3.4.16 \(\int \frac {(d \sec (e+f x))^{5/2}}{\sqrt {b \tan (e+f x)}} \, dx\) [316]

Optimal. Leaf size=92 \[ \frac {d^2 F\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right ) \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}{f \sqrt {b \tan (e+f x)}}+\frac {d^2 \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}}{b f} \]

[Out]

-d^2*(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)/f/(b*tan(f*x+e))^(1/2)+d^2*(d*sec(f*x+e))^(1/2)*(b*tan(f*x+e))^(1/2)/b
/f

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Rubi [A]
time = 0.08, antiderivative size = 92, 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 = {2693, 2696, 2721, 2720} \begin {gather*} \frac {d^2 \sqrt {b \tan (e+f x)} \sqrt {d \sec (e+f x)}}{b f}+\frac {d^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)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2693

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 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 {(d \sec (e+f x))^{5/2}}{\sqrt {b \tan (e+f x)}} \, dx &=\frac {d^2 \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}}{b f}+\frac {1}{2} d^2 \int \frac {\sqrt {d \sec (e+f x)}}{\sqrt {b \tan (e+f x)}} \, dx\\ &=\frac {d^2 \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}}{b f}+\frac {\left (d^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 {d^2 \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}}{b f}+\frac {\left (d^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 {d^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 {d^2 \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}}{b f}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 2.43, size = 83, normalized size = 0.90 \begin {gather*} \frac {d^2 \sqrt {d \sec (e+f x)} \left (\cos (e+f x) \, _2F_1\left (\frac {1}{4},\frac {3}{4};\frac {5}{4};-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{3/4} \sin (e+f x)+\tan (e+f x)\right )}{f \sqrt {b \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains complex when optimal does not.
time = 0.37, size = 208, normalized size = 2.26

method	result	size
default	\(-\frac {\left (i \sqrt {-\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticF \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )-\cos \left (f x +e \right ) \sqrt {2}+\sqrt {2}\right ) \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \sin \left (f x +e \right ) \cos \left (f x +e \right ) \sqrt {2}}{2 f \left (\cos \left (f x +e \right )-1\right ) \sqrt {\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}}}\)	\(208\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^(5/2)/(b*tan(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/2/f*(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)-cos(f*x+e)*2^(1/2)+2^(1/2))*(d/cos(f*x+e))^(5/2)*sin(f*x+e)*cos(f*x+e)/(cos(f*x+e)-1)/(b*sin(f*x
+e)/cos(f*x+e))^(1/2)*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((d*sec(f*x+e))^(5/2)/(b*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((d*sec(f*x + e))^(5/2)/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 = 0.10, size = 107, normalized size = 1.16 \begin {gather*} \frac {\sqrt {-2 i \, b d} d^{2} {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + \sqrt {2 i \, b d} d^{2} {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) + 2 \, d^{2} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{2 \, b f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(5/2)/(b*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

1/2*(sqrt(-2*I*b*d)*d^2*weierstrassPInverse(4, 0, cos(f*x + e) + I*sin(f*x + e)) + sqrt(2*I*b*d)*d^2*weierstra
ssPInverse(4, 0, cos(f*x + e) - I*sin(f*x + e)) + 2*d^2*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))
)/(b*f)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))**(5/2)/(b*tan(f*x+e))**(1/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4371 deep

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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((d*sec(f*x+e))^(5/2)/(b*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate((d*sec(f*x + e))^(5/2)/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 {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{\sqrt {b\,\mathrm {tan}\left (e+f\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d/cos(e + f*x))^(5/2)/(b*tan(e + f*x))^(1/2),x)

[Out]

int((d/cos(e + f*x))^(5/2)/(b*tan(e + f*x))^(1/2), x)

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