3.286 \(\int \frac{1}{(a+a \tan ^2(c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=118 \[ \frac{16 \tan (c+d x)}{35 a^3 d \sqrt{a \sec ^2(c+d x)}}+\frac{8 \tan (c+d x)}{35 a^2 d \left (a \sec ^2(c+d x)\right )^{3/2}}+\frac{6 \tan (c+d x)}{35 a d \left (a \sec ^2(c+d x)\right )^{5/2}}+\frac{\tan (c+d x)}{7 d \left (a \sec ^2(c+d x)\right )^{7/2}} \]

[Out]

Tan[c + d*x]/(7*d*(a*Sec[c + d*x]^2)^(7/2)) + (6*Tan[c + d*x])/(35*a*d*(a*Sec[c + d*x]^2)^(5/2)) + (8*Tan[c +
d*x])/(35*a^2*d*(a*Sec[c + d*x]^2)^(3/2)) + (16*Tan[c + d*x])/(35*a^3*d*Sqrt[a*Sec[c + d*x]^2])

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Rubi [A]  time = 0.0532144, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3657, 4122, 192, 191} \[ \frac{16 \tan (c+d x)}{35 a^3 d \sqrt{a \sec ^2(c+d x)}}+\frac{8 \tan (c+d x)}{35 a^2 d \left (a \sec ^2(c+d x)\right )^{3/2}}+\frac{6 \tan (c+d x)}{35 a d \left (a \sec ^2(c+d x)\right )^{5/2}}+\frac{\tan (c+d x)}{7 d \left (a \sec ^2(c+d x)\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Tan[c + d*x]^2)^(-7/2),x]

[Out]

Tan[c + d*x]/(7*d*(a*Sec[c + d*x]^2)^(7/2)) + (6*Tan[c + d*x])/(35*a*d*(a*Sec[c + d*x]^2)^(5/2)) + (8*Tan[c +
d*x])/(35*a^2*d*(a*Sec[c + d*x]^2)^(3/2)) + (16*Tan[c + d*x])/(35*a^3*d*Sqrt[a*Sec[c + d*x]^2])

Rule 3657

Int[(u_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*sec[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a, b]

Rule 4122

Int[((b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(b*ff)
/f, Subst[Int[(b + b*ff^2*x^2)^(p - 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{b, e, f, p}, x] &&  !IntegerQ[p
]

Rule 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -1]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

\begin{align*} \int \frac{1}{\left (a+a \tan ^2(c+d x)\right )^{7/2}} \, dx &=\int \frac{1}{\left (a \sec ^2(c+d x)\right )^{7/2}} \, dx\\ &=\frac{a \operatorname{Subst}\left (\int \frac{1}{\left (a+a x^2\right )^{9/2}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\tan (c+d x)}{7 d \left (a \sec ^2(c+d x)\right )^{7/2}}+\frac{6 \operatorname{Subst}\left (\int \frac{1}{\left (a+a x^2\right )^{7/2}} \, dx,x,\tan (c+d x)\right )}{7 d}\\ &=\frac{\tan (c+d x)}{7 d \left (a \sec ^2(c+d x)\right )^{7/2}}+\frac{6 \tan (c+d x)}{35 a d \left (a \sec ^2(c+d x)\right )^{5/2}}+\frac{24 \operatorname{Subst}\left (\int \frac{1}{\left (a+a x^2\right )^{5/2}} \, dx,x,\tan (c+d x)\right )}{35 a d}\\ &=\frac{\tan (c+d x)}{7 d \left (a \sec ^2(c+d x)\right )^{7/2}}+\frac{6 \tan (c+d x)}{35 a d \left (a \sec ^2(c+d x)\right )^{5/2}}+\frac{8 \tan (c+d x)}{35 a^2 d \left (a \sec ^2(c+d x)\right )^{3/2}}+\frac{16 \operatorname{Subst}\left (\int \frac{1}{\left (a+a x^2\right )^{3/2}} \, dx,x,\tan (c+d x)\right )}{35 a^2 d}\\ &=\frac{\tan (c+d x)}{7 d \left (a \sec ^2(c+d x)\right )^{7/2}}+\frac{6 \tan (c+d x)}{35 a d \left (a \sec ^2(c+d x)\right )^{5/2}}+\frac{8 \tan (c+d x)}{35 a^2 d \left (a \sec ^2(c+d x)\right )^{3/2}}+\frac{16 \tan (c+d x)}{35 a^3 d \sqrt{a \sec ^2(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.184002, size = 62, normalized size = 0.53 \[ \frac{\left (-5 \sin ^6(c+d x)+21 \sin ^4(c+d x)-35 \sin ^2(c+d x)+35\right ) \tan (c+d x)}{35 a^3 d \sqrt{a \sec ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((35 - 35*Sin[c + d*x]^2 + 21*Sin[c + d*x]^4 - 5*Sin[c + d*x]^6)*Tan[c + d*x])/(35*a^3*d*Sqrt[a*Sec[c + d*x]^2
])

