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

Optimal. Leaf size=78 \[ -\frac{3 \sin (a+b x) E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{b d^2 \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}}-\frac{2 \sin (a+b x)}{b d (d \tan (a+b x))^{3/2}} \]

[Out]

(-2*Sin[a + b*x])/(b*d*(d*Tan[a + b*x])^(3/2)) - (3*EllipticE[a - Pi/4 + b*x, 2]*Sin[a + b*x])/(b*d^2*Sqrt[Sin
[2*a + 2*b*x]]*Sqrt[d*Tan[a + b*x]])

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Rubi [A]  time = 0.0865126, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2597, 2601, 2572, 2639} \[ -\frac{3 \sin (a+b x) E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{b d^2 \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}}-\frac{2 \sin (a+b x)}{b d (d \tan (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sin[a + b*x])/(b*d*(d*Tan[a + b*x])^(3/2)) - (3*EllipticE[a - Pi/4 + b*x, 2]*Sin[a + b*x])/(b*d^2*Sqrt[Sin
[2*a + 2*b*x]]*Sqrt[d*Tan[a + b*x]])

Rule 2597

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 + n + 1)), x] - Dist[(n + 1)/(b^2*(m + n + 1)), Int[(a*Sin[e + f*x])
^m*(b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && LtQ[n, -1] && NeQ[m + n + 1, 0] && Integer
sQ[2*m, 2*n] &&  !(EqQ[n, -3/2] && EqQ[m, 1])

Rule 2601

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 2572

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 2639

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

Rubi steps

\begin{align*} \int \frac{\sin (a+b x)}{(d \tan (a+b x))^{5/2}} \, dx &=-\frac{2 \sin (a+b x)}{b d (d \tan (a+b x))^{3/2}}-\frac{3 \int \frac{\sin (a+b x)}{\sqrt{d \tan (a+b x)}} \, dx}{d^2}\\ &=-\frac{2 \sin (a+b x)}{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}{d^2 \sqrt{\cos (a+b x)} \sqrt{d \tan (a+b x)}}\\ &=-\frac{2 \sin (a+b x)}{b d (d \tan (a+b x))^{3/2}}-\frac{(3 \sin (a+b x)) \int \sqrt{\sin (2 a+2 b x)} \, dx}{d^2 \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}}\\ &=-\frac{2 \sin (a+b x)}{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)}{b d^2 \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}}\\ \end{align*}

Mathematica [C]  time = 0.376484, size = 69, normalized size = 0.88 \[ -\frac{2 \cos (a+b x) \left (\tan ^2(a+b x) \sqrt{\sec ^2(a+b x)} \, _2F_1\left (\frac{3}{4},\frac{3}{2};\frac{7}{4};-\tan ^2(a+b x)\right )+1\right )}{b d^2 \sqrt{d \tan (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cos[a + b*x]*(1 + Hypergeometric2F1[3/4, 3/2, 7/4, -Tan[a + b*x]^2]*Sqrt[Sec[a + b*x]^2]*Tan[a + b*x]^2))/
(b*d^2*Sqrt[d*Tan[a + b*x]])

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Maple [B]  time = 0.138, size = 511, normalized size = 6.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/b*2^(1/2)*(6*cos(b*x+a)*EllipticE((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*((cos(b*x+a)-
1)/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)-1-sin(b*x+a))/sin(b*x+a))^(1/2
)-3*cos(b*x+a)*EllipticF((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*((cos(b*x+a)-1)/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)-1-sin(b*x+a))/sin(b*x+a))^(1/2)+6*EllipticE
((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*((cos(b*x+a)-1)/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+s
in(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((-(cos(b*x+a)-1-sin(b*x
+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*((cos(b*x+a)-1)/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)-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))*sin(b*x+a)^2/cos
(b*x+a)^3/(d*sin(b*x+a)/cos(b*x+a))^(5/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (b x + a\right )}{\left (d \tan \left (b x + a\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{d \tan \left (b x + a\right )} \sin \left (b x + a\right )}{d^{3} \tan \left (b x + a\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(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)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (b x + a\right )}{\left (d \tan \left (b x + a\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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