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

Optimal. Leaf size=104 \[ \frac{4 \cos (a+b x) (d \tan (a+b x))^{3/2}}{b d^3}-\frac{4 \cos (a+b x) E\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (a+b x)}}{b d^2 \sqrt{\sin (2 a+2 b x)}}-\frac{2 \sec (a+b x)}{b d \sqrt{d \tan (a+b x)}} \]

[Out]

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

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Rubi [A]  time = 0.133749, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 5, 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} \[ \frac{4 \cos (a+b x) (d \tan (a+b x))^{3/2}}{b d^3}-\frac{4 \cos (a+b x) E\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (a+b x)}}{b d^2 \sqrt{\sin (2 a+2 b x)}}-\frac{2 \sec (a+b x)}{b d \sqrt{d \tan (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sec[a + b*x])/(b*d*Sqrt[d*Tan[a + b*x]]) - (4*Cos[a + b*x]*EllipticE[a - Pi/4 + b*x, 2]*Sqrt[d*Tan[a + b*x
]])/(b*d^2*Sqrt[Sin[2*a + 2*b*x]]) + (4*Cos[a + b*x]*(d*Tan[a + b*x])^(3/2))/(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 ^3(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx &=-\frac{2 \sec (a+b x)}{b d \sqrt{d \tan (a+b x)}}+\frac{2 \int \sec (a+b x) \sqrt{d \tan (a+b x)} \, dx}{d^2}\\ &=-\frac{2 \sec (a+b x)}{b d \sqrt{d \tan (a+b x)}}+\frac{4 \cos (a+b x) (d \tan (a+b x))^{3/2}}{b d^3}-\frac{4 \int \cos (a+b x) \sqrt{d \tan (a+b x)} \, dx}{d^2}\\ &=-\frac{2 \sec (a+b x)}{b d \sqrt{d \tan (a+b x)}}+\frac{4 \cos (a+b x) (d \tan (a+b x))^{3/2}}{b d^3}-\frac{\left (4 \sqrt{\cos (a+b x)} \sqrt{d \tan (a+b x)}\right ) \int \sqrt{\cos (a+b x)} \sqrt{\sin (a+b x)} \, dx}{d^2 \sqrt{\sin (a+b x)}}\\ &=-\frac{2 \sec (a+b x)}{b d \sqrt{d \tan (a+b x)}}+\frac{4 \cos (a+b x) (d \tan (a+b x))^{3/2}}{b d^3}-\frac{\left (4 \cos (a+b x) \sqrt{d \tan (a+b x)}\right ) \int \sqrt{\sin (2 a+2 b x)} \, dx}{d^2 \sqrt{\sin (2 a+2 b x)}}\\ &=-\frac{2 \sec (a+b x)}{b d \sqrt{d \tan (a+b x)}}-\frac{4 \cos (a+b x) E\left (\left .a-\frac{\pi }{4}+b x\right |2\right ) \sqrt{d \tan (a+b x)}}{b d^2 \sqrt{\sin (2 a+2 b x)}}+\frac{4 \cos (a+b x) (d \tan (a+b x))^{3/2}}{b d^3}\\ \end{align*}

Mathematica [C]  time = 0.447907, size = 93, normalized size = 0.89 \[ -\frac{2 \csc (a+b x) \sqrt{d \tan (a+b x)} \left (4 \tan ^2(a+b x) \, _2F_1\left (\frac{3}{4},\frac{3}{2};\frac{7}{4};-\tan ^2(a+b x)\right )+3 \cos (2 (a+b x)) \sqrt{\sec ^2(a+b x)}\right )}{3 b d^2 \sqrt{\sec ^2(a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Csc[a + b*x]*Sqrt[d*Tan[a + b*x]]*(3*Cos[2*(a + b*x)]*Sqrt[Sec[a + b*x]^2] + 4*Hypergeometric2F1[3/4, 3/2,
7/4, -Tan[a + b*x]^2]*Tan[a + b*x]^2))/(3*b*d^2*Sqrt[Sec[a + b*x]^2])

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Maple [B]  time = 0.177, size = 491, normalized size = 4.7 \begin{align*}{\frac{\sqrt{2}\sin \left ( bx+a \right ) }{b \left ( \cos \left ( bx+a \right ) \right ) ^{2}} \left ( 4\,\cos \left ( bx+a \right ){\it EllipticE} \left ( \sqrt{{\frac{1-\cos \left ( bx+a \right ) +\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}},1/2\,\sqrt{2} \right ) \sqrt{{\frac{\cos \left ( bx+a \right ) -1}{\sin \left ( bx+a \right ) }}}\sqrt{{\frac{1-\cos \left ( bx+a \right ) +\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}\sqrt{{\frac{\cos \left ( bx+a \right ) -1+\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}-2\,\cos \left ( bx+a \right ){\it EllipticF} \left ( \sqrt{{\frac{1-\cos \left ( bx+a \right ) +\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}},1/2\,\sqrt{2} \right ) \sqrt{{\frac{\cos \left ( bx+a \right ) -1}{\sin \left ( bx+a \right ) }}}\sqrt{{\frac{1-\cos \left ( bx+a \right ) +\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}\sqrt{{\frac{\cos \left ( bx+a \right ) -1+\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}+4\,{\it EllipticE} \left ( \sqrt{{\frac{1-\cos \left ( bx+a \right ) +\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}},1/2\,\sqrt{2} \right ) \sqrt{{\frac{\cos \left ( bx+a \right ) -1}{\sin \left ( bx+a \right ) }}}\sqrt{{\frac{1-\cos \left ( bx+a \right ) +\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}\sqrt{{\frac{\cos \left ( bx+a \right ) -1+\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}-2\,{\it EllipticF} \left ( \sqrt{{\frac{1-\cos \left ( bx+a \right ) +\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}},1/2\,\sqrt{2} \right ) \sqrt{{\frac{\cos \left ( bx+a \right ) -1}{\sin \left ( bx+a \right ) }}}\sqrt{{\frac{1-\cos \left ( bx+a \right ) +\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}\sqrt{{\frac{\cos \left ( bx+a \right ) -1+\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}-2\,\cos \left ( bx+a \right ) \sqrt{2}+\sqrt{2} \right ) \left ({\frac{d\sin \left ( bx+a \right ) }{\cos \left ( bx+a \right ) }} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*2^(1/2)*(4*cos(b*x+a)*EllipticE(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*((cos(b*x+a)-1)/
sin(b*x+a))^(1/2)*((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)-2*
cos(b*x+a)*EllipticF(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*((cos(b*x+a)-1)/sin(b*x+a))^(1/
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)+4*EllipticE(((1-c
os(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*((cos(b*x+a)-1)/sin(b*x+a))^(1/2)*((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)-2*EllipticF(((1-cos(b*x+a)+sin(b*x+a))/sin(
b*x+a))^(1/2),1/2*2^(1/2))*((cos(b*x+a)-1)/sin(b*x+a))^(1/2)*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((co
s(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)-2*cos(b*x+a)*2^(1/2)+2^(1/2))*sin(b*x+a)/cos(b*x+a)^2/(d*sin(b*x+a)/c
os(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 )^{3}}{\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)^3/(d*tan(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

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

[Out]

integral(sqrt(d*tan(b*x + a))*sec(b*x + a)^3/(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 ^{3}{\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)**3/(d*tan(b*x+a))**(3/2),x)

[Out]

Integral(sec(a + b*x)**3/(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 )^{3}}{\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)^3/(d*tan(b*x+a))^(3/2),x, algorithm="giac")

[Out]

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