\frac {\left (2 a^{4}+11 a^{2} b^{2}+2 b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3}}}{d}\)	\(258\)
risch	\(\frac {i \left (24 i a^{4} b \,{\mathrm e}^{3 i \left (d x +c \right )}+2 i a^{2} b^{3} {\mathrm e}^{3 i \left (d x +c \right )}-11 i a^{2} b^{3} {\mathrm e}^{5 i \left (d x +c \right )}+29 i a^{2} b^{3} {\mathrm e}^{i \left (d x +c \right )}-2 i b^{5} {\mathrm e}^{i \left (d x +c \right )}-2 i b^{5} {\mathrm e}^{5 i \left (d x +c \right )}+6 a^{5} {\mathrm e}^{4 i \left (d x +c \right )}+33 a^{3} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+6 a \,b^{4} {\mathrm e}^{4 i \left (d x +c \right )}+18 i a^{4} b \,{\mathrm e}^{i \left (d x +c \right )}-2 i a^{4} b \,{\mathrm e}^{5 i \left (d x +c \right )}+4 i b^{5} {\mathrm e}^{3 i \left (d x +c \right )}+14 a^{5} {\mathrm e}^{2 i \left (d x +c \right )}+12 a^{3} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+4 a \,b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-5 a^{3} b^{2}-10 a \,b^{4}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right )^{2} d \left (a^{2}-b^{2}\right )^{3}}+\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}+\frac {11 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{2}}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{4}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}-\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}-\frac {11 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{2}}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{4}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}\)	\(808\)

Input:

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

Output:

1/d*(-1/(a-b)^3/(tan(1/2*d*x+1/2*c)+1)-1/(a+b)^3/(tan(1/2*d*x+1/2*c)-1)-2/ 
(a-b)^3/(a+b)^3*(((5/2*a^3*b^2+a*b^4)*tan(1/2*d*x+1/2*c)^3+1/2*b*(4*a^4+11 
*a^2*b^2+6*b^4)*tan(1/2*d*x+1/2*c)^2+1/2*a*b^2*(11*a^2+10*b^2)*tan(1/2*d*x 
+1/2*c)+1/2*a^2*b*(4*a^2+3*b^2))/(tan(1/2*d*x+1/2*c)^2*a+2*b*tan(1/2*d*x+1 
/2*c)+a)^2+1/2*(2*a^4+11*a^2*b^2+2*b^4)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*ta 
n(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 934, normalized size of antiderivative = 2.67 \[ \int \frac {\tan ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/4*(4*a^6*b - 12*a^4*b^3 + 12*a^2*b^5 - 4*b^7 + 2*(8*a^6*b + a^4*b^3 - 1 
1*a^2*b^5 + 2*b^7)*cos(d*x + c)^2 + ((2*a^4*b^2 + 11*a^2*b^4 + 2*b^6)*cos( 
d*x + c)^3 - 2*(2*a^5*b + 11*a^3*b^3 + 2*a*b^5)*cos(d*x + c)*sin(d*x + c) 
- (2*a^6 + 13*a^4*b^2 + 13*a^2*b^4 + 2*b^6)*cos(d*x + c))*sqrt(-a^2 + b^2) 
*log(((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 + 2*(a 
*cos(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d* 
x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)) - 2*(2*a^7 - 6*a^5*b^2 + 6*a^3 
*b^4 - 2*a*b^6 - 5*(a^5*b^2 + a^3*b^4 - 2*a*b^6)*cos(d*x + c)^2)*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*cos(d*x + c)^ 
3 - 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)*s 
in(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)), 1/2*(2*a^6*b - 6*a^4*b^3 + 6*a^2*b^5 - 2*b^7 + (8*a^6*b 
 + a^4*b^3 - 11*a^2*b^5 + 2*b^7)*cos(d*x + c)^2 + ((2*a^4*b^2 + 11*a^2*b^4 
 + 2*b^6)*cos(d*x + c)^3 - 2*(2*a^5*b + 11*a^3*b^3 + 2*a*b^5)*cos(d*x + c) 
*sin(d*x + c) - (2*a^6 + 13*a^4*b^2 + 13*a^2*b^4 + 2*b^6)*cos(d*x + c))*sq 
rt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) 
 - (2*a^7 - 6*a^5*b^2 + 6*a^3*b^4 - 2*a*b^6 - 5*(a^5*b^2 + a^3*b^4 - 2*a*b 
^6)*cos(d*x + c)^2)*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*cos(d*x + c)^3 - 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^10 - 3*a^8*b^2 + 2*a^6*b^...
 

