\(\int \frac {\tan ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) [187]

Optimal result

Integrand size = 21, antiderivative size = 333 \[ \int \frac {\tan ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {2 a^5 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2} d}+\frac {8 a^3 b^2 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2} d}+\frac {\cos (c+d x)}{12 (a+b)^2 d (1-\sin (c+d x))^2}+\frac {\cos (c+d x)}{12 (a+b)^2 d (1-\sin (c+d x))}-\frac {(3 a+b) \cos (c+d x)}{4 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{12 (a-b)^2 d (1+\sin (c+d x))^2}-\frac {\cos (c+d x)}{12 (a-b)^2 d (1+\sin (c+d x))}+\frac {(3 a-b) \cos (c+d x)}{4 (a-b)^3 d (1+\sin (c+d x))}+\frac {a^4 b \cos (c+d x)}{\left (a^2-b^2\right )^3 d (a+b \sin (c+d x))} \] Output:

2*a^5*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/(a^2-b^2)^(7/2)/d+8 
*a^3*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+1/12*cos(d*x+c)/(a+b)^2/d/(1-sin(d*x+c))^2+1/12*cos(d*x+c)/(a+b)^2/d/(1- 
sin(d*x+c))-1/4*(3*a+b)*cos(d*x+c)/(a+b)^3/d/(1-sin(d*x+c))-1/12*cos(d*x+c 
)/(a-b)^2/d/(1+sin(d*x+c))^2-1/12*cos(d*x+c)/(a-b)^2/d/(1+sin(d*x+c))+1/4* 
(3*a-b)*cos(d*x+c)/(a-b)^3/d/(1+sin(d*x+c))+a^4*b*cos(d*x+c)/(a^2-b^2)^3/d 
/(a+b*sin(d*x+c))
 

Mathematica [A] (verified)

Time = 1.61 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.02 \[ \int \frac {\tan ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {24 a^3 \left (a^2+4 b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2}}+\frac {1}{(a+b)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {2 \sin \left (\frac {1}{2} (c+d x)\right )}{(a+b)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}-\frac {4 (4 a+b) \sin \left (\frac {1}{2} (c+d x)\right )}{(a+b)^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {2 \sin \left (\frac {1}{2} (c+d x)\right )}{(a-b)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}-\frac {1}{(a-b)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 (-4 a+b) \sin \left (\frac {1}{2} (c+d x)\right )}{(a-b)^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {12 a^4 b \cos (c+d x)}{(a-b)^3 (a+b)^3 (a+b \sin (c+d x))}}{12 d} \] Input:

Integrate[Tan[c + d*x]^4/(a + b*Sin[c + d*x])^2,x]
 

Output:

((24*a^3*(a^2 + 4*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/( 
a^2 - b^2)^(7/2) + 1/((a + b)^2*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + 
 (2*Sin[(c + d*x)/2])/((a + b)^2*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3) 
- (4*(4*a + b)*Sin[(c + d*x)/2])/((a + b)^3*(Cos[(c + d*x)/2] - Sin[(c + d 
*x)/2])) + (2*Sin[(c + d*x)/2])/((a - b)^2*(Cos[(c + d*x)/2] + Sin[(c + d* 
x)/2])^3) - 1/((a - b)^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (4*(-4 
*a + b)*Sin[(c + d*x)/2])/((a - b)^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) 
) + (12*a^4*b*Cos[c + d*x])/((a - b)^3*(a + b)^3*(a + b*Sin[c + d*x])))/(1 
2*d)
 

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 333, 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.

\(\displaystyle \int \frac {\tan ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x)^4}{(a+b \sin (c+d x))^2}dx\)

