3.4.5 \(\int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{7/2}} \, dx\) [305]

Optimal. Leaf size=131 \[ \frac {4 b^2 F\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right ) \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}{21 d^4 f \sqrt {b \tan (e+f x)}}-\frac {2 b \sqrt {b \tan (e+f x)}}{7 f (d \sec (e+f x))^{7/2}}+\frac {2 b \sqrt {b \tan (e+f x)}}{21 d^2 f (d \sec (e+f x))^{3/2}} \]

[Out]

-4/21*b^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)/d^4/f/(b*tan(f*x+e))^(1/2)-2/7*b*(b*tan(f*x+e))^(1/2)/f/(d*sec(f*
x+e))^(7/2)+2/21*b*(b*tan(f*x+e))^(1/2)/d^2/f/(d*sec(f*x+e))^(3/2)

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Rubi [A]
time = 0.12, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2690, 2692, 2696, 2721, 2720} \begin {gather*} \frac {4 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)}}{21 d^4 f \sqrt {b \tan (e+f x)}}+\frac {2 b \sqrt {b \tan (e+f x)}}{21 d^2 f (d \sec (e+f x))^{3/2}}-\frac {2 b \sqrt {b \tan (e+f x)}}{7 f (d \sec (e+f x))^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Tan[e + f*x])^(3/2)/(d*Sec[e + f*x])^(7/2),x]

[Out]

(4*b^2*EllipticF[(e - Pi/2 + f*x)/2, 2]*Sqrt[d*Sec[e + f*x]]*Sqrt[Sin[e + f*x]])/(21*d^4*f*Sqrt[b*Tan[e + f*x]
]) - (2*b*Sqrt[b*Tan[e + f*x]])/(7*f*(d*Sec[e + f*x])^(7/2)) + (2*b*Sqrt[b*Tan[e + f*x]])/(21*d^2*f*(d*Sec[e +
f*x])^(3/2))

Rule 2690

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)), x] - Dist[b^2*((n - 1)/(a^2*m)), Int[(a*Sec[e + f*x])^(m + 2)*(b*Tan
[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[n, 1] && (LtQ[m, -1] || (EqQ[m, -1] && EqQ[n, 3/2]
)) && IntegersQ[2*m, 2*n]

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 {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{7/2}} \, dx &=-\frac {2 b \sqrt {b \tan (e+f x)}}{7 f (d \sec (e+f x))^{7/2}}+\frac {b^2 \int \frac {1}{(d \sec (e+f x))^{3/2} \sqrt {b \tan (e+f x)}} \, dx}{7 d^2}\\ &=-\frac {2 b \sqrt {b \tan (e+f x)}}{7 f (d \sec (e+f x))^{7/2}}+\frac {2 b \sqrt {b \tan (e+f x)}}{21 d^2 f (d \sec (e+f x))^{3/2}}+\frac {\left (2 b^2\right ) \int \frac {\sqrt {d \sec (e+f x)}}{\sqrt {b \tan (e+f x)}} \, dx}{21 d^4}\\ &=-\frac {2 b \sqrt {b \tan (e+f x)}}{7 f (d \sec (e+f x))^{7/2}}+\frac {2 b \sqrt {b \tan (e+f x)}}{21 d^2 f (d \sec (e+f x))^{3/2}}+\frac {\left (2 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}{21 d^4 \sqrt {b \tan (e+f x)}}\\ &=-\frac {2 b \sqrt {b \tan (e+f x)}}{7 f (d \sec (e+f x))^{7/2}}+\frac {2 b \sqrt {b \tan (e+f x)}}{21 d^2 f (d \sec (e+f x))^{3/2}}+\frac {\left (2 b^2 \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}\right ) \int \frac {1}{\sqrt {\sin (e+f x)}} \, dx}{21 d^4 \sqrt {b \tan (e+f x)}}\\ &=\frac {4 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)}}{21 d^4 f \sqrt {b \tan (e+f x)}}-\frac {2 b \sqrt {b \tan (e+f x)}}{7 f (d \sec (e+f x))^{7/2}}+\frac {2 b \sqrt {b \tan (e+f x)}}{21 d^2 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 = 1.42, size = 105, normalized size = 0.80 \begin {gather*} -\frac {b \sqrt {b \tan (e+f x)} \left (4 \, _2F_1\left (\frac {1}{4},\frac {3}{4};\frac {5}{4};\sec ^2(e+f x)\right ) \sec ^2(e+f x)+(1+3 \cos (2 (e+f x))) \sqrt [4]{-\tan ^2(e+f x)}\right )}{21 d^2 f (d \sec (e+f x))^{3/2} \sqrt [4]{-\tan ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[e + f*x])^(3/2)/(d*Sec[e + f*x])^(7/2),x]

[Out]

-1/21*(b*Sqrt[b*Tan[e + f*x]]*(4*Hypergeometric2F1[1/4, 3/4, 5/4, Sec[e + f*x]^2]*Sec[e + f*x]^2 + (1 + 3*Cos[
2*(e + f*x)])*(-Tan[e + f*x]^2)^(1/4)))/(d^2*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 = 0.44, size = 241, normalized size = 1.84

method	result	size
default	\(-\frac {\left (2 i \sqrt {-\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticF \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) \sin \left (f x +e \right )+3 \sqrt {2}\, \left (\cos ^{4}\left (f x +e \right )\right )-3 \left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {2}-\left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {2}+\cos \left (f x +e \right ) \sqrt {2}\right ) \left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \sqrt {2}}{21 f \left (\cos \left (f x +e \right )-1\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )^{2} \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {7}{2}}}\)	\(241\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(f*x+e))^(3/2)/(d*sec(f*x+e))^(7/2),x,method=_RETURNVERBOSE)

[Out]

-1/21/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)+3*2^(1/2)*cos(f*x+e)^4-3*2^(1/2)*cos(f*x+e)^3-cos(f*x+e)^2*2^(1/2)+cos(f*x+e)*2^(1/2))*(b*sin(f*x+e)/cos
(f*x+e))^(3/2)/(cos(f*x+e)-1)/sin(f*x+e)/cos(f*x+e)^2/(d/cos(f*x+e))^(7/2)*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((b*tan(f*x+e))^(3/2)/(d*sec(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e))^(3/2)/(d*sec(f*x + e))^(7/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.12, size = 125, normalized size = 0.95 \begin {gather*} \frac {2 \, {\left (\sqrt {-2 i \, b d} b {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + \sqrt {2 i \, b d} b {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) - {\left (3 \, b \cos \left (f x + e\right )^{4} - b \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}\right )}}{21 \, d^{4} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(3/2)/(d*sec(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

2/21*(sqrt(-2*I*b*d)*b*weierstrassPInverse(4, 0, cos(f*x + e) + I*sin(f*x + e)) + sqrt(2*I*b*d)*b*weierstrassP
Inverse(4, 0, cos(f*x + e) - I*sin(f*x + e)) - (3*b*cos(f*x + e)^4 - b*cos(f*x + e)^2)*sqrt(b*sin(f*x + e)/cos
(f*x + e))*sqrt(d/cos(f*x + e)))/(d^4*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((b*tan(f*x+e))**(3/2)/(d*sec(f*x+e))**(7/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 6191 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((b*tan(f*x+e))^(3/2)/(d*sec(f*x+e))^(7/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e))^(3/2)/(d*sec(f*x + e))^(7/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(e + f*x))^(3/2)/(d/cos(e + f*x))^(7/2),x)

[Out]

int((b*tan(e + f*x))^(3/2)/(d/cos(e + f*x))^(7/2), x)

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