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

Optimal. Leaf size=82 \[ -\frac{\sqrt{\sin (2 a+2 b x)} \sec (a+b x) F\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{3 b d^2 \sqrt{d \tan (a+b x)}}-\frac{2 \sec (a+b x)}{3 b d (d \tan (a+b x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0873308, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2609, 2614, 2573, 2641} \[ -\frac{\sqrt{\sin (2 a+2 b x)} \sec (a+b x) F\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{3 b d^2 \sqrt{d \tan (a+b x)}}-\frac{2 \sec (a+b x)}{3 b d (d \tan (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2609

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

Rule 2614

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[e + f*x]]/(Sqrt[Co
s[e + f*x]]*Sqrt[b*Tan[e + f*x]]), Int[1/(Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]]), x], x] /; FreeQ[{b, e, f}, x
]

Rule 2573

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]

Rule 2641

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

Rubi steps

\begin{align*} \int \frac{\sec (a+b x)}{(d \tan (a+b x))^{5/2}} \, dx &=-\frac{2 \sec (a+b x)}{3 b d (d \tan (a+b x))^{3/2}}-\frac{\int \frac{\sec (a+b x)}{\sqrt{d \tan (a+b x)}} \, dx}{3 d^2}\\ &=-\frac{2 \sec (a+b x)}{3 b d (d \tan (a+b x))^{3/2}}-\frac{\sqrt{\sin (a+b x)} \int \frac{1}{\sqrt{\cos (a+b x)} \sqrt{\sin (a+b x)}} \, dx}{3 d^2 \sqrt{\cos (a+b x)} \sqrt{d \tan (a+b x)}}\\ &=-\frac{2 \sec (a+b x)}{3 b d (d \tan (a+b x))^{3/2}}-\frac{\left (\sec (a+b x) \sqrt{\sin (2 a+2 b x)}\right ) \int \frac{1}{\sqrt{\sin (2 a+2 b x)}} \, dx}{3 d^2 \sqrt{d \tan (a+b x)}}\\ &=-\frac{2 \sec (a+b x)}{3 b d (d \tan (a+b x))^{3/2}}-\frac{F\left (\left .a-\frac{\pi }{4}+b x\right |2\right ) \sec (a+b x) \sqrt{\sin (2 a+2 b x)}}{3 b d^2 \sqrt{d \tan (a+b x)}}\\ \end{align*}

Mathematica [C]  time = 0.694892, size = 113, normalized size = 1.38 \[ \frac{2 \cos (2 (a+b x)) \csc (a+b x) \sqrt{\sec ^2(a+b x)} \left (\sqrt{\sec ^2(a+b x)}-\sqrt [4]{-1} \tan ^{\frac{3}{2}}(a+b x) F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-1} \sqrt{\tan (a+b x)}\right )\right |-1\right )\right )}{3 b d^2 \left (\tan ^2(a+b x)-1\right ) \sqrt{d \tan (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[2*(a + b*x)]*Csc[a + b*x]*Sqrt[Sec[a + b*x]^2]*(Sqrt[Sec[a + b*x]^2] - (-1)^(1/4)*EllipticF[I*ArcSinh[(
-1)^(1/4)*Sqrt[Tan[a + b*x]]], -1]*Tan[a + b*x]^(3/2)))/(3*b*d^2*Sqrt[d*Tan[a + b*x]]*(-1 + Tan[a + b*x]^2))

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Maple [B]  time = 0.135, size = 302, normalized size = 3.7 \begin{align*} -{\frac{\sqrt{2} \left ( \cos \left ( bx+a \right ) -1 \right ) ^{2} \left ( \cos \left ( bx+a \right ) +1 \right ) ^{2}}{3\,b \left ( \sin \left ( bx+a \right ) \right ) ^{3} \left ( \cos \left ( bx+a \right ) \right ) ^{3}} \left ( \sin \left ( bx+a \right ) \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 ) }}},{\frac{\sqrt{2}}{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 ) }}}+\sin \left ( bx+a \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 ) }}}{\it EllipticF} \left ( \sqrt{{\frac{1-\cos \left ( bx+a \right ) +\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}},{\frac{\sqrt{2}}{2}} \right ) +\cos \left ( bx+a \right ) \sqrt{2} \right ) \left ({\frac{d\sin \left ( bx+a \right ) }{\cos \left ( bx+a \right ) }} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/b*2^(1/2)*(cos(b*x+a)-1)^2*(sin(b*x+a)*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)+sin(b*x+a)*((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)*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)*2^(1/2))*(cos(b*x+a)+1)^2/sin(b*x+a)^3/cos(b*x+a)^3/(d*sin(b*x+a)/cos(b*x+a))^(5/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sec(b*x + a)/(d*tan(b*x + a))^(5/2), 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 )}{d^{3} \tan \left (b x + a\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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