\(\int \frac {\tan ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx\) [381]

Optimal result

Integrand size = 23, antiderivative size = 97 \[ \int \frac {\tan ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {a^{5/2} \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{(a+b)^{7/2} d}+\frac {a^2 \tan (c+d x)}{(a+b)^3 d}-\frac {a \tan ^3(c+d x)}{3 (a+b)^2 d}+\frac {\tan ^5(c+d x)}{5 (a+b) d} \] Output:

-a^(5/2)*arctan((a+b)^(1/2)*tan(d*x+c)/a^(1/2))/(a+b)^(7/2)/d+a^2*tan(d*x+ 
c)/(a+b)^3/d-1/3*a*tan(d*x+c)^3/(a+b)^2/d+1/5*tan(d*x+c)^5/(a+b)/d
 

Mathematica [A] (verified)

Time = 1.26 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.14 \[ \int \frac {\tan ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {-15 a^{5/2} \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )+\sqrt {a+b} \left (23 a^2+11 a b+3 b^2-\left (11 a^2+17 a b+6 b^2\right ) \sec ^2(c+d x)+3 (a+b)^2 \sec ^4(c+d x)\right ) \tan (c+d x)}{15 (a+b)^{7/2} d} \] Input:

Integrate[Tan[c + d*x]^6/(a + b*Sin[c + d*x]^2),x]
 

Output:

(-15*a^(5/2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]] + Sqrt[a + b]*(23* 
a^2 + 11*a*b + 3*b^2 - (11*a^2 + 17*a*b + 6*b^2)*Sec[c + d*x]^2 + 3*(a + b 
)^2*Sec[c + d*x]^4)*Tan[c + d*x])/(15*(a + b)^(7/2)*d)
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.92, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3674, 254, 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.

\(\displaystyle \int \frac {\tan ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x)^6}{a+b \sin (c+d x)^2}dx\)

\(\Big \downarrow \) 3674

\(\displaystyle \frac {\int \frac {\tan ^6(c+d x)}{(a+b) \tan ^2(c+d x)+a}d\tan (c+d x)}{d}\)

\(\Big \downarrow \) 254

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

\(\Big \downarrow \) 2009

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

Input:

Int[Tan[c + d*x]^6/(a + b*Sin[c + d*x]^2),x]
 

Output:

(-((a^(5/2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(a + b)^(7/2)) + ( 
a^2*Tan[c + d*x])/(a + b)^3 - (a*Tan[c + d*x]^3)/(3*(a + b)^2) + Tan[c + d 
*x]^5/(5*(a + b)))/d
 

Defintions of rubi rules used

rule 254

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, 
 a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
 

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 3674

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

Maple [A] (verified)

Time = 3.28 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.25

method	result	size
derivativedivides	\(\frac {\frac {\frac {a^{2} \tan \left (d x +c \right )^{5}}{5}+\frac {2 a \tan \left (d x +c \right )^{5} b}{5}+\frac {b^{2} \tan \left (d x +c \right )^{5}}{5}-\frac {a^{2} \tan \left (d x +c \right )^{3}}{3}-\frac {a \tan \left (d x +c \right )^{3} b}{3}+a^{2} \tan \left (d x +c \right )}{\left (a +b \right )^{3}}-\frac {a^{3} \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{\left (a +b \right )^{3} \sqrt {a \left (a +b \right )}}}{d}\)	\(121\)
default	\(\frac {\frac {\frac {a^{2} \tan \left (d x +c \right )^{5}}{5}+\frac {2 a \tan \left (d x +c \right )^{5} b}{5}+\frac {b^{2} \tan \left (d x +c \right )^{5}}{5}-\frac {a^{2} \tan \left (d x +c \right )^{3}}{3}-\frac {a \tan \left (d x +c \right )^{3} b}{3}+a^{2} \tan \left (d x +c \right )}{\left (a +b \right )^{3}}-\frac {a^{3} \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{\left (a +b \right )^{3} \sqrt {a \left (a +b \right )}}}{d}\)	\(121\)
risch	\(\frac {2 i \left (45 a^{2} {\mathrm e}^{8 i \left (d x +c \right )}+45 \,{\mathrm e}^{8 i \left (d x +c \right )} a b +15 b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+90 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+30 \,{\mathrm e}^{6 i \left (d x +c \right )} a b +140 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+80 \,{\mathrm e}^{4 i \left (d x +c \right )} a b +30 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+70 \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2}+10 \,{\mathrm e}^{2 i \left (d x +c \right )} a b +23 a^{2}+11 a b +3 b^{2}\right )}{15 d \left (a +b \right )^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {\sqrt {-a \left (a +b \right )}\, a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a \left (a +b \right )}-2 a -b}{b}\right )}{2 \left (a +b \right )^{4} d}+\frac {\sqrt {-a \left (a +b \right )}\, a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a \left (a +b \right )}+2 a +b}{b}\right )}{2 \left (a +b \right )^{4} d}\)	\(284\)

