3.3.99 \(\int (d \sec (e+f x))^{5/2} (b \tan (e+f x))^{3/2} \, dx\) [299]

Optimal. Leaf size=131 \[ -\frac {b^2 d^2 F\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right ) \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}{6 f \sqrt {b \tan (e+f x)}}-\frac {b d^2 \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}}{6 f}+\frac {b (d \sec (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}{3 f} \]

[Out]

1/6*b^2*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)+1/3*b*(d*sec(f*x+e))^(5/2)*(b*tan(f*x+e)
)^(1/2)/f-1/6*b*d^2*(d*sec(f*x+e))^(1/2)*(b*tan(f*x+e))^(1/2)/f

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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 = {2691, 2693, 2696, 2721, 2720} \begin {gather*} -\frac {b^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)}}{6 f \sqrt {b \tan (e+f x)}}-\frac {b d^2 \sqrt {b \tan (e+f x)} \sqrt {d \sec (e+f x)}}{6 f}+\frac {b \sqrt {b \tan (e+f x)} (d \sec (e+f x))^{5/2}}{3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*Sec[e + f*x])^(5/2)*(b*Tan[e + f*x])^(3/2),x]

[Out]

-1/6*(b^2*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]
]) - (b*d^2*Sqrt[d*Sec[e + f*x]]*Sqrt[b*Tan[e + f*x]])/(6*f) + (b*(d*Sec[e + f*x])^(5/2)*Sqrt[b*Tan[e + f*x]])
/(3*f)

Rule 2691

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 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 (d \sec (e+f x))^{5/2} (b \tan (e+f x))^{3/2} \, dx &=\frac {b (d \sec (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}{3 f}-\frac {1}{6} b^2 \int \frac {(d \sec (e+f x))^{5/2}}{\sqrt {b \tan (e+f x)}} \, dx\\ &=-\frac {b d^2 \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}}{6 f}+\frac {b (d \sec (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}{3 f}-\frac {1}{12} \left (b^2 d^2\right ) \int \frac {\sqrt {d \sec (e+f x)}}{\sqrt {b \tan (e+f x)}} \, dx\\ &=-\frac {b d^2 \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}}{6 f}+\frac {b (d \sec (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}{3 f}-\frac {\left (b^2 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}{12 \sqrt {b \tan (e+f x)}}\\ &=-\frac {b d^2 \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}}{6 f}+\frac {b (d \sec (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}{3 f}-\frac {\left (b^2 d^2 \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}\right ) \int \frac {1}{\sqrt {\sin (e+f x)}} \, dx}{12 \sqrt {b \tan (e+f x)}}\\ &=-\frac {b^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)}}{6 f \sqrt {b \tan (e+f x)}}-\frac {b d^2 \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}}{6 f}+\frac {b (d \sec (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}{3 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 = 7.14, size = 95, normalized size = 0.73 \begin {gather*} \frac {b d^2 \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)} \left (\, _2F_1\left (\frac {1}{4},\frac {3}{4};\frac {5}{4};\sec ^2(e+f x)\right )+\left (-1+2 \sec ^2(e+f x)\right ) \sqrt [4]{-\tan ^2(e+f x)}\right )}{6 f \sqrt [4]{-\tan ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d*Sec[e + f*x])^(5/2)*(b*Tan[e + f*x])^(3/2),x]

[Out]

(b*d^2*Sqrt[d*Sec[e + f*x]]*Sqrt[b*Tan[e + f*x]]*(Hypergeometric2F1[1/4, 3/4, 5/4, Sec[e + f*x]^2] + (-1 + 2*S
ec[e + f*x]^2)*(-Tan[e + f*x]^2)^(1/4)))/(6*f*(-Tan[e + f*x]^2)^(1/4))

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Maple [C] Result contains complex when optimal does not.
time = 0.40, size = 239, normalized size = 1.82

method	result	size
default	\(\frac {\left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \cos \left (f x +e \right ) \left (i \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticF \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}}\, \sin \left (f x +e \right ) \left (\cos ^{3}\left (f x +e \right )\right )-\left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {2}+\left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {2}+2 \cos \left (f x +e \right ) \sqrt {2}-2 \sqrt {2}\right ) \sqrt {2}}{12 f \left (\cos \left (f x +e \right )-1\right ) \sin \left (f x +e \right )}\)	\(239\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^(5/2)*(b*tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/12/f*(b*sin(f*x+e)/cos(f*x+e))^(3/2)*(d/cos(f*x+e))^(5/2)*cos(f*x+e)*(I*((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)*sin(f*x+e)*cos(f*x+e)^3-2^(1/2)*cos(f*x+e)^3+cos(f*x+e
)^2*2^(1/2)+2*cos(f*x+e)*2^(1/2)-2*2^(1/2))/(cos(f*x+e)-1)/sin(f*x+e)*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))^(3/2),x, algorithm="maxima")

[Out]

integrate((d*sec(f*x + e))^(5/2)*(b*tan(f*x + e))^(3/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.17, size = 151, normalized size = 1.15 \begin {gather*} -\frac {\sqrt {-2 i \, b d} b d^{2} \cos \left (f x + e\right )^{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} b d^{2} \cos \left (f x + e\right )^{2} {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) + 2 \, {\left (b d^{2} \cos \left (f x + e\right )^{2} - 2 \, b d^{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 )}}}{12 \, f \cos \left (f x + e\right )^{2}} \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))^(3/2),x, algorithm="fricas")

[Out]

-1/12*(sqrt(-2*I*b*d)*b*d^2*cos(f*x + e)^2*weierstrassPInverse(4, 0, cos(f*x + e) + I*sin(f*x + e)) + sqrt(2*I
*b*d)*b*d^2*cos(f*x + e)^2*weierstrassPInverse(4, 0, cos(f*x + e) - I*sin(f*x + e)) + 2*(b*d^2*cos(f*x + e)^2
- 2*b*d^2)*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e)))/(f*cos(f*x + e)^2)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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))**(3/2),x)

[Out]

Timed out

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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))^(3/2),x, algorithm="giac")

[Out]

integrate((d*sec(f*x + e))^(5/2)*(b*tan(f*x + e))^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/2} \,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))^(5/2),x)

[Out]

int((b*tan(e + f*x))^(3/2)*(d/cos(e + f*x))^(5/2), x)

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