\(\int \frac {\tan ^7(c+d x)}{a+b \sin ^2(c+d x)} \, dx\) [372]

Optimal result

Integrand size = 23, antiderivative size = 128 \[ \int \frac {\tan ^7(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {a^3 \log (\cos (c+d x))}{(a+b)^4 d}-\frac {a^3 \log \left (a+b \sin ^2(c+d x)\right )}{2 (a+b)^4 d}+\frac {\left (3 a^2+3 a b+b^2\right ) \sec ^2(c+d x)}{2 (a+b)^3 d}-\frac {(3 a+2 b) \sec ^4(c+d x)}{4 (a+b)^2 d}+\frac {\sec ^6(c+d x)}{6 (a+b) d} \] Output:

a^3*ln(cos(d*x+c))/(a+b)^4/d-1/2*a^3*ln(a+b*sin(d*x+c)^2)/(a+b)^4/d+1/2*(3 
*a^2+3*a*b+b^2)*sec(d*x+c)^2/(a+b)^3/d-1/4*(3*a+2*b)*sec(d*x+c)^4/(a+b)^2/ 
d+1/6*sec(d*x+c)^6/(a+b)/d
 

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.88 \[ \int \frac {\tan ^7(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {\frac {12 a^3 \log (\cos (c+d x))}{(a+b)^4}-\frac {6 a^3 \log \left (a+b \sin ^2(c+d x)\right )}{(a+b)^4}+\frac {6 \left (3 a^2+3 a b+b^2\right ) \sec ^2(c+d x)}{(a+b)^3}-\frac {3 (3 a+2 b) \sec ^4(c+d x)}{(a+b)^2}+\frac {2 \sec ^6(c+d x)}{a+b}}{12 d} \] Input:

Integrate[Tan[c + d*x]^7/(a + b*Sin[c + d*x]^2),x]
 

Output:

((12*a^3*Log[Cos[c + d*x]])/(a + b)^4 - (6*a^3*Log[a + b*Sin[c + d*x]^2])/ 
(a + b)^4 + (6*(3*a^2 + 3*a*b + b^2)*Sec[c + d*x]^2)/(a + b)^3 - (3*(3*a + 
 2*b)*Sec[c + d*x]^4)/(a + b)^2 + (2*Sec[c + d*x]^6)/(a + b))/(12*d)
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3673, 99, 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.

\(\displaystyle \int \frac {\tan ^7(c+d x)}{a+b \sin ^2(c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x)^7}{a+b \sin (c+d x)^2}dx\)

\(\Big \downarrow \) 3673

\(\displaystyle \frac {\int \frac {\sin ^6(c+d x)}{\left (1-\sin ^2(c+d x)\right )^4 \left (b \sin ^2(c+d x)+a\right )}d\sin ^2(c+d x)}{2 d}\)

\(\Big \downarrow \) 99

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

\(\Big \downarrow \) 2009

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

Input:

Int[Tan[c + d*x]^7/(a + b*Sin[c + d*x]^2),x]
 

Output:

((a^3*Log[1 - Sin[c + d*x]^2])/(a + b)^4 - (a^3*Log[a + b*Sin[c + d*x]^2]) 
/(a + b)^4 + 1/(3*(a + b)*(1 - Sin[c + d*x]^2)^3) - (3*a + 2*b)/(2*(a + b) 
^2*(1 - Sin[c + d*x]^2)^2) + (3*a^2 + 3*a*b + b^2)/((a + b)^3*(1 - Sin[c + 
 d*x]^2)))/(2*d)
 

Defintions of rubi rules used

rule 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | 
| (GtQ[m, 0] && GeQ[n, -1]))
 

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 3673

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^ 
(m_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x]^2, x]}, Simp[ff^((m 
 + 1)/2)/(2*f)   Subst[Int[x^((m - 1)/2)*((a + b*ff*x)^p/(1 - ff*x)^((m + 1 
)/2)), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && Integ 
erQ[(m - 1)/2]
 

Maple [A] (verified)

