3.201 \(\int \frac{\tan ^6(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=127 \[ -\frac{2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{a^{5/2} d}+\frac{2 \tan (c+d x)}{a^2 d \sqrt{a \sec (c+d x)+a}}+\frac{2 \tan ^5(c+d x)}{5 d (a \sec (c+d x)+a)^{5/2}}-\frac{2 \tan ^3(c+d x)}{3 a d (a \sec (c+d x)+a)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0865185, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3887, 302, 203} \[ -\frac{2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{a^{5/2} d}+\frac{2 \tan (c+d x)}{a^2 d \sqrt{a \sec (c+d x)+a}}+\frac{2 \tan ^5(c+d x)}{5 d (a \sec (c+d x)+a)^{5/2}}-\frac{2 \tan ^3(c+d x)}{3 a d (a \sec (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^6/(a + a*Sec[c + d*x])^(5/2),x]

[Out]

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

Rule 3887

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.), x_Symbol] :> Dist[(-2*a^(m/2 +
n + 1/2))/d, Subst[Int[(x^m*(2 + a*x^2)^(m/2 + n - 1/2))/(1 + a*x^2), x], x, Cot[c + d*x]/Sqrt[a + b*Csc[c +
d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m/2] && IntegerQ[n - 1/2]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\tan ^6(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx &=-\frac{(2 a) \operatorname{Subst}\left (\int \frac{x^6}{1+a x^2} \, dx,x,-\frac{\tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{d}\\ &=-\frac{(2 a) \operatorname{Subst}\left (\int \left (\frac{1}{a^3}-\frac{x^2}{a^2}+\frac{x^4}{a}-\frac{1}{a^3 \left (1+a x^2\right )}\right ) \, dx,x,-\frac{\tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{d}\\ &=\frac{2 \tan (c+d x)}{a^2 d \sqrt{a+a \sec (c+d x)}}-\frac{2 \tan ^3(c+d x)}{3 a d (a+a \sec (c+d x))^{3/2}}+\frac{2 \tan ^5(c+d x)}{5 d (a+a \sec (c+d x))^{5/2}}+\frac{2 \operatorname{Subst}\left (\int \frac{1}{1+a x^2} \, dx,x,-\frac{\tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{a^2 d}\\ &=-\frac{2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{a^{5/2} d}+\frac{2 \tan (c+d x)}{a^2 d \sqrt{a+a \sec (c+d x)}}-\frac{2 \tan ^3(c+d x)}{3 a d (a+a \sec (c+d x))^{3/2}}+\frac{2 \tan ^5(c+d x)}{5 d (a+a \sec (c+d x))^{5/2}}\\ \end{align*}

Mathematica [B]  time = 6.06771, size = 447, normalized size = 3.52 \[ \frac{\sqrt{2} \sqrt{\tan ^2\left (\frac{1}{2} (c+d x)\right )+1} \left (\frac{2 \tan ^2\left (\frac{1}{2} (c+d x)\right )}{\tan ^2\left (\frac{1}{2} (c+d x)\right )+1}-1\right )^3 \tan ^7(c+d x) \cot ^8\left (\frac{1}{2} (c+d x)\right ) \left (\frac{1}{\sec (c+d x)+1}\right )^{9/2} \left (\frac{8 \tan ^6\left (\frac{1}{2} (c+d x)\right )}{5 \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )+1\right )^3 \left (\frac{2 \tan ^2\left (\frac{1}{2} (c+d x)\right )}{\tan ^2\left (\frac{1}{2} (c+d x)\right )+1}-1\right )^3}+\frac{4 \tan ^4\left (\frac{1}{2} (c+d x)\right )}{3 \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )+1\right )^2 \left (\frac{2 \tan ^2\left (\frac{1}{2} (c+d x)\right )}{\tan ^2\left (\frac{1}{2} (c+d x)\right )+1}-1\right )^2}+\frac{2 \tan ^2\left (\frac{1}{2} (c+d x)\right )}{\left (\tan ^2\left (\frac{1}{2} (c+d x)\right )+1\right ) \left (\frac{2 \tan ^2\left (\frac{1}{2} (c+d x)\right )}{\tan ^2\left (\frac{1}{2} (c+d x)\right )+1}-1\right )}+\frac{\sqrt{2} \tan \left (\frac{1}{2} (c+d x)\right ) \sin ^{-1}\left (\frac{\sqrt{2} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{\tan ^2\left (\frac{1}{2} (c+d x)\right )+1}}\right )}{\sqrt{\tan ^2\left (\frac{1}{2} (c+d x)\right )+1} \sqrt{1-\frac{2 \tan ^2\left (\frac{1}{2} (c+d x)\right )}{\tan ^2\left (\frac{1}{2} (c+d x)\right )+1}}}\right )}{d \left (1-\frac{2 \tan ^2\left (\frac{1}{2} (c+d x)\right )}{\tan ^2\left (\frac{1}{2} (c+d x)\right )+1}\right )^{5/2} (a (\sec (c+d x)+1))^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Tan[c + d*x]^6/(a + a*Sec[c + d*x])^(5/2),x]

[Out]