\(\Big \downarrow \) 3210

\(\displaystyle \int \left (\frac {a^4}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}+\frac {4 a^3 b^2}{\left (a^2-b^2\right )^3 (a+b \sin (c+d x))}+\frac {3 a+b}{4 (a+b)^3 (\sin (c+d x)-1)}+\frac {b-3 a}{4 (a-b)^3 (\sin (c+d x)+1)}+\frac {1}{4 (a+b)^2 (\sin (c+d x)-1)^2}+\frac {1}{4 (a-b)^2 (\sin (c+d x)+1)^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 a^5 \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{7/2}}+\frac {a^4 b \cos (c+d x)}{d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}+\frac {8 a^3 b^2 \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{7/2}}-\frac {(3 a+b) \cos (c+d x)}{4 d (a+b)^3 (1-\sin (c+d x))}+\frac {\cos (c+d x)}{12 d (a+b)^2 (1-\sin (c+d x))}+\frac {(3 a-b) \cos (c+d x)}{4 d (a-b)^3 (\sin (c+d x)+1)}-\frac {\cos (c+d x)}{12 d (a-b)^2 (\sin (c+d x)+1)}+\frac {\cos (c+d x)}{12 d (a+b)^2 (1-\sin (c+d x))^2}-\frac {\cos (c+d x)}{12 d (a-b)^2 (\sin (c+d x)+1)^2}\)

Input:

Int[Tan[c + d*x]^4/(a + b*Sin[c + d*x])^2,x]
 

Output:

(2*a^5*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(7/2 
)*d) + (8*a^3*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]/(12*(a + b)^2*d*(1 - Sin[c + d*x])^2) + Cos 
[c + d*x]/(12*(a + b)^2*d*(1 - Sin[c + d*x])) - ((3*a + b)*Cos[c + d*x])/( 
4*(a + b)^3*d*(1 - Sin[c + d*x])) - Cos[c + d*x]/(12*(a - b)^2*d*(1 + Sin[ 
c + d*x])^2) - Cos[c + d*x]/(12*(a - b)^2*d*(1 + Sin[c + d*x])) + ((3*a - 
b)*Cos[c + d*x])/(4*(a - b)^3*d*(1 + Sin[c + d*x])) + (a^4*b*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 = 1.48 (sec) , antiderivative size = 248, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(\frac {\frac {2 a^{3} \left (\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a b}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {\left (a^{2}+4 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3}}-\frac {1}{3 \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{2 \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {a}{\left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{3 \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{2 \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {a}{\left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\)	\(248\)
default	\(\frac {\frac {2 a^{3} \left (\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a b}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {\left (a^{2}+4 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3}}-\frac {1}{3 \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{2 \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {a}{\left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{3 \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{2 \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {a}{\left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\)	\(248\)
risch	\(\frac {\frac {22 a^{5} {\mathrm e}^{i \left (d x +c \right )}}{3}+6 i a^{2} b^{3}+\frac {64 a^{3} b^{2} {\mathrm e}^{3 i \left (d x +c \right )}}{3}+\frac {4 a \,b^{4} {\mathrm e}^{5 i \left (d x +c \right )}}{3}+6 i a^{2} b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+\frac {14 i a^{2} b^{3} {\mathrm e}^{4 i \left (d x +c \right )}}{3}+2 i b^{5} {\mathrm e}^{6 i \left (d x +c \right )}+\frac {14 i b \,a^{4}}{3}+\frac {82 i a^{4} b \,{\mathrm e}^{4 i \left (d x +c \right )}}{3}-2 i b^{5} {\mathrm e}^{4 i \left (d x +c \right )}+4 a^{3} b^{2} {\mathrm e}^{i \left (d x +c \right )}+14 i a^{4} b \,{\mathrm e}^{6 i \left (d x +c \right )}+2 a^{5} {\mathrm e}^{7 i \left (d x +c \right )}-\frac {4 a \,b^{4} {\mathrm e}^{i \left (d x +c \right )}}{3}-\frac {16 a \,b^{4} {\mathrm e}^{3 i \left (d x +c \right )}}{3}+14 a^{5} {\mathrm e}^{3 i \left (d x +c \right )}+14 a^{5} {\mathrm e}^{5 i \left (d x +c \right )}-6 i a^{2} b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+\frac {70 i a^{4} b \,{\mathrm e}^{2 i \left (d x +c \right )}}{3}+8 a^{3} b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+\frac {44 a^{3} b^{2} {\mathrm e}^{5 i \left (d x +c \right )}}{3}+\frac {2 i b^{5} {\mathrm e}^{2 i \left (d x +c \right )}}{3}-\frac {2 i b^{5}}{3}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )} b -b +2 i {\mathrm e}^{i \left (d x +c \right )} a \right ) \left (a^{2}-b^{2}\right )^{3} d}-\frac {a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}-\frac {4 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) b^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}+\frac {a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}+\frac {4 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) b^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}\)	\(729\)