Sympy [F]

\[ \int \frac {\tan ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\int \frac {\tan ^{2}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{3}}\, dx \] Input:

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

Output:

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

Maxima [F(-2)]

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

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f 
or more de
 

Giac [A] (verification not implemented)

Time = 0.52 (sec) , antiderivative size = 384, normalized size of antiderivative = 1.10 \[ \int \frac {\tan ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {\frac {{\left (2 \, a^{4} + 11 \, a^{2} b^{2} + 2 \, b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} + \frac {2 \, {\left (a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{2} b - b^{3}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}} + \frac {5 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 11 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 6 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 11 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 10 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, a^{4} b + 3 \, a^{2} b^{3}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}^{2}}}{d} \] Input:

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

Output:

-((2*a^4 + 11*a^2*b^2 + 2*b^4)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + 
arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))/((a^6 - 3*a^4*b^2 + 
3*a^2*b^4 - b^6)*sqrt(a^2 - b^2)) + 2*(a^3*tan(1/2*d*x + 1/2*c) + 3*a*b^2* 
tan(1/2*d*x + 1/2*c) - 3*a^2*b - b^3)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) 
*(tan(1/2*d*x + 1/2*c)^2 - 1)) + (5*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 + 2*a*b 
^4*tan(1/2*d*x + 1/2*c)^3 + 4*a^4*b*tan(1/2*d*x + 1/2*c)^2 + 11*a^2*b^3*ta 
n(1/2*d*x + 1/2*c)^2 + 6*b^5*tan(1/2*d*x + 1/2*c)^2 + 11*a^3*b^2*tan(1/2*d 
*x + 1/2*c) + 10*a*b^4*tan(1/2*d*x + 1/2*c) + 4*a^4*b + 3*a^2*b^3)/((a^6 - 
 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x 
+ 1/2*c) + a)^2))/d
 

Mupad [B] (verification not implemented)

Time = 22.11 (sec) , antiderivative size = 627, normalized size of antiderivative = 1.79 \[ \int \frac {\tan ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {\frac {5\,\left (2\,a^4\,b+a^2\,b^3\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,a^5+a^3\,b^2+12\,a\,b^4\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^4\,b+6\,a^2\,b^3+7\,b^5\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}-\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (2\,a^4\,b+11\,a^2\,b^3+2\,b^5\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}+\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (-2\,a^4+29\,a^2\,b^2+18\,b^4\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,a^4+11\,a^2\,b^2+2\,b^4\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-a^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^2+4\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2+4\,b^2\right )+4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}-\frac {\mathrm {atan}\left (\frac {\frac {\left (2\,a^4+11\,a^2\,b^2+2\,b^4\right )\,\left (2\,a^6\,b-6\,a^4\,b^3+6\,a^2\,b^5-2\,b^7\right )}{2\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}+\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^4+11\,a^2\,b^2+2\,b^4\right )\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}{{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}}{2\,a^4+11\,a^2\,b^2+2\,b^4}\right )\,\left (2\,a^4+11\,a^2\,b^2+2\,b^4\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}} \] Input:

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

Output:

