3.5.18 \(\int \frac {\sin ^4(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx\) [418]

Optimal. Leaf size=95 \[ \frac {8 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{15 f \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}-\frac {4 b \sin (e+f x)}{15 f (b \sec (e+f x))^{3/2}}-\frac {2 b \sin ^3(e+f x)}{9 f (b \sec (e+f x))^{3/2}} \]

[Out]

-4/15*b*sin(f*x+e)/f/(b*sec(f*x+e))^(3/2)-2/9*b*sin(f*x+e)^3/f/(b*sec(f*x+e))^(3/2)+8/15*(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)

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Rubi [A]
time = 0.07, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2707, 3856, 2719} \begin {gather*} -\frac {2 b \sin ^3(e+f x)}{9 f (b \sec (e+f x))^{3/2}}-\frac {4 b \sin (e+f x)}{15 f (b \sec (e+f x))^{3/2}}+\frac {8 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{15 f \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^4/Sqrt[b*Sec[e + f*x]],x]

[Out]

(8*EllipticE[(e + f*x)/2, 2])/(15*f*Sqrt[Cos[e + f*x]]*Sqrt[b*Sec[e + f*x]]) - (4*b*Sin[e + f*x])/(15*f*(b*Sec
[e + f*x])^(3/2)) - (2*b*Sin[e + f*x]^3)/(9*f*(b*Sec[e + f*x])^(3/2))

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 \frac {\sin ^4(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx &=-\frac {2 b \sin ^3(e+f x)}{9 f (b \sec (e+f x))^{3/2}}+\frac {2}{3} \int \frac {\sin ^2(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx\\ &=-\frac {4 b \sin (e+f x)}{15 f (b \sec (e+f x))^{3/2}}-\frac {2 b \sin ^3(e+f x)}{9 f (b \sec (e+f x))^{3/2}}+\frac {4}{15} \int \frac {1}{\sqrt {b \sec (e+f x)}} \, dx\\ &=-\frac {4 b \sin (e+f x)}{15 f (b \sec (e+f x))^{3/2}}-\frac {2 b \sin ^3(e+f x)}{9 f (b \sec (e+f x))^{3/2}}+\frac {4 \int \sqrt {\cos (e+f x)} \, dx}{15 \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}\\ &=\frac {8 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{15 f \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}-\frac {4 b \sin (e+f x)}{15 f (b \sec (e+f x))^{3/2}}-\frac {2 b \sin ^3(e+f x)}{9 f (b \sec (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.24, size = 63, normalized size = 0.66 \begin {gather*} \frac {\frac {192 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{\sqrt {\cos (e+f x)}}-68 \sin (2 (e+f x))+10 \sin (4 (e+f x))}{360 f \sqrt {b \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^4/Sqrt[b*Sec[e + f*x]],x]

[Out]

((192*EllipticE[(e + f*x)/2, 2])/Sqrt[Cos[e + f*x]] - 68*Sin[2*(e + f*x)] + 10*Sin[4*(e + f*x)])/(360*f*Sqrt[b
*Sec[e + f*x]])

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Maple [C] Result contains complex when optimal does not.
time = 0.23, size = 328, normalized size = 3.45

method	result	size
default	\(\frac {2 \left (12 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )-12 i \EllipticE \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}-5 \left (\cos ^{6}\left (f x +e \right )\right )+12 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right )-12 i \sin \left (f x +e \right ) \EllipticE \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}+16 \left (\cos ^{4}\left (f x +e \right )\right )-23 \left (\cos ^{2}\left (f x +e \right )\right )+12 \cos \left (f x +e \right )\right ) \sqrt {\frac {b}{\cos \left (f x +e \right )}}}{45 f \sin \left (f x +e \right ) b}\)	\(328\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^4/(b*sec(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/45/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)-5*cos(f*x+e)^6+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)*(co
s(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)+16*cos(f*x+e)^4-23*cos(f*x+e)^2+12*co
s(f*x+e))*(b/cos(f*x+e))^(1/2)/sin(f*x+e)/b

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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(sin(f*x+e)^4/(b*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.12, size = 115, normalized size = 1.21 \begin {gather*} \frac {2 \, {\left ({\left (5 \, \cos \left (f x + e\right )^{4} - 11 \, \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) + 6 i \, \sqrt {2} \sqrt {b} {\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} \sqrt {b} {\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 )\right )}}{45 \, b f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4/(b*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

2/45*((5*cos(f*x + e)^4 - 11*cos(f*x + e)^2)*sqrt(b/cos(f*x + e))*sin(f*x + e) + 6*I*sqrt(2)*sqrt(b)*weierstra
ssZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e))) - 6*I*sqrt(2)*sqrt(b)*weierstrassZeta
(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e))))/(b*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sin ^{4}{\left (e + f x \right )}}{\sqrt {b \sec {\left (e + f x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)**4/(b*sec(f*x+e))**(1/2),x)

[Out]

Integral(sin(e + f*x)**4/sqrt(b*sec(e + f*x)), x)

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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(sin(f*x+e)^4/(b*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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

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

[Out]

int(sin(e + f*x)^4/(b/cos(e + f*x))^(1/2), x)

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