Input:

int(tan(d*x+c)^4/(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 815, normalized size of antiderivative = 2.45 \[ \int \frac {\tan ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx =\text {Too large to display} \] Input:

integrate(tan(d*x+c)^4/(a+b*sin(d*x+c))^2,x, algorithm="fricas")
 

Output:

[-1/6*(2*a^6*b - 6*a^4*b^3 + 6*a^2*b^5 - 2*b^7 - 2*(7*a^6*b + 2*a^4*b^3 - 
10*a^2*b^5 + b^7)*cos(d*x + c)^4 - 2*(7*a^6*b - 16*a^4*b^3 + 11*a^2*b^5 - 
2*b^7)*cos(d*x + c)^2 - 3*((a^5*b + 4*a^3*b^3)*cos(d*x + c)^3*sin(d*x + c) 
 + (a^6 + 4*a^4*b^2)*cos(d*x + c)^3)*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*(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6 - (4*a^7 
- 7*a^5*b^2 + 2*a^3*b^4 + a*b^6)*cos(d*x + c)^2)*sin(d*x + c))/((a^8*b - 4 
*a^6*b^3 + 6*a^4*b^5 - 4*a^2*b^7 + b^9)*d*cos(d*x + c)^3*sin(d*x + c) + (a 
^9 - 4*a^7*b^2 + 6*a^5*b^4 - 4*a^3*b^6 + a*b^8)*d*cos(d*x + c)^3), -1/3*(a 
^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7 - (7*a^6*b + 2*a^4*b^3 - 10*a^2*b^5 + b 
^7)*cos(d*x + c)^4 - (7*a^6*b - 16*a^4*b^3 + 11*a^2*b^5 - 2*b^7)*cos(d*x + 
 c)^2 + 3*((a^5*b + 4*a^3*b^3)*cos(d*x + c)^3*sin(d*x + c) + (a^6 + 4*a^4* 
b^2)*cos(d*x + c)^3)*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^ 
2 - b^2)*cos(d*x + c))) - (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6 - (4*a^7 - 
7*a^5*b^2 + 2*a^3*b^4 + a*b^6)*cos(d*x + c)^2)*sin(d*x + c))/((a^8*b - 4*a 
^6*b^3 + 6*a^4*b^5 - 4*a^2*b^7 + b^9)*d*cos(d*x + c)^3*sin(d*x + c) + (a^9 
 - 4*a^7*b^2 + 6*a^5*b^4 - 4*a^3*b^6 + a*b^8)*d*cos(d*x + c)^3)]
 

Sympy [F]

\[ \int \frac {\tan ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\int \frac {\tan ^{4}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \] Input:

integrate(tan(d*x+c)**4/(a+b*sin(d*x+c))**2,x)
 

Output:

Integral(tan(c + d*x)**4/(a + b*sin(c + d*x))**2, x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(tan(d*x+c)^4/(a+b*sin(d*x+c))^2,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 = 0.25 (sec) , antiderivative size = 406, normalized size of antiderivative = 1.22 \[ \int \frac {\tan ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {2 \, {\left (\frac {3 \, {\left (a^{5} + 4 \, a^{3} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} + \frac {3 \, {\left (a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{4} b\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}} + \frac {3 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 6 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 10 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 18 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 10 \, a^{3} b - 2 \, a b^{3}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}\right )}}{3 \, d} \] Input:

integrate(tan(d*x+c)^4/(a+b*sin(d*x+c))^2,x, algorithm="giac")
 

Output:

2/3*(3*(a^5 + 4*a^3*b^2)*(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)) + 3*(a^3*b^2*tan(1/2*d*x + 1/2*c) + a^4*b)/((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)) + (3*a^4*tan(1/2*d*x + 1/2*c)^5 + 9*a^2*b^2*tan(1/2*d*x 
 + 1/2*c)^5 - 6*a^3*b*tan(1/2*d*x + 1/2*c)^4 - 6*a*b^3*tan(1/2*d*x + 1/2*c 
)^4 - 10*a^4*tan(1/2*d*x + 1/2*c)^3 - 18*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 
4*b^4*tan(1/2*d*x + 1/2*c)^3 + 24*a^3*b*tan(1/2*d*x + 1/2*c)^2 + 3*a^4*tan 
(1/2*d*x + 1/2*c) + 9*a^2*b^2*tan(1/2*d*x + 1/2*c) - 10*a^3*b - 2*a*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)^3))/d
 