(Sqrt[2]*Cot[(c + d*x)/2]^8*((1 + Sec[c + d*x])^(-1))^(9/2)*Sqrt[1 + Tan[(c + d*x)/2]^2]*(-1 + (2*Tan[(c + d*x
)/2]^2)/(1 + Tan[(c + d*x)/2]^2))^3*((Sqrt[2]*ArcSin[(Sqrt[2]*Tan[(c + d*x)/2])/Sqrt[1 + Tan[(c + d*x)/2]^2]]*
Tan[(c + d*x)/2])/(Sqrt[1 + Tan[(c + d*x)/2]^2]*Sqrt[1 - (2*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)]) + (
8*Tan[(c + d*x)/2]^6)/(5*(1 + Tan[(c + d*x)/2]^2)^3*(-1 + (2*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2))^3)
+ (4*Tan[(c + d*x)/2]^4)/(3*(1 + Tan[(c + d*x)/2]^2)^2*(-1 + (2*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2))^
2) + (2*Tan[(c + d*x)/2]^2)/((1 + Tan[(c + d*x)/2]^2)*(-1 + (2*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2))))
*Tan[c + d*x]^7)/(d*(a*(1 + Sec[c + d*x]))^(5/2)*(1 - (2*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2))^(5/2))

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Maple [B]  time = 0.214, size = 302, normalized size = 2.4 \begin{align*}{\frac{1}{60\,d{a}^{3}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( 15\,{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{5/2}\sqrt{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+30\,{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{5/2}\sqrt{2}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) +15\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{5/2}\sin \left ( dx+c \right ) -184\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}+272\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-112\,\cos \left ( dx+c \right ) +24 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^6/(a+a*sec(d*x+c))^(5/2),x)

[Out]

1/60/d/a^3*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(15*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*si
n(d*x+c)/cos(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*2^(1/2)*sin(d*x+c)*cos(d*x+c)^2+30*arctanh(1/2*2^(1/
2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*2^(1/2)*si
n(d*x+c)*cos(d*x+c)+15*2^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))
*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*sin(d*x+c)-184*cos(d*x+c)^3+272*cos(d*x+c)^2-112*cos(d*x+c)+24)/sin(d*x+
c)/cos(d*x+c)^2

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^6/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 2.03001, size = 849, normalized size = 6.69 \begin{align*} \left [-\frac{15 \,{\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \sqrt{-a} \log \left (\frac{2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt{-a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) - 2 \,{\left (23 \, \cos \left (d x + c\right )^{2} - 11 \, \cos \left (d x + c\right ) + 3\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}}, \frac{2 \,{\left (15 \,{\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \sqrt{a} \arctan \left (\frac{\sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt{a} \sin \left (d x + c\right )}\right ) +{\left (23 \, \cos \left (d x + c\right )^{2} - 11 \, \cos \left (d x + c\right ) + 3\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{15 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^6/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[-1/15*(15*(cos(d*x + c)^3 + cos(d*x + c)^2)*sqrt(-a)*log((2*a*cos(d*x + c)^2 - 2*sqrt(-a)*sqrt((a*cos(d*x + c
) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) - 2*(23*cos(d*x + c)^
2 - 11*cos(d*x + c) + 3)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + a^3*d*c
os(d*x + c)^2), 2/15*(15*(cos(d*x + c)^3 + cos(d*x + c)^2)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x +
c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) + (23*cos(d*x + c)^2 - 11*cos(d*x + c) + 3)*sqrt((a*cos(d*x + c) + a)
/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{6}{\left (c + d x \right )}}{\left (a \left (\sec{\left (c + d x \right )} + 1\right )\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**6/(a+a*sec(d*x+c))**(5/2),x)

[Out]

Integral(tan(c + d*x)**6/(a*(sec(c + d*x) + 1))**(5/2), x)

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Giac [B]  time = 16.5845, size = 394, normalized size = 3.1 \begin{align*} -\frac{15 \, \sqrt{-a}{\left (\frac{\log \left ({\left |{\left (\sqrt{-a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} - a{\left (2 \, \sqrt{2} + 3\right )} \right |}\right )}{a^{3} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} - \frac{\log \left ({\left |{\left (\sqrt{-a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} + a{\left (2 \, \sqrt{2} - 3\right )} \right |}\right )}{a^{3} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}\right )} + \frac{2 \,{\left ({\left (\frac{37 \, \sqrt{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} - \frac{40 \, \sqrt{2}}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + \frac{15 \, \sqrt{2}}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^6/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

-1/15*(15*sqrt(-a)*(log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a*(2*sqr
t(2) + 3)))/(a^3*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) - log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d
*x + 1/2*c)^2 + a))^2 + a*(2*sqrt(2) - 3)))/(a^3*sgn(tan(1/2*d*x + 1/2*c)^2 - 1))) + 2*((37*sqrt(2)*tan(1/2*d*
x + 1/2*c)^2/sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - 40*sqrt(2)/sgn(tan(1/2*d*x + 1/2*c)^2 - 1))*tan(1/2*d*x + 1/2*c
)^2 + 15*sqrt(2)/sgn(tan(1/2*d*x + 1/2*c)^2 - 1))*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^2*sqrt(
-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d