Input:

int(tan(d*x+c)^6/(a+b*sin(d*x+c)^2),x,method=_RETURNVERBOSE)
 

Output:

1/d*(1/(a+b)^3*(1/5*a^2*tan(d*x+c)^5+2/5*a*tan(d*x+c)^5*b+1/5*b^2*tan(d*x+ 
c)^5-1/3*a^2*tan(d*x+c)^3-1/3*a*tan(d*x+c)^3*b+a^2*tan(d*x+c))-a^3/(a+b)^3 
/(a*(a+b))^(1/2)*arctan((a+b)*tan(d*x+c)/(a*(a+b))^(1/2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 181 vs. \(2 (85) = 170\).

Time = 0.12 (sec) , antiderivative size = 472, normalized size of antiderivative = 4.87 \[ \int \frac {\tan ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\left [\frac {15 \, a^{2} \sqrt {-\frac {a}{a + b}} \cos \left (d x + c\right )^{5} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left ({\left (2 \, a^{2} + 3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {-\frac {a}{a + b}} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) + 4 \, {\left ({\left (23 \, a^{2} + 11 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - {\left (11 \, a^{2} + 17 \, a b + 6 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}\right )} \sin \left (d x + c\right )}{60 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d \cos \left (d x + c\right )^{5}}, \frac {15 \, a^{2} \sqrt {\frac {a}{a + b}} \arctan \left (\frac {{\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt {\frac {a}{a + b}}}{2 \, a \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{5} + 2 \, {\left ({\left (23 \, a^{2} + 11 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - {\left (11 \, a^{2} + 17 \, a b + 6 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d \cos \left (d x + c\right )^{5}}\right ] \] Input:

integrate(tan(d*x+c)^6/(a+b*sin(d*x+c)^2),x, algorithm="fricas")
 

Output:

[1/60*(15*a^2*sqrt(-a/(a + b))*cos(d*x + c)^5*log(((8*a^2 + 8*a*b + b^2)*c 
os(d*x + c)^4 - 2*(4*a^2 + 5*a*b + b^2)*cos(d*x + c)^2 + 4*((2*a^2 + 3*a*b 
 + b^2)*cos(d*x + c)^3 - (a^2 + 2*a*b + b^2)*cos(d*x + c))*sqrt(-a/(a + b) 
)*sin(d*x + c) + a^2 + 2*a*b + b^2)/(b^2*cos(d*x + c)^4 - 2*(a*b + b^2)*co 
s(d*x + c)^2 + a^2 + 2*a*b + b^2)) + 4*((23*a^2 + 11*a*b + 3*b^2)*cos(d*x 
+ c)^4 - (11*a^2 + 17*a*b + 6*b^2)*cos(d*x + c)^2 + 3*a^2 + 6*a*b + 3*b^2) 
*sin(d*x + c))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*cos(d*x + c)^5), 1/30*(1 
5*a^2*sqrt(a/(a + b))*arctan(1/2*((2*a + b)*cos(d*x + c)^2 - a - b)*sqrt(a 
/(a + b))/(a*cos(d*x + c)*sin(d*x + c)))*cos(d*x + c)^5 + 2*((23*a^2 + 11* 
a*b + 3*b^2)*cos(d*x + c)^4 - (11*a^2 + 17*a*b + 6*b^2)*cos(d*x + c)^2 + 3 
*a^2 + 6*a*b + 3*b^2)*sin(d*x + c))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*cos 
(d*x + c)^5)]
 