Mupad [B] (verification not implemented)

Time = 19.76 (sec) , antiderivative size = 722, normalized size of antiderivative = 2.17 \[ \int \frac {\tan ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx =\text {Too large to display} \] Input:

int(tan(c + d*x)^4/(a + b*sin(c + d*x))^2,x)
 

Output:

((2*(13*a^4*b + 2*a^2*b^3))/(3*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) + (2*t 
an(c/2 + (d*x)/2)*(4*a*b^4 - 3*a^5 + 14*a^3*b^2))/(3*(a^6 - b^6 + 3*a^2*b^ 
4 - 3*a^4*b^2)) - (2*tan(c/2 + (d*x)/2)^6*(a^4*b + 4*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)^2*(29*a^4*b + 16*a^2*b^3) 
)/(3*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) + (2*tan(c/2 + (d*x)/2)^5*(8*a*b 
^4 + 7*a^5 + 30*a^3*b^2))/(3*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) - (2*tan 
(c/2 + (d*x)/2)^3*(4*a*b^4 - 7*a^5 + 48*a^3*b^2))/(3*(a^6 - b^6 + 3*a^2*b^ 
4 - 3*a^4*b^2)) + (2*tan(c/2 + (d*x)/2)^4*(11*a^4*b - 8*b^5 + 42*a^2*b^3)) 
/(3*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) - (2*a^3*tan(c/2 + (d*x)/2)^7*(a^ 
2 + 4*b^2))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2))/(d*(a + 2*b*tan(c/2 + (d* 
x)/2) - 2*a*tan(c/2 + (d*x)/2)^2 + 2*a*tan(c/2 + (d*x)/2)^6 - a*tan(c/2 + 
(d*x)/2)^8 - 6*b*tan(c/2 + (d*x)/2)^3 + 6*b*tan(c/2 + (d*x)/2)^5 - 2*b*tan 
(c/2 + (d*x)/2)^7)) + (2*a^3*atan(((a^3*(a^2 + 4*b^2)*(2*a^6*b - 2*b^7 + 6 
*a^2*b^5 - 6*a^4*b^3))/((a + b)^(7/2)*(a - b)^(7/2)) + (2*a^4*tan(c/2 + (d 
*x)/2)*(a^2 + 4*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^5 + 8*a^3*b^2))*(a^2 + 4*b^2))/(d*(a + b)^(7/2)*(a - b 
)^(7/2))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 1255, normalized size of antiderivative = 3.77 \[ \int \frac {\tan ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx =\text {Too large to display} \] Input:

int(tan(d*x+c)^4/(a+b*sin(d*x+c))^2,x)
 

Output:

(6*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)**3*a**5*b**2 + 24*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)**3*a**3*b**4 + 
6*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**6*b + 24*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**4*b**3 - 6*sq 
rt(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 - 24*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**3*b**4 - 6*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 - 24*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2) 
)*cos(c + d*x)*a**4*b**3 - 3*cos(c + d*x)*sin(c + d*x)**3*a**7*b + 4*cos(c 
 + d*x)*sin(c + d*x)**3*a**5*b**3 + 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 - 3*cos(c + d*x)*sin(c + d*x)**2* 
a**8 + 4*cos(c + d*x)*sin(c + d*x)**2*a**6*b**2 + cos(c + d*x)*sin(c + d*x 
)**2*a**4*b**4 - 2*cos(c + d*x)*sin(c + d*x)**2*a**2*b**6 + 3*cos(c + d*x) 
*sin(c + d*x)*a**7*b - 4*cos(c + d*x)*sin(c + d*x)*a**5*b**3 - cos(c + d*x 
)*sin(c + d*x)*a**3*b**5 + 2*cos(c + d*x)*sin(c + d*x)*a*b**7 + 3*cos(c + 
d*x)*a**8 - 4*cos(c + d*x)*a**6*b**2 - cos(c + d*x)*a**4*b**4 + 2*cos(c + 
d*x)*a**2*b**6 - 7*sin(c + d*x)**4*a**6*b**2 - 2*sin(c + d*x)**4*a**4*b...