3.2.9 \(\int \frac {\sin ^5(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx\) [109]

Optimal. Leaf size=114 \[ -\frac {\sin ^3(a+b x)}{10 b d (d \tan (a+b x))^{3/2}}+\frac {\sin ^5(a+b x)}{5 b d (d \tan (a+b x))^{3/2}}+\frac {3 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sin (a+b x)}{20 b d^2 \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}} \]

[Out]

-3/20*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticE(cos(a+1/4*Pi+b*x),2^(1/2))*sin(b*x+a)/b/d^2/sin(
2*b*x+2*a)^(1/2)/(d*tan(b*x+a))^(1/2)-1/10*sin(b*x+a)^3/b/d/(d*tan(b*x+a))^(3/2)+1/5*sin(b*x+a)^5/b/d/(d*tan(b
*x+a))^(3/2)

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Rubi [A]
time = 0.10, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2676, 2678, 2681, 2652, 2719} \begin {gather*} \frac {3 \sin (a+b x) E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{20 b d^2 \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}+\frac {\sin ^5(a+b x)}{5 b d (d \tan (a+b x))^{3/2}}-\frac {\sin ^3(a+b x)}{10 b d (d \tan (a+b x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/10*Sin[a + b*x]^3/(b*d*(d*Tan[a + b*x])^(3/2)) + Sin[a + b*x]^5/(5*b*d*(d*Tan[a + b*x])^(3/2)) + (3*Ellipti
cE[a - Pi/4 + b*x, 2]*Sin[a + b*x])/(20*b*d^2*Sqrt[Sin[2*a + 2*b*x]]*Sqrt[d*Tan[a + b*x]])

Rule 2652

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a*Sin[e +
f*x]]*(Sqrt[b*Cos[e + f*x]]/Sqrt[Sin[2*e + 2*f*x]]), Int[Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f},
x]

Rule 2676

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

Rule 2678

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[(-b)*(a*Sin
[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*m)), x] + Dist[a^2*((m + n - 1)/m), Int[(a*Sin[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])) && IntegersQ
[2*m, 2*n]

Rule 2681

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[Cos[e + f*x]
^n*((b*Tan[e + f*x])^n/(a*Sin[e + f*x])^n), Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b
, e, f, m, n}, x] &&  !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(-1)]) || IntegersQ[m - 1/2, n -
1/2])

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]

Rubi steps

\begin {align*} \int \frac {\sin ^5(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx &=\frac {\sin ^5(a+b x)}{5 b d (d \tan (a+b x))^{3/2}}+\frac {3 \int \frac {\sin ^3(a+b x)}{\sqrt {d \tan (a+b x)}} \, dx}{10 d^2}\\ &=-\frac {\sin ^3(a+b x)}{10 b d (d \tan (a+b x))^{3/2}}+\frac {\sin ^5(a+b x)}{5 b d (d \tan (a+b x))^{3/2}}+\frac {3 \int \frac {\sin (a+b x)}{\sqrt {d \tan (a+b x)}} \, dx}{20 d^2}\\ &=-\frac {\sin ^3(a+b x)}{10 b d (d \tan (a+b x))^{3/2}}+\frac {\sin ^5(a+b x)}{5 b d (d \tan (a+b x))^{3/2}}+\frac {\left (3 \sqrt {\sin (a+b x)}\right ) \int \sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)} \, dx}{20 d^2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}\\ &=-\frac {\sin ^3(a+b x)}{10 b d (d \tan (a+b x))^{3/2}}+\frac {\sin ^5(a+b x)}{5 b d (d \tan (a+b x))^{3/2}}+\frac {(3 \sin (a+b x)) \int \sqrt {\sin (2 a+2 b x)} \, dx}{20 d^2 \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}\\ &=-\frac {\sin ^3(a+b x)}{10 b d (d \tan (a+b x))^{3/2}}+\frac {\sin ^5(a+b x)}{5 b d (d \tan (a+b x))^{3/2}}+\frac {3 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sin (a+b x)}{20 b d^2 \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 1.13, size = 100, normalized size = 0.88 \begin {gather*} \frac {\sqrt {d \tan (a+b x)} \left (-\sqrt {\sec ^2(a+b x)} (\sin (3 (a+b x))+\sin (5 (a+b x)))+8 \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {7}{4};-\tan ^2(a+b x)\right ) \sec (a+b x) \tan (a+b x)\right )}{80 b d^3 \sqrt {\sec ^2(a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d*Tan[a + b*x]]*(-(Sqrt[Sec[a + b*x]^2]*(Sin[3*(a + b*x)] + Sin[5*(a + b*x)])) + 8*Hypergeometric2F1[3/4
, 3/2, 7/4, -Tan[a + b*x]^2]*Sec[a + b*x]*Tan[a + b*x]))/(80*b*d^3*Sqrt[Sec[a + b*x]^2])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(549\) vs. \(2(125)=250\).
time = 0.33, size = 550, normalized size = 4.82

method	result	size
default	\(\frac {\left (-1+\cos \left (b x +a \right )\right )^{2} \left (4 \left (\cos ^{6}\left (b x +a \right )\right ) \sqrt {2}-6 \left (\cos ^{4}\left (b x +a \right )\right ) \sqrt {2}-6 \cos \left (b x +a \right ) \EllipticE \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}+3 \cos \left (b x +a \right ) \EllipticF \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}-6 \EllipticE \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}+3 \EllipticF \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}-\left (\cos ^{2}\left (b x +a \right )\right ) \sqrt {2}+3 \cos \left (b x +a \right ) \sqrt {2}\right ) \left (\cos \left (b x +a \right )+1\right )^{2} \sqrt {2}}{40 b \cos \left (b x +a \right )^{3} \sin \left (b x +a \right )^{2} \left (\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}\right )^{\frac {5}{2}}}\)	\(550\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40/b*(-1+cos(b*x+a))^2*(4*cos(b*x+a)^6*2^(1/2)-6*cos(b*x+a)^4*2^(1/2)-6*cos(b*x+a)*EllipticE(((1-cos(b*x+a)+
sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*((1-cos(b*x+a)+sin(b*x+a))/sin(b
*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)+3*cos(b*x+a)*EllipticF(((1-cos(b*x+a)+sin(b*x+a))/si
n(b*x+a))^(1/2),1/2*2^(1/2))*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*(
(cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)-6*EllipticE(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/
2))*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))
/sin(b*x+a))^(1/2)+3*EllipticF(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*((-1+cos(b*x+a))/sin(
b*x+a))^(1/2)*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)-cos(b*
x+a)^2*2^(1/2)+3*cos(b*x+a)*2^(1/2))*(cos(b*x+a)+1)^2/cos(b*x+a)^3/sin(b*x+a)^2/(d*sin(b*x+a)/cos(b*x+a))^(5/2
)*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(sin(b*x+a)^5/(d*tan(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [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)^5/(d*tan(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

integral((cos(b*x + a)^4 - 2*cos(b*x + a)^2 + 1)*sqrt(d*tan(b*x + a))*sin(b*x + a)/(d^3*tan(b*x + a)^3), x)

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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)**5/(d*tan(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)^5/(d*tan(b*x+a))^(5/2),x, algorithm="giac")

[Out]

integrate(sin(b*x + a)^5/(d*tan(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 )}^5}{{\left (d\,\mathrm {tan}\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)^5/(d*tan(a + b*x))^(5/2),x)

[Out]

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

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