3.301 \(\int \sqrt{d \sec (e+f x)} (b \tan (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=88 \[ \frac{b \sqrt{b \tan (e+f x)} \sqrt{d \sec (e+f x)}}{f}-\frac{b^2 \sqrt{\sin (e+f x)} F\left (\left .\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{d \sec (e+f x)}}{f \sqrt{b \tan (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.115462, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2611, 2616, 2642, 2641} \[ \frac{b \sqrt{b \tan (e+f x)} \sqrt{d \sec (e+f x)}}{f}-\frac{b^2 \sqrt{\sin (e+f x)} F\left (\left .\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{d \sec (e+f x)}}{f \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2611

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

Rule 2616

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

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, 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 \sqrt{d \sec (e+f x)} (b \tan (e+f x))^{3/2} \, dx &=\frac{b \sqrt{d \sec (e+f x)} \sqrt{b \tan (e+f x)}}{f}-\frac{1}{2} b^2 \int \frac{\sqrt{d \sec (e+f x)}}{\sqrt{b \tan (e+f x)}} \, dx\\ &=\frac{b \sqrt{d \sec (e+f x)} \sqrt{b \tan (e+f x)}}{f}-\frac{\left (b^2 \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}\right ) \int \frac{1}{\sqrt{b \sin (e+f x)}} \, dx}{2 \sqrt{b \tan (e+f x)}}\\ &=\frac{b \sqrt{d \sec (e+f x)} \sqrt{b \tan (e+f x)}}{f}-\frac{\left (b^2 \sqrt{d \sec (e+f x)} \sqrt{\sin (e+f x)}\right ) \int \frac{1}{\sqrt{\sin (e+f x)}} \, dx}{2 \sqrt{b \tan (e+f x)}}\\ &=-\frac{b^2 F\left (\left .\frac{1}{2} \left (e-\frac{\pi }{2}+f x\right )\right |2\right ) \sqrt{d \sec (e+f x)} \sqrt{\sin (e+f x)}}{f \sqrt{b \tan (e+f x)}}+\frac{b \sqrt{d \sec (e+f x)} \sqrt{b \tan (e+f x)}}{f}\\ \end{align*}

Mathematica [C]  time = 0.73783, size = 105, normalized size = 1.19 \[ \frac{b \sqrt{b \tan (e+f x)} \sqrt{d \sec (e+f x)} \left (\sec ^{\frac{3}{2}}(e+f x)-\frac{\sec ^2\left (\frac{1}{2} (e+f x)\right ) \sqrt{\sec (e+f x)+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )}{\sqrt{2}}\right )}{f \sec ^{\frac{3}{2}}(e+f x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sqrt[d*Sec[e + f*x]]*(Sec[e + f*x]^(3/2) - (Hypergeometric2F1[1/4, 1/2, 5/4, -Tan[(e + f*x)/2]^2]*Sec[(e +
f*x)/2]^2*Sqrt[1 + Sec[e + f*x]])/Sqrt[2])*Sqrt[b*Tan[e + f*x]])/(f*Sec[e + f*x]^(3/2))

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Maple [C]  time = 0.226, size = 211, normalized size = 2.4 \begin{align*}{\frac{\cos \left ( fx+e \right ) \sqrt{2}}{2\,f \left ( \cos \left ( fx+e \right ) -1 \right ) \sin \left ( fx+e \right ) } \left ({\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{{\frac{3}{2}}}\sqrt{{\frac{d}{\cos \left ( fx+e \right ) }}} \left ( i\cos \left ( fx+e \right ) \sin \left ( fx+e \right ) \sqrt{{\frac{i\cos \left ( fx+e \right ) -i+\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}\sqrt{-{\frac{i\cos \left ( fx+e \right ) -i-\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}{\it EllipticF} \left ( \sqrt{{\frac{i\cos \left ( fx+e \right ) -i+\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}},{\frac{\sqrt{2}}{2}} \right ) \sqrt{{\frac{-i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }}}+\cos \left ( fx+e \right ) \sqrt{2}-\sqrt{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^(1/2)*(b*tan(f*x+e))^(3/2),x)

[Out]

1/2/f*2^(1/2)*(b*sin(f*x+e)/cos(f*x+e))^(3/2)*(d/cos(f*x+e))^(1/2)*cos(f*x+e)*(I*cos(f*x+e)*sin(f*x+e)*((I*cos
(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(I*cos(f*x+e)-I-sin(f*x+e))/sin(f*x+e))^(1/2)*EllipticF(((I*cos(f*x+
e)-I+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*(-I*(cos(f*x+e)-1)/sin(f*x+e))^(1/2)+cos(f*x+e)*2^(1/2)-2^(1/2
))/(cos(f*x+e)-1)/sin(f*x+e)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(1/2)*(b*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate(sqrt(d*sec(f*x + e))*(b*tan(f*x + e))^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(1/2)*(b*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*sec(f*x + e))*sqrt(b*tan(f*x + e))*b*tan(f*x + e), x)

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Sympy [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((d*sec(f*x+e))**(1/2)*(b*tan(f*x+e))**(3/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(1/2)*(b*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate(sqrt(d*sec(f*x + e))*(b*tan(f*x + e))^(3/2), x)