Time = 4.66 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.89

method	result	size
derivativedivides	\(\frac {\frac {a^{3} \ln \left (\cos \left (d x +c \right )\right )}{\left (a +b \right )^{4}}-\frac {3 a +2 b}{4 \left (a +b \right )^{2} \cos \left (d x +c \right )^{4}}-\frac {-3 a^{2}-3 a b -b^{2}}{2 \left (a +b \right )^{3} \cos \left (d x +c \right )^{2}}+\frac {1}{6 \left (a +b \right ) \cos \left (d x +c \right )^{6}}-\frac {a^{3} \ln \left (a +b -b \cos \left (d x +c \right )^{2}\right )}{2 \left (a +b \right )^{4}}}{d}\)	\(114\)
default	\(\frac {\frac {a^{3} \ln \left (\cos \left (d x +c \right )\right )}{\left (a +b \right )^{4}}-\frac {3 a +2 b}{4 \left (a +b \right )^{2} \cos \left (d x +c \right )^{4}}-\frac {-3 a^{2}-3 a b -b^{2}}{2 \left (a +b \right )^{3} \cos \left (d x +c \right )^{2}}+\frac {1}{6 \left (a +b \right ) \cos \left (d x +c \right )^{6}}-\frac {a^{3} \ln \left (a +b -b \cos \left (d x +c \right )^{2}\right )}{2 \left (a +b \right )^{4}}}{d}\)	\(114\)
risch	\(\frac {6 a^{2} {\mathrm e}^{10 i \left (d x +c \right )}+6 \,{\mathrm e}^{10 i \left (d x +c \right )} a b +2 b^{2} {\mathrm e}^{10 i \left (d x +c \right )}+12 a^{2} {\mathrm e}^{8 i \left (d x +c \right )}+4 \,{\mathrm e}^{8 i \left (d x +c \right )} a b +\frac {68 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}}{3}+\frac {52 \,{\mathrm e}^{6 i \left (d x +c \right )} a b}{3}+\frac {20 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}}{3}+12 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+4 \,{\mathrm e}^{4 i \left (d x +c \right )} a b +6 \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2}+6 \,{\mathrm e}^{2 i \left (d x +c \right )} a b +2 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{d \left (a +b \right )^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{6}}+\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d \left (a^{4}+4 a^{3} b +6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}-\frac {a^{3} \ln \left ({\mathrm e}^{4 i \left (d x +c \right )}-\frac {2 \left (2 a +b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{b}+1\right )}{2 d \left (a^{4}+4 a^{3} b +6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )}\)	\(319\)

Input:

int(tan(d*x+c)^7/(a+b*sin(d*x+c)^2),x,method=_RETURNVERBOSE)
 

Output:

1/d*(a^3/(a+b)^4*ln(cos(d*x+c))-1/4*(3*a+2*b)/(a+b)^2/cos(d*x+c)^4-1/2*(-3 
*a^2-3*a*b-b^2)/(a+b)^3/cos(d*x+c)^2+1/6/(a+b)/cos(d*x+c)^6-1/2*a^3/(a+b)^ 
4*ln(a+b-b*cos(d*x+c)^2))
 

Fricas [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.40 \[ \int \frac {\tan ^7(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {6 \, a^{3} \cos \left (d x + c\right )^{6} \log \left (-b \cos \left (d x + c\right )^{2} + a + b\right ) - 12 \, a^{3} \cos \left (d x + c\right )^{6} \log \left (-\cos \left (d x + c\right )\right ) - 6 \, {\left (3 \, a^{3} + 6 \, a^{2} b + 4 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{4} - 2 \, a^{3} - 6 \, a^{2} b - 6 \, a b^{2} - 2 \, b^{3} + 3 \, {\left (3 \, a^{3} + 8 \, a^{2} b + 7 \, a b^{2} + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}}{12 \, {\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} d \cos \left (d x + c\right )^{6}} \] Input:

integrate(tan(d*x+c)^7/(a+b*sin(d*x+c)^2),x, algorithm="fricas")
 

Output:

-1/12*(6*a^3*cos(d*x + c)^6*log(-b*cos(d*x + c)^2 + a + b) - 12*a^3*cos(d* 
x + c)^6*log(-cos(d*x + c)) - 6*(3*a^3 + 6*a^2*b + 4*a*b^2 + b^3)*cos(d*x 
+ c)^4 - 2*a^3 - 6*a^2*b - 6*a*b^2 - 2*b^3 + 3*(3*a^3 + 8*a^2*b + 7*a*b^2 
+ 2*b^3)*cos(d*x + c)^2)/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*d*co 
s(d*x + c)^6)
 

Sympy [F]

\[ \int \frac {\tan ^7(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\int \frac {\tan ^{7}{\left (c + d x \right )}}{a + b \sin ^{2}{\left (c + d x \right )}}\, dx \] Input:

integrate(tan(d*x+c)**7/(a+b*sin(d*x+c)**2),x)
 

Output:

Integral(tan(c + d*x)**7/(a + b*sin(c + d*x)**2), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 273 vs. \(2 (120) = 240\).

