3.4.20 \(\int \frac {1}{(d \sec (e+f x))^{3/2} \sqrt {b \tan (e+f x)}} \, dx\) [320]

Optimal. Leaf size=95 \[ \frac {4 F\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right ) \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}{3 d^2 f \sqrt {b \tan (e+f x)}}+\frac {2 \sqrt {b \tan (e+f x)}}{3 b f (d \sec (e+f x))^{3/2}} \]

[Out]

-4/3*(sin(1/2*e+1/4*Pi+1/2*f*x)^2)^(1/2)/sin(1/2*e+1/4*Pi+1/2*f*x)*EllipticF(cos(1/2*e+1/4*Pi+1/2*f*x),2^(1/2)
)*(d*sec(f*x+e))^(1/2)*sin(f*x+e)^(1/2)/d^2/f/(b*tan(f*x+e))^(1/2)+2/3*(b*tan(f*x+e))^(1/2)/b/f/(d*sec(f*x+e))
^(3/2)

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Rubi [A]
time = 0.09, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2692, 2696, 2721, 2720} \begin {gather*} \frac {4 \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)}}{3 d^2 f \sqrt {b \tan (e+f x)}}+\frac {2 \sqrt {b \tan (e+f x)}}{3 b f (d \sec (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2692

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*m)), x] + Dist[(m + n + 1)/(a^2*m), Int[(a*Sec[e + f*x])^(m + 2)*(b*Ta
n[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (LtQ[m, -1] || (EqQ[m, -1] && EqQ[n, -2^(-1)])) && Integ
ersQ[2*m, 2*n]

Rule 2696

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 2720

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

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rubi steps

\begin {align*} \int \frac {1}{(d \sec (e+f x))^{3/2} \sqrt {b \tan (e+f x)}} \, dx &=\frac {2 \sqrt {b \tan (e+f x)}}{3 b f (d \sec (e+f x))^{3/2}}+\frac {2 \int \frac {\sqrt {d \sec (e+f x)}}{\sqrt {b \tan (e+f x)}} \, dx}{3 d^2}\\ &=\frac {2 \sqrt {b \tan (e+f x)}}{3 b f (d \sec (e+f x))^{3/2}}+\frac {\left (2 \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}\right ) \int \frac {1}{\sqrt {b \sin (e+f x)}} \, dx}{3 d^2 \sqrt {b \tan (e+f x)}}\\ &=\frac {2 \sqrt {b \tan (e+f x)}}{3 b f (d \sec (e+f x))^{3/2}}+\frac {\left (2 \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}\right ) \int \frac {1}{\sqrt {\sin (e+f x)}} \, dx}{3 d^2 \sqrt {b \tan (e+f x)}}\\ &=\frac {4 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)}}{3 d^2 f \sqrt {b \tan (e+f x)}}+\frac {2 \sqrt {b \tan (e+f x)}}{3 b f (d \sec (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 1.12, size = 91, normalized size = 0.96 \begin {gather*} \frac {2 \sqrt {b \tan (e+f x)} \left (-2 \, _2F_1\left (\frac {1}{4},\frac {3}{4};\frac {5}{4};\sec ^2(e+f x)\right ) \sec ^2(e+f x)+\sqrt [4]{-\tan ^2(e+f x)}\right )}{3 b f (d \sec (e+f x))^{3/2} \sqrt [4]{-\tan ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains complex when optimal does not.
time = 0.37, size = 213, normalized size = 2.24

method	result	size
default	\(-\frac {\left (2 i \sqrt {-\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticF \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) \sin \left (f x +e \right )-\left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {2}+\cos \left (f x +e \right ) \sqrt {2}\right ) \sin \left (f x +e \right ) \sqrt {2}}{3 f \left (\cos \left (f x +e \right )-1\right ) \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \sqrt {\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}}\, \cos \left (f x +e \right )^{2}}\)	\(213\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 109, normalized size = 1.15 \begin {gather*} \frac {2 \, {\left (\sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} + \sqrt {-2 i \, b d} {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + \sqrt {2 i \, b d} {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right )}}{3 \, b d^{2} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*cos(f*x + e)^2 + sqrt(-2*I*b*d)*weierstrassPInvers
e(4, 0, cos(f*x + e) + I*sin(f*x + e)) + sqrt(2*I*b*d)*weierstrassPInverse(4, 0, cos(f*x + e) - I*sin(f*x + e)
))/(b*d^2*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {b \tan {\left (e + f x \right )}} \left (d \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\sqrt {b\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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