3.4.29 \(\int \frac {(d \sec (e+f x))^{5/2}}{(b \tan (e+f x))^{5/2}} \, dx\) [329]

Optimal. Leaf size=101 \[ -\frac {2 d^2 \sqrt {d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}}+\frac {2 d^2 F\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right ) \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}{3 b^2 f \sqrt {b \tan (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 101, 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 = {2688, 2696, 2721, 2720} \begin {gather*} \frac {2 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)}}{3 b^2 f \sqrt {b \tan (e+f x)}}-\frac {2 d^2 \sqrt {d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2688

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 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}}{(b \tan (e+f x))^{5/2}} \, dx &=-\frac {2 d^2 \sqrt {d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}}+\frac {d^2 \int \frac {\sqrt {d \sec (e+f x)}}{\sqrt {b \tan (e+f x)}} \, dx}{3 b^2}\\ &=-\frac {2 d^2 \sqrt {d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}}+\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}{3 b^2 \sqrt {b \tan (e+f x)}}\\ &=-\frac {2 d^2 \sqrt {d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}}+\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}{3 b^2 \sqrt {b \tan (e+f x)}}\\ &=-\frac {2 d^2 \sqrt {d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}}+\frac {2 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)}}{3 b^2 f \sqrt {b \tan (e+f x)}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.49, size = 116, normalized size = 1.15 \begin {gather*} \frac {2 d^3 \left (\sqrt {2} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {\sec (e+f x)}-\cot (e+f x) \csc (e+f x) \sqrt {1+\sec (e+f x)}\right ) \sqrt {b \tan (e+f x)}}{3 b^3 f \sqrt {d \sec (e+f x)} \sqrt {1+\sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains complex when optimal does not.
time = 0.35, size = 314, normalized size = 3.11

method	result	size
default	\(-\frac {\left (-i \sqrt {-\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticF \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )-i \sqrt {-\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticF \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) \sin \left (f x +e \right )+\cos \left (f x +e \right ) \sqrt {2}\right ) \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \sin \left (f x +e \right ) \sqrt {2}}{3 f \left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}}}\)	\(314\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/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)-I*(-I*(cos(f*x+e)-1)/sin(f*x+e))^(1/2)*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))+cos(f*x+e)*2^(1/2))*(d/cos(f*x+e))^(5/2)*sin(f*x+e)/(b*sin(f*x+e)/cos(f*x+e))^(5/2)*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))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 165, normalized size = 1.63 \begin {gather*} \frac {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 )}} \cos \left (f x + e\right )^{2} + {\left (d^{2} \cos \left (f x + e\right )^{2} - d^{2}\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 ) + {\left (d^{2} \cos \left (f x + e\right )^{2} - d^{2}\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 )}{3 \, {\left (b^{3} f \cos \left (f x + e\right )^{2} - b^{3} f\right )}} \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))^(5/2),x, algorithm="fricas")

[Out]

1/3*(2*d^2*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*cos(f*x + e)^2 + (d^2*cos(f*x + e)^2 - d^2)*
sqrt(-2*I*b*d)*weierstrassPInverse(4, 0, cos(f*x + e) + I*sin(f*x + e)) + (d^2*cos(f*x + e)^2 - d^2)*sqrt(2*I*
b*d)*weierstrassPInverse(4, 0, cos(f*x + e) - I*sin(f*x + e)))/(b^3*f*cos(f*x + e)^2 - b^3*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))**(5/2),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))^(5/2),x, algorithm="giac")

[Out]

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

[Out]

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

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