3.3.18 \(\int \frac {\sin ^4(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx\) [218]

Optimal. Leaf size=102 \[ -\frac {8 \sqrt {\cos (a+b x)} F\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{3 b d^2 \sqrt {d \cos (a+b x)}}+\frac {4 \sqrt {d \cos (a+b x)} \sin (a+b x)}{3 b d^3}+\frac {2 \sin ^3(a+b x)}{3 b d (d \cos (a+b x))^{3/2}} \]

[Out]

2/3*sin(b*x+a)^3/b/d/(d*cos(b*x+a))^(3/2)-8/3*(cos(1/2*a+1/2*b*x)^2)^(1/2)/cos(1/2*a+1/2*b*x)*EllipticF(sin(1/
2*a+1/2*b*x),2^(1/2))*cos(b*x+a)^(1/2)/b/d^2/(d*cos(b*x+a))^(1/2)+4/3*sin(b*x+a)*(d*cos(b*x+a))^(1/2)/b/d^3

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Rubi [A]
time = 0.08, antiderivative size = 102, 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 = {2646, 2648, 2721, 2720} \begin {gather*} \frac {4 \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b d^3}-\frac {8 \sqrt {\cos (a+b x)} F\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{3 b d^2 \sqrt {d \cos (a+b x)}}+\frac {2 \sin ^3(a+b x)}{3 b d (d \cos (a+b x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^4/(d*Cos[a + b*x])^(5/2),x]

[Out]

(-8*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2])/(3*b*d^2*Sqrt[d*Cos[a + b*x]]) + (4*Sqrt[d*Cos[a + b*x]]*Sin
[a + b*x])/(3*b*d^3) + (2*Sin[a + b*x]^3)/(3*b*d*(d*Cos[a + b*x])^(3/2))

Rule 2646

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-a)*(a*Sin[e
+ f*x])^(m - 1)*((b*Cos[e + f*x])^(n + 1)/(b*f*(n + 1))), x] + Dist[a^2*((m - 1)/(b^2*(n + 1))), Int[(a*Sin[e
+ f*x])^(m - 2)*(b*Cos[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[m, 1] && LtQ[n, -1] && (Int
egersQ[2*m, 2*n] || EqQ[m + n, 0])

Rule 2648

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

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 {\sin ^4(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx &=\frac {2 \sin ^3(a+b x)}{3 b d (d \cos (a+b x))^{3/2}}-\frac {2 \int \frac {\sin ^2(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx}{d^2}\\ &=\frac {4 \sqrt {d \cos (a+b x)} \sin (a+b x)}{3 b d^3}+\frac {2 \sin ^3(a+b x)}{3 b d (d \cos (a+b x))^{3/2}}-\frac {4 \int \frac {1}{\sqrt {d \cos (a+b x)}} \, dx}{3 d^2}\\ &=\frac {4 \sqrt {d \cos (a+b x)} \sin (a+b x)}{3 b d^3}+\frac {2 \sin ^3(a+b x)}{3 b d (d \cos (a+b x))^{3/2}}-\frac {\left (4 \sqrt {\cos (a+b x)}\right ) \int \frac {1}{\sqrt {\cos (a+b x)}} \, dx}{3 d^2 \sqrt {d \cos (a+b x)}}\\ &=-\frac {8 \sqrt {\cos (a+b x)} F\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{3 b d^2 \sqrt {d \cos (a+b x)}}+\frac {4 \sqrt {d \cos (a+b x)} \sin (a+b x)}{3 b d^3}+\frac {2 \sin ^3(a+b x)}{3 b d (d \cos (a+b 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 = 0.05, size = 60, normalized size = 0.59 \begin {gather*} \frac {\cos ^2(a+b x)^{3/4} \, _2F_1\left (\frac {7}{4},\frac {5}{2};\frac {7}{2};\sin ^2(a+b x)\right ) \sin ^5(a+b x)}{5 b d (d \cos (a+b x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^4/(d*Cos[a + b*x])^(5/2),x]

[Out]

((Cos[a + b*x]^2)^(3/4)*Hypergeometric2F1[7/4, 5/2, 7/2, Sin[a + b*x]^2]*Sin[a + b*x]^5)/(5*b*d*(d*Cos[a + b*x
])^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(285\) vs. \(2(114)=228\).
time = 0.21, size = 286, normalized size = 2.80

method	result	size
default	\(-\frac {8 \left (-2 \left (\sin ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+2 \sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right ) \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-\sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )-\left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \cos \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \sqrt {d \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}}{3 d^{2} \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right ) \sqrt {-d \left (2 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-\left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )\right )}\, \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {d \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right )}\, b}\)	\(286\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)^4/(d*cos(b*x+a))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-8/3*(-2*sin(1/2*b*x+1/2*a)^6*cos(1/2*b*x+1/2*a)+2*(sin(1/2*b*x+1/2*a)^2)^(1/2)*(2*sin(1/2*b*x+1/2*a)^2-1)^(1/
2)*EllipticF(cos(1/2*b*x+1/2*a),2^(1/2))*sin(1/2*b*x+1/2*a)^2+2*cos(1/2*b*x+1/2*a)*sin(1/2*b*x+1/2*a)^4-(sin(1
/2*b*x+1/2*a)^2)^(1/2)*(2*sin(1/2*b*x+1/2*a)^2-1)^(1/2)*EllipticF(cos(1/2*b*x+1/2*a),2^(1/2))-sin(1/2*b*x+1/2*
a)^2*cos(1/2*b*x+1/2*a))/d^2*(d*(2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)/(2*cos(1/2*b*x+1/2*a)^2
-1)/(-d*(2*sin(1/2*b*x+1/2*a)^4-sin(1/2*b*x+1/2*a)^2))^(1/2)/sin(1/2*b*x+1/2*a)/(d*(2*cos(1/2*b*x+1/2*a)^2-1))
^(1/2)/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(b*x+a)^4/(d*cos(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

integrate(sin(b*x + a)^4/(d*cos(b*x + a))^(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 = 113, normalized size = 1.11 \begin {gather*} -\frac {2 \, {\left (-2 i \, \sqrt {2} \sqrt {d} \cos \left (b x + a\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + 2 i \, \sqrt {2} \sqrt {d} \cos \left (b x + a\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) - \sqrt {d \cos \left (b x + a\right )} {\left (\cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right )\right )}}{3 \, b d^{3} \cos \left (b x + a\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^4/(d*cos(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

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

[Out]

integrate(sin(b*x + a)^4/(d*cos(b*x + a))^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\sin \left (a+b\,x\right )}^4}{{\left (d\,\cos \left (a+b\,x\right )\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^4/(d*cos(a + b*x))^(5/2),x)

[Out]

int(sin(a + b*x)^4/(d*cos(a + b*x))^(5/2), x)

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