3.189 \(\int \frac{\tan ^6(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=157 \[ \frac{2 a^2 \tan ^7(c+d x)}{7 d (a \sec (c+d x)+a)^{7/2}}-\frac{2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{a^{3/2} d}+\frac{2 a \tan ^5(c+d x)}{5 d (a \sec (c+d x)+a)^{5/2}}-\frac{2 \tan ^3(c+d x)}{3 d (a \sec (c+d x)+a)^{3/2}}+\frac{2 \tan (c+d x)}{a d \sqrt{a \sec (c+d x)+a}} \]

[Out]

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

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Rubi [A]  time = 0.0970172, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3887, 459, 302, 203} \[ \frac{2 a^2 \tan ^7(c+d x)}{7 d (a \sec (c+d x)+a)^{7/2}}-\frac{2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{a^{3/2} d}+\frac{2 a \tan ^5(c+d x)}{5 d (a \sec (c+d x)+a)^{5/2}}-\frac{2 \tan ^3(c+d x)}{3 d (a \sec (c+d x)+a)^{3/2}}+\frac{2 \tan (c+d x)}{a d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^6/(a + a*Sec[c + d*x])^(3/2),x]

[Out]

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

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 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

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))^{3/2}} \, dx &=-\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{x^6 \left (2+a x^2\right )}{1+a x^2} \, dx,x,-\frac{\tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{d}\\ &=\frac{2 a^2 \tan ^7(c+d x)}{7 d (a+a \sec (c+d x))^{7/2}}-\frac{\left (2 a^2\right ) \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^2 \tan ^7(c+d x)}{7 d (a+a \sec (c+d x))^{7/2}}-\frac{\left (2 a^2\right ) \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 d \sqrt{a+a \sec (c+d x)}}-\frac{2 \tan ^3(c+d x)}{3 d (a+a \sec (c+d x))^{3/2}}+\frac{2 a \tan ^5(c+d x)}{5 d (a+a \sec (c+d x))^{5/2}}+\frac{2 a^2 \tan ^7(c+d x)}{7 d (a+a \sec (c+d x))^{7/2}}+\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 d}\\ &=-\frac{2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{a^{3/2} d}+\frac{2 \tan (c+d x)}{a d \sqrt{a+a \sec (c+d x)}}-\frac{2 \tan ^3(c+d x)}{3 d (a+a \sec (c+d x))^{3/2}}+\frac{2 a \tan ^5(c+d x)}{5 d (a+a \sec (c+d x))^{5/2}}+\frac{2 a^2 \tan ^7(c+d x)}{7 d (a+a \sec (c+d x))^{7/2}}\\ \end{align*}

Mathematica [C]  time = 2.44781, size = 248, normalized size = 1.58 \[ \frac{32 \sqrt{2} \tan ^7(c+d x) \left (\frac{1}{\sec (c+d x)+1}\right )^{11/2} \left (\frac{\cos (c+d x) (7 \cos (c+d x)+11) \csc ^8\left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \left ((-198 \cos (c+d x)+61 \cos (2 (c+d x))-44 \cos (3 (c+d x))+76) \sqrt{1-\sec (c+d x)}+105 \cos ^3(c+d x) \tanh ^{-1}\left (\sqrt{1-\sec (c+d x)}\right )\right )}{3360 \sqrt{1-\sec (c+d x)}}-\frac{4}{11} \tan ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \text{Hypergeometric2F1}\left (2,\frac{11}{2},\frac{13}{2},-2 \sin ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)\right )\right )}{7 d \left (1-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )^{9/2} (a (\sec (c+d x)+1))^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Tan[c + d*x]^6/(a + a*Sec[c + d*x])^(3/2),x]

[Out]

(32*Sqrt[2]*((1 + Sec[c + d*x])^(-1))^(11/2)*((Cos[c + d*x]*(11 + 7*Cos[c + d*x])*Csc[(c + d*x)/2]^8*Sec[(c +
d*x)/2]^2*(105*ArcTanh[Sqrt[1 - Sec[c + d*x]]]*Cos[c + d*x]^3 + (76 - 198*Cos[c + d*x] + 61*Cos[2*(c + d*x)] -
44*Cos[3*(c + d*x)])*Sqrt[1 - Sec[c + d*x]]))/(3360*Sqrt[1 - Sec[c + d*x]]) - (4*Hypergeometric2F1[2, 11/2, 1
3/2, -2*Sec[c + d*x]*Sin[(c + d*x)/2]^2]*Sec[c + d*x]*Tan[(c + d*x)/2]^2)/11)*Tan[c + d*x]^7)/(7*d*(a*(1 + Sec
[c + d*x]))^(3/2)*(1 - Tan[(c + d*x)/2]^2)^(9/2))

