3.263 \(\int \frac{\sec ^5(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx\)

Optimal. Leaf size=138 \[ \frac{24 \cos (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}+\frac{12 \sec (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}-\frac{24 \cos (a+b x) E\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (a+b x)}}{5 b d^2 \sqrt{\sin (2 a+2 b x)}}-\frac{2 \sec ^3(a+b x)}{b d \sqrt{d \tan (a+b x)}} \]

[Out]

(-2*Sec[a + b*x]^3)/(b*d*Sqrt[d*Tan[a + b*x]]) - (24*Cos[a + b*x]*EllipticE[a - Pi/4 + b*x, 2]*Sqrt[d*Tan[a +
b*x]])/(5*b*d^2*Sqrt[Sin[2*a + 2*b*x]]) + (24*Cos[a + b*x]*(d*Tan[a + b*x])^(3/2))/(5*b*d^3) + (12*Sec[a + b*x
]*(d*Tan[a + b*x])^(3/2))/(5*b*d^3)

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Rubi [A]  time = 0.174187, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2608, 2613, 2615, 2572, 2639} \[ \frac{24 \cos (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}+\frac{12 \sec (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}-\frac{24 \cos (a+b x) E\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (a+b x)}}{5 b d^2 \sqrt{\sin (2 a+2 b x)}}-\frac{2 \sec ^3(a+b x)}{b d \sqrt{d \tan (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^5/(d*Tan[a + b*x])^(3/2),x]

[Out]

(-2*Sec[a + b*x]^3)/(b*d*Sqrt[d*Tan[a + b*x]]) - (24*Cos[a + b*x]*EllipticE[a - Pi/4 + b*x, 2]*Sqrt[d*Tan[a +
b*x]])/(5*b*d^2*Sqrt[Sin[2*a + 2*b*x]]) + (24*Cos[a + b*x]*(d*Tan[a + b*x])^(3/2))/(5*b*d^3) + (12*Sec[a + b*x
]*(d*Tan[a + b*x])^(3/2))/(5*b*d^3)

Rule 2608

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a^2*(a*Sec[
e + f*x])^(m - 2)*(b*Tan[e + f*x])^(n + 1))/(b*f*(n + 1)), x] - Dist[(a^2*(m - 2))/(b^2*(n + 1)), Int[(a*Sec[e
+ f*x])^(m - 2)*(b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && LtQ[n, -1] && (GtQ[m, 1] || (Eq
Q[m, 1] && EqQ[n, -3/2])) && IntegersQ[2*m, 2*n]

Rule 2613

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a^2*(a*Sec[
e + f*x])^(m - 2)*(b*Tan[e + f*x])^(n + 1))/(b*f*(m + n - 1)), x] + Dist[(a^2*(m - 2))/(m + n - 1), Int[(a*Sec
[e + f*x])^(m - 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (GtQ[m, 1] || (EqQ[m, 1] && EqQ[
n, 1/2])) && NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]

Rule 2615

Int[Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]]/sec[(e_.) + (f_.)*(x_)], x_Symbol] :> Dist[(Sqrt[Cos[e + f*x]]*Sqrt[b*
Tan[e + f*x]])/Sqrt[Sin[e + f*x]], Int[Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]], x], x] /; FreeQ[{b, e, f}, x]

Rule 2572

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(Sqrt[a*Sin[e +
f*x]]*Sqrt[b*Cos[e + f*x]])/Sqrt[Sin[2*e + 2*f*x]], Int[Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f},
x]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rubi steps