Time = 0.04 (sec) , antiderivative size = 273, normalized size of antiderivative = 2.13 \[ \int \frac {\tan ^7(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {\frac {6 \, a^{3} \log \left (b \sin \left (d x + c\right )^{2} + a\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} - \frac {6 \, a^{3} \log \left (\sin \left (d x + c\right )^{2} - 1\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} + \frac {6 \, {\left (3 \, a^{2} + 3 \, a b + b^{2}\right )} \sin \left (d x + c\right )^{4} - 3 \, {\left (9 \, a^{2} + 7 \, a b + 2 \, b^{2}\right )} \sin \left (d x + c\right )^{2} + 11 \, a^{2} + 7 \, a b + 2 \, b^{2}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sin \left (d x + c\right )^{6} - 3 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sin \left (d x + c\right )^{4} - a^{3} - 3 \, a^{2} b - 3 \, a b^{2} - b^{3} + 3 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sin \left (d x + c\right )^{2}}}{12 \, d} \] Input:

integrate(tan(d*x+c)^7/(a+b*sin(d*x+c)^2),x, algorithm="maxima")
 

Output:

-1/12*(6*a^3*log(b*sin(d*x + c)^2 + a)/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^ 
3 + b^4) - 6*a^3*log(sin(d*x + c)^2 - 1)/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a* 
b^3 + b^4) + (6*(3*a^2 + 3*a*b + b^2)*sin(d*x + c)^4 - 3*(9*a^2 + 7*a*b + 
2*b^2)*sin(d*x + c)^2 + 11*a^2 + 7*a*b + 2*b^2)/((a^3 + 3*a^2*b + 3*a*b^2 
+ b^3)*sin(d*x + c)^6 - 3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sin(d*x + c)^4 - 
 a^3 - 3*a^2*b - 3*a*b^2 - b^3 + 3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sin(d*x 
 + c)^2))/d
 

Giac [A] (verification not implemented)

Time = 0.56 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.70 \[ \int \frac {\tan ^7(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {a^{3} b \log \left ({\left | b \cos \left (d x + c\right )^{2} - a - b \right |}\right )}{2 \, {\left (a^{4} b d + 4 \, a^{3} b^{2} d + 6 \, a^{2} b^{3} d + 4 \, a b^{4} d + b^{5} d\right )}} + \frac {a^{3} \log \left ({\left | \cos \left (d x + c\right ) \right |}\right )}{a^{4} d + 4 \, a^{3} b d + 6 \, a^{2} b^{2} d + 4 \, a b^{3} d + b^{4} d} + \frac {6 \, {\left (3 \, a^{3} + 6 \, a^{2} b + 4 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{4} + 2 \, a^{3} + 6 \, a^{2} b + 6 \, a b^{2} + 2 \, b^{3} - 3 \, {\left (3 \, a^{3} + 8 \, a^{2} b + 7 \, a b^{2} + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}}{12 \, {\left (a + b\right )}^{4} d \cos \left (d x + c\right )^{6}} \] Input:

integrate(tan(d*x+c)^7/(a+b*sin(d*x+c)^2),x, algorithm="giac")
 

Output:

-1/2*a^3*b*log(abs(b*cos(d*x + c)^2 - a - b))/(a^4*b*d + 4*a^3*b^2*d + 6*a 
^2*b^3*d + 4*a*b^4*d + b^5*d) + a^3*log(abs(cos(d*x + c)))/(a^4*d + 4*a^3* 
b*d + 6*a^2*b^2*d + 4*a*b^3*d + b^4*d) + 1/12*(6*(3*a^3 + 6*a^2*b + 4*a*b^ 
2 + b^3)*cos(d*x + c)^4 + 2*a^3 + 6*a^2*b + 6*a*b^2 + 2*b^3 - 3*(3*a^3 + 8 
*a^2*b + 7*a*b^2 + 2*b^3)*cos(d*x + c)^2)/((a + b)^4*d*cos(d*x + c)^6)
 