Sympy [F]

\[ \int \frac {\tan ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\int \frac {\tan ^{6}{\left (c + d x \right )}}{a + b \sin ^{2}{\left (c + d x \right )}}\, dx \] Input:

integrate(tan(d*x+c)**6/(a+b*sin(d*x+c)**2),x)
 

Output:

Integral(tan(c + d*x)**6/(a + b*sin(c + d*x)**2), x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.34 \[ \int \frac {\tan ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {\frac {15 \, a^{3} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {{\left (a + b\right )} a}} - \frac {3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (d x + c\right )^{5} - 5 \, {\left (a^{2} + a b\right )} \tan \left (d x + c\right )^{3} + 15 \, a^{2} \tan \left (d x + c\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}}}{15 \, d} \] Input:

integrate(tan(d*x+c)^6/(a+b*sin(d*x+c)^2),x, algorithm="maxima")
 

Output:

-1/15*(15*a^3*arctan((a + b)*tan(d*x + c)/sqrt((a + b)*a))/((a^3 + 3*a^2*b 
 + 3*a*b^2 + b^3)*sqrt((a + b)*a)) - (3*(a^2 + 2*a*b + b^2)*tan(d*x + c)^5 
 - 5*(a^2 + a*b)*tan(d*x + c)^3 + 15*a^2*tan(d*x + c))/(a^3 + 3*a^2*b + 3* 
a*b^2 + b^3))/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 330 vs. \(2 (85) = 170\).

Time = 0.75 (sec) , antiderivative size = 330, normalized size of antiderivative = 3.40 \[ \int \frac {\tan ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {a^{3} \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )}{{\left (a^{3} d + 3 \, a^{2} b d + 3 \, a b^{2} d + b^{3} d\right )} \sqrt {a^{2} + a b}} + \frac {3 \, a^{4} d^{4} \tan \left (d x + c\right )^{5} + 12 \, a^{3} b d^{4} \tan \left (d x + c\right )^{5} + 18 \, a^{2} b^{2} d^{4} \tan \left (d x + c\right )^{5} + 12 \, a b^{3} d^{4} \tan \left (d x + c\right )^{5} + 3 \, b^{4} d^{4} \tan \left (d x + c\right )^{5} - 5 \, a^{4} d^{4} \tan \left (d x + c\right )^{3} - 15 \, a^{3} b d^{4} \tan \left (d x + c\right )^{3} - 15 \, a^{2} b^{2} d^{4} \tan \left (d x + c\right )^{3} - 5 \, a b^{3} d^{4} \tan \left (d x + c\right )^{3} + 15 \, a^{4} d^{4} \tan \left (d x + c\right ) + 30 \, a^{3} b d^{4} \tan \left (d x + c\right ) + 15 \, a^{2} b^{2} d^{4} \tan \left (d x + c\right )}{15 \, {\left (a^{5} d^{5} + 5 \, a^{4} b d^{5} + 10 \, a^{3} b^{2} d^{5} + 10 \, a^{2} b^{3} d^{5} + 5 \, a b^{4} d^{5} + b^{5} d^{5}\right )}} \] Input:

integrate(tan(d*x+c)^6/(a+b*sin(d*x+c)^2),x, algorithm="giac")
 

Output:

-a^3*arctan((a*tan(d*x + c) + b*tan(d*x + c))/sqrt(a^2 + a*b))/((a^3*d + 3 
*a^2*b*d + 3*a*b^2*d + b^3*d)*sqrt(a^2 + a*b)) + 1/15*(3*a^4*d^4*tan(d*x + 
 c)^5 + 12*a^3*b*d^4*tan(d*x + c)^5 + 18*a^2*b^2*d^4*tan(d*x + c)^5 + 12*a 
*b^3*d^4*tan(d*x + c)^5 + 3*b^4*d^4*tan(d*x + c)^5 - 5*a^4*d^4*tan(d*x + c 
)^3 - 15*a^3*b*d^4*tan(d*x + c)^3 - 15*a^2*b^2*d^4*tan(d*x + c)^3 - 5*a*b^ 
3*d^4*tan(d*x + c)^3 + 15*a^4*d^4*tan(d*x + c) + 30*a^3*b*d^4*tan(d*x + c) 
 + 15*a^2*b^2*d^4*tan(d*x + c))/(a^5*d^5 + 5*a^4*b*d^5 + 10*a^3*b^2*d^5 + 
10*a^2*b^3*d^5 + 5*a*b^4*d^5 + b^5*d^5)
 

Mupad [B] (verification not implemented)

Time = 34.63 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.15 \[ \int \frac {\tan ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {{\mathrm {tan}\left (c+d\,x\right )}^5}{5\,d\,\left (a+b\right )}-\frac {a^{5/2}\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (2\,a+2\,b\right )\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}{2\,\sqrt {a}\,{\left (a+b\right )}^{7/2}}\right )}{d\,{\left (a+b\right )}^{7/2}}-\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,d\,{\left (a+b\right )}^2}+\frac {a^2\,\mathrm {tan}\left (c+d\,x\right )}{d\,{\left (a+b\right )}^3} \] Input:

int(tan(c + d*x)^6/(a + b*sin(c + d*x)^2),x)
 

Output:

tan(c + d*x)^5/(5*d*(a + b)) - (a^(5/2)*atan((tan(c + d*x)*(2*a + 2*b)*(3* 
a*b^2 + 3*a^2*b + a^3 + b^3))/(2*a^(1/2)*(a + b)^(7/2))))/(d*(a + b)^(7/2) 
) - (a*tan(c + d*x)^3)/(3*d*(a + b)^2) + (a^2*tan(c + d*x))/(d*(a + b)^3)
 

Reduce [B] (verification not implemented)

Time = 1.03 (sec) , antiderivative size = 2081, normalized size of antiderivative = 21.45 \[ \int \frac {\tan ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx =\text {Too large to display} \] Input:

int(tan(d*x+c)^6/(a+b*sin(d*x+c)^2),x)
 

Output:

(30*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan 
((tan((c + d*x)/2)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*cos 
(c + d*x)*sin(c + d*x)**4*a - 60*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b 
)*sqrt(a + b) + a + 2*b)*atan((tan((c + d*x)/2)*a)/(sqrt(a)*sqrt(2*sqrt(b) 
*sqrt(a + b) + a + 2*b)))*cos(c + d*x)*sin(c + d*x)**2*a + 30*sqrt(b)*sqrt 
(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((tan((c + d*x)/ 
2)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*cos(c + d*x)*a - 30 
*sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((tan((c + d*x)/2)*a)/( 
sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*cos(c + d*x)*sin(c + d*x)* 
*4*a**2 - 30*sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((tan((c + 
d*x)/2)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*cos(c + d*x)*s 
in(c + d*x)**4*a*b + 60*sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan 
((tan((c + d*x)/2)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*cos 
(c + d*x)*sin(c + d*x)**2*a**2 + 60*sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a 
 + 2*b)*atan((tan((c + d*x)/2)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a 
+ 2*b)))*cos(c + d*x)*sin(c + d*x)**2*a*b - 30*sqrt(a)*sqrt(2*sqrt(b)*sqrt 
(a + b) + a + 2*b)*atan((tan((c + d*x)/2)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt( 
a + b) + a + 2*b)))*cos(c + d*x)*a**2 - 30*sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + 
 b) + a + 2*b)*atan((tan((c + d*x)/2)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + 
b) + a + 2*b)))*cos(c + d*x)*a*b + 15*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(...