((5*(2*a^4*b + a^2*b^3))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) - (2*tan(c/2 
+ (d*x)/2)^3*(12*a*b^4 + 2*a^5 + a^3*b^2))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4* 
b^2) + (2*tan(c/2 + (d*x)/2)^2*(2*a^4*b + 7*b^5 + 6*a^2*b^3))/(a^6 - b^6 + 
 3*a^2*b^4 - 3*a^4*b^2) - (3*tan(c/2 + (d*x)/2)^4*(2*a^4*b + 2*b^5 + 11*a^ 
2*b^3))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) + (a*tan(c/2 + (d*x)/2)*(18*b^ 
4 - 2*a^4 + 29*a^2*b^2))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) - (a*tan(c/2 
+ (d*x)/2)^5*(2*a^4 + 2*b^4 + 11*a^2*b^2))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4* 
b^2))/(d*(a^2*tan(c/2 + (d*x)/2)^6 - a^2 - tan(c/2 + (d*x)/2)^2*(a^2 + 4*b 
^2) + tan(c/2 + (d*x)/2)^4*(a^2 + 4*b^2) + 4*a*b*tan(c/2 + (d*x)/2)^5 - 4* 
a*b*tan(c/2 + (d*x)/2))) - (atan((((2*a^4 + 2*b^4 + 11*a^2*b^2)*(2*a^6*b - 
 2*b^7 + 6*a^2*b^5 - 6*a^4*b^3))/(2*(a + b)^(7/2)*(a - b)^(7/2)) + (a*tan( 
c/2 + (d*x)/2)*(2*a^4 + 2*b^4 + 11*a^2*b^2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4 
*b^2))/((a + b)^(7/2)*(a - b)^(7/2)))/(2*a^4 + 2*b^4 + 11*a^2*b^2))*(2*a^4 
 + 2*b^4 + 11*a^2*b^2))/(d*(a + b)^(7/2)*(a - b)^(7/2))
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 1084, normalized size of antiderivative = 3.10 \[ \int \frac {\tan ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 8*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*c 
os(c + d*x)*sin(c + d*x)**2*a**4*b**3 - 44*sqrt(a**2 - b**2)*atan((tan((c 
+ d*x)/2)*a + b)/sqrt(a**2 - b**2))*cos(c + d*x)*sin(c + d*x)**2*a**2*b**5 
 - 8*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*co 
s(c + d*x)*sin(c + d*x)**2*b**7 - 16*sqrt(a**2 - b**2)*atan((tan((c + d*x) 
/2)*a + b)/sqrt(a**2 - b**2))*cos(c + d*x)*sin(c + d*x)*a**5*b**2 - 88*sqr 
t(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*cos(c + d* 
x)*sin(c + d*x)*a**3*b**4 - 16*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a 
+ b)/sqrt(a**2 - b**2))*cos(c + d*x)*sin(c + d*x)*a*b**6 - 8*sqrt(a**2 - b 
**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*cos(c + d*x)*a**6*b 
- 44*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*co 
s(c + d*x)*a**4*b**3 - 8*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/s 
qrt(a**2 - b**2))*cos(c + d*x)*a**2*b**5 + 2*cos(c + d*x)*sin(c + d*x)**2* 
a**6*b**2 - 11*cos(c + d*x)*sin(c + d*x)**2*a**4*b**4 + cos(c + d*x)*sin(c 
 + d*x)**2*a**2*b**6 + 8*cos(c + d*x)*sin(c + d*x)**2*b**8 + 4*cos(c + d*x 
)*sin(c + d*x)*a**7*b - 22*cos(c + d*x)*sin(c + d*x)*a**5*b**3 + 2*cos(c + 
 d*x)*sin(c + d*x)*a**3*b**5 + 16*cos(c + d*x)*sin(c + d*x)*a*b**7 + 2*cos 
(c + d*x)*a**8 - 11*cos(c + d*x)*a**6*b**2 + cos(c + d*x)*a**4*b**4 + 8*co 
s(c + d*x)*a**2*b**6 + 10*sin(c + d*x)**3*a**5*b**3 + 10*sin(c + d*x)**3*a 
**3*b**5 - 20*sin(c + d*x)**3*a*b**7 + 16*sin(c + d*x)**2*a**6*b**2 + 2...