\begin{align*} \int \frac{\sec ^5(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx &=-\frac{2 \sec ^3(a+b x)}{b d \sqrt{d \tan (a+b x)}}+\frac{6 \int \sec ^3(a+b x) \sqrt{d \tan (a+b x)} \, dx}{d^2}\\ &=-\frac{2 \sec ^3(a+b x)}{b d \sqrt{d \tan (a+b x)}}+\frac{12 \sec (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}+\frac{12 \int \sec (a+b x) \sqrt{d \tan (a+b x)} \, dx}{5 d^2}\\ &=-\frac{2 \sec ^3(a+b x)}{b d \sqrt{d \tan (a+b x)}}+\frac{24 \cos (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}+\frac{12 \sec (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}-\frac{24 \int \cos (a+b x) \sqrt{d \tan (a+b x)} \, dx}{5 d^2}\\ &=-\frac{2 \sec ^3(a+b x)}{b d \sqrt{d \tan (a+b x)}}+\frac{24 \cos (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}+\frac{12 \sec (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}-\frac{\left (24 \sqrt{\cos (a+b x)} \sqrt{d \tan (a+b x)}\right ) \int \sqrt{\cos (a+b x)} \sqrt{\sin (a+b x)} \, dx}{5 d^2 \sqrt{\sin (a+b x)}}\\ &=-\frac{2 \sec ^3(a+b x)}{b d \sqrt{d \tan (a+b x)}}+\frac{24 \cos (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}+\frac{12 \sec (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}-\frac{\left (24 \cos (a+b x) \sqrt{d \tan (a+b x)}\right ) \int \sqrt{\sin (2 a+2 b x)} \, dx}{5 d^2 \sqrt{\sin (2 a+2 b x)}}\\ &=-\frac{2 \sec ^3(a+b x)}{b d \sqrt{d \tan (a+b x)}}-\frac{24 \cos (a+b x) E\left (\left .a-\frac{\pi }{4}+b x\right |2\right ) \sqrt{d \tan (a+b x)}}{5 b d^2 \sqrt{\sin (2 a+2 b x)}}+\frac{24 \cos (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}+\frac{12 \sec (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}\\ \end{align*}

Mathematica [C]  time = 0.868208, size = 104, normalized size = 0.75 \[ \frac{2 \csc (a+b x) \sqrt{d \tan (a+b x)} \left (\sqrt{\sec ^2(a+b x)} \left (12 \sin ^2(a+b x)+\tan ^2(a+b x)-5\right )-8 \tan ^2(a+b x) \, _2F_1\left (\frac{3}{4},\frac{3}{2};\frac{7}{4};-\tan ^2(a+b x)\right )\right )}{5 b d^2 \sqrt{\sec ^2(a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^5/(d*Tan[a + b*x])^(3/2),x]

[Out]

(2*Csc[a + b*x]*Sqrt[d*Tan[a + b*x]]*(-8*Hypergeometric2F1[3/4, 3/2, 7/4, -Tan[a + b*x]^2]*Tan[a + b*x]^2 + Sq
rt[Sec[a + b*x]^2]*(-5 + 12*Sin[a + b*x]^2 + Tan[a + b*x]^2)))/(5*b*d^2*Sqrt[Sec[a + b*x]^2])

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Maple [B]  time = 0.179, size = 529, normalized size = 3.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^5/(d*tan(b*x+a))^(3/2),x)

[Out]

-1/5/b*2^(1/2)*(12*cos(b*x+a)^3*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*
x+a))^(1/2)*((cos(b*x+a)-1)/sin(b*x+a))^(1/2)*EllipticF(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/
2))-24*cos(b*x+a)^3*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*
((cos(b*x+a)-1)/sin(b*x+a))^(1/2)*EllipticE(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))+12*cos(b
*x+a)^2*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)
-1)/sin(b*x+a))^(1/2)*EllipticF(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))-24*cos(b*x+a)^2*((1-
cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1)/sin(b*x+
a))^(1/2)*EllipticE(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))+12*cos(b*x+a)^3*2^(1/2)-6*cos(b*
x+a)^2*2^(1/2)-2^(1/2))*sin(b*x+a)/cos(b*x+a)^4/(d*sin(b*x+a)/cos(b*x+a))^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (b x + a\right )^{5}}{\left (d \tan \left (b x + a\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^5/(d*tan(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

integrate(sec(b*x + a)^5/(d*tan(b*x + a))^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{d \tan \left (b x + a\right )} \sec \left (b x + a\right )^{5}}{d^{2} \tan \left (b x + a\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^5/(d*tan(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*tan(b*x + a))*sec(b*x + a)^5/(d^2*tan(b*x + a)^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{5}{\left (a + b x \right )}}{\left (d \tan{\left (a + b x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**5/(d*tan(b*x+a))**(3/2),x)

[Out]

Integral(sec(a + b*x)**5/(d*tan(a + b*x))**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (b x + a\right )^{5}}{\left (d \tan \left (b x + a\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^5/(d*tan(b*x+a))^(3/2),x, algorithm="giac")

[Out]

integrate(sec(b*x + a)^5/(d*tan(b*x + a))^(3/2), x)