Mupad [B] (verification not implemented)

Time = 34.81 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.90 \[ \int \frac {\tan ^7(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {{\mathrm {tan}\left (c+d\,x\right )}^6}{6\,d\,\left (a+b\right )}+\frac {a^2\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2\,d\,{\left (a+b\right )}^3}-\frac {a^3\,\ln \left (\left (a+b\right )\,{\mathrm {tan}\left (c+d\,x\right )}^2+a\right )}{d\,\left (2\,a^4+8\,a^3\,b+12\,a^2\,b^2+8\,a\,b^3+2\,b^4\right )}-\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^4}{4\,d\,{\left (a+b\right )}^2} \] Input:

int(tan(c + d*x)^7/(a + b*sin(c + d*x)^2),x)
 

Output:

tan(c + d*x)^6/(6*d*(a + b)) + (a^2*tan(c + d*x)^2)/(2*d*(a + b)^3) - (a^3 
*log(a + tan(c + d*x)^2*(a + b)))/(d*(8*a*b^3 + 8*a^3*b + 2*a^4 + 2*b^4 + 
12*a^2*b^2)) - (a*tan(c + d*x)^4)/(4*d*(a + b)^2)
 

Reduce [B] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 1041, normalized size of antiderivative = 8.13 \[ \int \frac {\tan ^7(c+d x)}{a+b \sin ^2(c+d x)} \, dx =\text {Too large to display} \] Input:

int(tan(d*x+c)^7/(a+b*sin(d*x+c)^2),x)
 

Output:

( - 6*log( - sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + sqrt(a)*tan((c + d*x) 
/2))*sin(c + d*x)**6*a**3 + 18*log( - sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b 
) + sqrt(a)*tan((c + d*x)/2))*sin(c + d*x)**4*a**3 - 18*log( - sqrt(2*sqrt 
(b)*sqrt(a + b) - a - 2*b) + sqrt(a)*tan((c + d*x)/2))*sin(c + d*x)**2*a** 
3 + 6*log( - sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + sqrt(a)*tan((c + d*x) 
/2))*a**3 - 6*log(sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + sqrt(a)*tan((c + 
 d*x)/2))*sin(c + d*x)**6*a**3 + 18*log(sqrt(2*sqrt(b)*sqrt(a + b) - a - 2 
*b) + sqrt(a)*tan((c + d*x)/2))*sin(c + d*x)**4*a**3 - 18*log(sqrt(2*sqrt( 
b)*sqrt(a + b) - a - 2*b) + sqrt(a)*tan((c + d*x)/2))*sin(c + d*x)**2*a**3 
 + 6*log(sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + sqrt(a)*tan((c + d*x)/2)) 
*a**3 + 12*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**6*a**3 - 36*log(tan((c 
+ d*x)/2) - 1)*sin(c + d*x)**4*a**3 + 36*log(tan((c + d*x)/2) - 1)*sin(c + 
 d*x)**2*a**3 - 12*log(tan((c + d*x)/2) - 1)*a**3 + 12*log(tan((c + d*x)/2 
) + 1)*sin(c + d*x)**6*a**3 - 36*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4 
*a**3 + 36*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**3 - 12*log(tan((c 
+ d*x)/2) + 1)*a**3 - 6*log(2*sqrt(b)*sqrt(a + b) + tan((c + d*x)/2)**2*a 
+ a + 2*b)*sin(c + d*x)**6*a**3 + 18*log(2*sqrt(b)*sqrt(a + b) + tan((c + 
d*x)/2)**2*a + a + 2*b)*sin(c + d*x)**4*a**3 - 18*log(2*sqrt(b)*sqrt(a + b 
) + tan((c + d*x)/2)**2*a + a + 2*b)*sin(c + d*x)**2*a**3 + 6*log(2*sqrt(b 
)*sqrt(a + b) + tan((c + d*x)/2)**2*a + a + 2*b)*a**3 - 11*sin(c + d*x)...