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Maple [A]  time = 0.021, size = 119, normalized size = 1. \begin{align*}{\frac{a}{d} \left ({\frac{\tan \left ( dx+c \right ) }{7\,a} \left ( a+a \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) ^{-{\frac{7}{2}}}}+{\frac{6}{7\,a} \left ({\frac{\tan \left ( dx+c \right ) }{5\,a} \left ( a+a \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{4}{5\,a} \left ({\frac{\tan \left ( dx+c \right ) }{3\,a} \left ( a+a \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,\tan \left ( dx+c \right ) }{3\,{a}^{2}}{\frac{1}{\sqrt{a+a \left ( \tan \left ( dx+c \right ) \right ) ^{2}}}}} \right ) } \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*tan(d*x+c)^2)^(7/2),x)

[Out]

1/d*a*(1/7/a*tan(d*x+c)/(a+a*tan(d*x+c)^2)^(7/2)+6/7/a*(1/5/a*tan(d*x+c)/(a+a*tan(d*x+c)^2)^(5/2)+4/5/a*(1/3/a
*tan(d*x+c)/(a+a*tan(d*x+c)^2)^(3/2)+2/3/a^2*tan(d*x+c)/(a+a*tan(d*x+c)^2)^(1/2))))

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Maxima [A]  time = 1.82782, size = 68, normalized size = 0.58 \begin{align*} \frac{5 \, \sin \left (7 \, d x + 7 \, c\right ) + 49 \, \sin \left (5 \, d x + 5 \, c\right ) + 245 \, \sin \left (3 \, d x + 3 \, c\right ) + 1225 \, \sin \left (d x + c\right )}{2240 \, a^{\frac{7}{2}} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*tan(d*x+c)^2)^(7/2),x, algorithm="maxima")

[Out]

1/2240*(5*sin(7*d*x + 7*c) + 49*sin(5*d*x + 5*c) + 245*sin(3*d*x + 3*c) + 1225*sin(d*x + c))/(a^(7/2)*d)

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Fricas [A]  time = 1.53594, size = 293, normalized size = 2.48 \begin{align*} \frac{{\left (16 \, \tan \left (d x + c\right )^{7} + 56 \, \tan \left (d x + c\right )^{5} + 70 \, \tan \left (d x + c\right )^{3} + 35 \, \tan \left (d x + c\right )\right )} \sqrt{a \tan \left (d x + c\right )^{2} + a}}{35 \,{\left (a^{4} d \tan \left (d x + c\right )^{8} + 4 \, a^{4} d \tan \left (d x + c\right )^{6} + 6 \, a^{4} d \tan \left (d x + c\right )^{4} + 4 \, a^{4} d \tan \left (d x + c\right )^{2} + a^{4} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*tan(d*x+c)^2)^(7/2),x, algorithm="fricas")

[Out]

1/35*(16*tan(d*x + c)^7 + 56*tan(d*x + c)^5 + 70*tan(d*x + c)^3 + 35*tan(d*x + c))*sqrt(a*tan(d*x + c)^2 + a)/
(a^4*d*tan(d*x + c)^8 + 4*a^4*d*tan(d*x + c)^6 + 6*a^4*d*tan(d*x + c)^4 + 4*a^4*d*tan(d*x + c)^2 + a^4*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a \tan ^{2}{\left (c + d x \right )} + a\right )^{\frac{7}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*tan(d*x+c)**2)**(7/2),x)

[Out]

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

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Giac [A]  time = 2.11193, size = 185, normalized size = 1.57 \begin{align*} -\frac{2 \,{\left (35 \, \sqrt{a}{\left (\frac{1}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )} + \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}^{6} - 140 \, \sqrt{a}{\left (\frac{1}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )} + \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}^{4} + 336 \, \sqrt{a}{\left (\frac{1}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )} + \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}^{2} - 320 \, \sqrt{a}\right )}}{35 \, a^{4} d{\left (\frac{1}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )} + \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}^{7} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*tan(d*x+c)^2)^(7/2),x, algorithm="giac")

[Out]

-2/35*(35*sqrt(a)*(1/tan(1/2*d*x + 1/2*c) + tan(1/2*d*x + 1/2*c))^6 - 140*sqrt(a)*(1/tan(1/2*d*x + 1/2*c) + ta
n(1/2*d*x + 1/2*c))^4 + 336*sqrt(a)*(1/tan(1/2*d*x + 1/2*c) + tan(1/2*d*x + 1/2*c))^2 - 320*sqrt(a))/(a^4*d*(1
/tan(1/2*d*x + 1/2*c) + tan(1/2*d*x + 1/2*c))^7*sgn(tan(1/2*d*x + 1/2*c)^4 - 1))