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Maple [B]  time = 0.229, size = 391, normalized size = 2.5 \begin{align*} -{\frac{1}{840\,d{a}^{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( 105\,\sqrt{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}{\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 ) ^{7/2}+315\,\sqrt{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{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 ) ^{7/2}+315\,\sqrt{2}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ){\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 ) ^{7/2}+105\,\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 ) ^{7/2}\sin \left ( dx+c \right ) +2336\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}-2848\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}+128\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}+624\,\cos \left ( dx+c \right ) -240 \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))^(3/2),x)

[Out]

-1/840/d/a^2*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(105*2^(1/2)*sin(d*x+c)*cos(d*x+c)^3*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))^(7/2)+315*2^(1/2)*sin(d
*x+c)*cos(d*x+c)^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))^(7/2)+315*2^(1/2)*sin(d*x+c)*cos(d*x+c)*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))^(7/2)+105*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))^(7/2)*sin(d*x+c)+2336*cos(d*x+
c)^4-2848*cos(d*x+c)^3+128*cos(d*x+c)^2+624*cos(d*x+c)-240)/sin(d*x+c)/cos(d*x+c)^3

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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))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.82474, size = 914, normalized size = 5.82 \begin{align*} \left [-\frac{105 \,{\left (\cos \left (d x + c\right )^{4} + \cos \left (d x + c\right )^{3}\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 (146 \, \cos \left (d x + c\right )^{3} - 32 \, \cos \left (d x + c\right )^{2} - 24 \, \cos \left (d x + c\right ) + 15\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \,{\left (a^{2} d \cos \left (d x + c\right )^{4} + a^{2} d \cos \left (d x + c\right )^{3}\right )}}, \frac{2 \,{\left (105 \,{\left (\cos \left (d x + c\right )^{4} + \cos \left (d x + c\right )^{3}\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 (146 \, \cos \left (d x + c\right )^{3} - 32 \, \cos \left (d x + c\right )^{2} - 24 \, \cos \left (d x + c\right ) + 15\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{105 \,{\left (a^{2} d \cos \left (d x + c\right )^{4} + a^{2} d \cos \left (d x + c\right )^{3}\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))^(3/2),x, algorithm="fricas")

[Out]

[-1/105*(105*(cos(d*x + c)^4 + cos(d*x + c)^3)*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*(146*cos(d*x +
c)^3 - 32*cos(d*x + c)^2 - 24*cos(d*x + c) + 15)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^2*d*
cos(d*x + c)^4 + a^2*d*cos(d*x + c)^3), 2/105*(105*(cos(d*x + c)^4 + cos(d*x + c)^3)*sqrt(a)*arctan(sqrt((a*co
s(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) + (146*cos(d*x + c)^3 - 32*cos(d*x + c)^2 -
24*cos(d*x + c) + 15)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^4 + a^2*d*cos
(d*x + c)^3)]

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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{3}{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))**(3/2),x)

[Out]

Integral(tan(c + d*x)**6/(a*(sec(c + d*x) + 1))**(3/2), x)

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Giac [B]  time = 15.6055, size = 456, normalized size = 2.9 \begin{align*} -\frac{105 \, \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^{2} \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^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}\right )} + \frac{2 \,{\left ({\left ({\left (\frac{139 \, \sqrt{2} a^{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{539 \, \sqrt{2} a^{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{385 \, \sqrt{2} a^{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{105 \, \sqrt{2} a^{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 )}^{3} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}}}{105 \, 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))^(3/2),x, algorithm="giac")

[Out]

-1/105*(105*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*s
qrt(2) + 3)))/(a^2*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^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1))) + 2*(((139*sqrt(2)*a^2*ta
n(1/2*d*x + 1/2*c)^2/sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - 539*sqrt(2)*a^2/sgn(tan(1/2*d*x + 1/2*c)^2 - 1))*tan(1/
2*d*x + 1/2*c)^2 + 385*sqrt(2)*a^2/sgn(tan(1/2*d*x + 1/2*c)^2 - 1))*tan(1/2*d*x + 1/2*c)^2 - 105*sqrt(2)*a^2/s
gn(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)^3*sqrt(-a*tan(1/2*d*x + 1
/2*c)^2 + a)))/d