3.4.92 \(\int (b \sec (e+f x))^{3/2} \sin ^4(e+f x) \, dx\) [392]

Optimal. Leaf size=98 \[ -\frac {24 b^2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 f \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}+\frac {12 b^3 \sin (e+f x)}{5 f (b \sec (e+f x))^{3/2}}+\frac {2 b \sqrt {b \sec (e+f x)} \sin ^3(e+f x)}{f} \]

[Out]

12/5*b^3*sin(f*x+e)/f/(b*sec(f*x+e))^(3/2)-24/5*b^2*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticE(
sin(1/2*f*x+1/2*e),2^(1/2))/f/cos(f*x+e)^(1/2)/(b*sec(f*x+e))^(1/2)+2*b*sin(f*x+e)^3*(b*sec(f*x+e))^(1/2)/f

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Rubi [A]
time = 0.08, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2704, 2707, 3856, 2719} \begin {gather*} \frac {12 b^3 \sin (e+f x)}{5 f (b \sec (e+f x))^{3/2}}-\frac {24 b^2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 f \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}+\frac {2 b \sin ^3(e+f x) \sqrt {b \sec (e+f x)}}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^4,x]

[Out]

(-24*b^2*EllipticE[(e + f*x)/2, 2])/(5*f*Sqrt[Cos[e + f*x]]*Sqrt[b*Sec[e + f*x]]) + (12*b^3*Sin[e + f*x])/(5*f
*(b*Sec[e + f*x])^(3/2)) + (2*b*Sqrt[b*Sec[e + f*x]]*Sin[e + f*x]^3)/f

Rule 2704

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a*Csc[e +
f*x])^(m + 1)*((b*Sec[e + f*x])^(n - 1)/(f*a*(n - 1))), x] + Dist[b^2*((m + 1)/(a^2*(n - 1))), Int[(a*Csc[e +
f*x])^(m + 2)*(b*Sec[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[n, 1] && LtQ[m, -1] && Integer
sQ[2*m, 2*n]

Rule 2707

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[b*(a*Csc[e +
f*x])^(m + 1)*((b*Sec[e + f*x])^(n - 1)/(a*f*(m + n))), x] + Dist[(m + 1)/(a^2*(m + n)), Int[(a*Csc[e + f*x])
^(m + 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, -1] && NeQ[m + n, 0] && IntegersQ[2
*m, 2*n]

Rule 2719

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

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rubi steps

\begin {align*} \int (b \sec (e+f x))^{3/2} \sin ^4(e+f x) \, dx &=\frac {2 b \sqrt {b \sec (e+f x)} \sin ^3(e+f x)}{f}-\left (6 b^2\right ) \int \frac {\sin ^2(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx\\ &=\frac {12 b^3 \sin (e+f x)}{5 f (b \sec (e+f x))^{3/2}}+\frac {2 b \sqrt {b \sec (e+f x)} \sin ^3(e+f x)}{f}-\frac {1}{5} \left (12 b^2\right ) \int \frac {1}{\sqrt {b \sec (e+f x)}} \, dx\\ &=\frac {12 b^3 \sin (e+f x)}{5 f (b \sec (e+f x))^{3/2}}+\frac {2 b \sqrt {b \sec (e+f x)} \sin ^3(e+f x)}{f}-\frac {\left (12 b^2\right ) \int \sqrt {\cos (e+f x)} \, dx}{5 \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}\\ &=-\frac {24 b^2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 f \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}+\frac {12 b^3 \sin (e+f x)}{5 f (b \sec (e+f x))^{3/2}}+\frac {2 b \sqrt {b \sec (e+f x)} \sin ^3(e+f x)}{f}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 60, normalized size = 0.61 \begin {gather*} \frac {b \sqrt {b \sec (e+f x)} \left (-48 \sqrt {\cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )+21 \sin (e+f x)+\sin (3 (e+f x))\right )}{10 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^4,x]

[Out]

(b*Sqrt[b*Sec[e + f*x]]*(-48*Sqrt[Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2] + 21*Sin[e + f*x] + Sin[3*(e + f*x)]
))/(10*f)

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Maple [C] Result contains complex when optimal does not.
time = 0.25, size = 320, normalized size = 3.27

method	result	size
default	\(-\frac {2 \left (12 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )-12 i \EllipticE \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}+12 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right )-12 i \sin \left (f x +e \right ) \EllipticE \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}+\cos ^{4}\left (f x +e \right )-8 \left (\cos ^{2}\left (f x +e \right )\right )+12 \cos \left (f x +e \right )-5\right ) \cos \left (f x +e \right ) \left (\frac {b}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}}}{5 f \sin \left (f x +e \right )}\)	\(320\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sec(f*x+e))^(3/2)*sin(f*x+e)^4,x,method=_RETURNVERBOSE)

[Out]

-2/5/f*(12*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e)
,I)*sin(f*x+e)*cos(f*x+e)-12*I*cos(f*x+e)*sin(f*x+e)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2
)*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)+12*I*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(1/(cos(f*x+e)+1))^
(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)-12*I*sin(f*x+e)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f
*x+e)+1))^(1/2)*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)+cos(f*x+e)^4-8*cos(f*x+e)^2+12*cos(f*x+e)-5)*cos(f*x
+e)*(b/cos(f*x+e))^(3/2)/sin(f*x+e)

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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*sec(f*x+e))^(3/2)*sin(f*x+e)^4,x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e))^(3/2)*sin(f*x + e)^4, x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 105, normalized size = 1.07 \begin {gather*} -\frac {2 \, {\left (6 i \, \sqrt {2} b^{\frac {3}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) - 6 i \, \sqrt {2} b^{\frac {3}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) - {\left (b \cos \left (f x + e\right )^{2} + 5 \, b\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )\right )}}{5 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^(3/2)*sin(f*x+e)^4,x, algorithm="fricas")

[Out]

-2/5*(6*I*sqrt(2)*b^(3/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e))) -
6*I*sqrt(2)*b^(3/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e))) - (b*cos
(f*x + e)^2 + 5*b)*sqrt(b/cos(f*x + e))*sin(f*x + e))/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*sec(f*x+e))**(3/2)*sin(f*x+e)**4,x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 8009 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*sec(f*x+e))^(3/2)*sin(f*x+e)^4,x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e))^(3/2)*sin(f*x + e)^4, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\sin \left (e+f\,x\right )}^4\,{\left (\frac {b}{\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(sin(e + f*x)^4*(b/cos(e + f*x))^(3/2),x)

[Out]

int(sin(e + f*x)^4*(b/cos(e + f*x))^(3/2), x)

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