3.89 \(\int \frac{\sin ^5(a+b x)}{\sqrt{d \tan (a+b x)}} \, dx\)

Optimal. Leaf size=107 \[ -\frac{d \sin ^5(a+b x)}{5 b (d \tan (a+b x))^{3/2}}-\frac{7 d \sin ^3(a+b x)}{30 b (d \tan (a+b x))^{3/2}}+\frac{7 \sin (a+b x) E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{20 b \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}} \]

[Out]

(-7*d*Sin[a + b*x]^3)/(30*b*(d*Tan[a + b*x])^(3/2)) - (d*Sin[a + b*x]^5)/(5*b*(d*Tan[a + b*x])^(3/2)) + (7*Ell
ipticE[a - Pi/4 + b*x, 2]*Sin[a + b*x])/(20*b*Sqrt[Sin[2*a + 2*b*x]]*Sqrt[d*Tan[a + b*x]])

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Rubi [A]  time = 0.132531, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2598, 2601, 2572, 2639} \[ -\frac{d \sin ^5(a+b x)}{5 b (d \tan (a+b x))^{3/2}}-\frac{7 d \sin ^3(a+b x)}{30 b (d \tan (a+b x))^{3/2}}+\frac{7 \sin (a+b x) E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{20 b \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^5/Sqrt[d*Tan[a + b*x]],x]

[Out]

(-7*d*Sin[a + b*x]^3)/(30*b*(d*Tan[a + b*x])^(3/2)) - (d*Sin[a + b*x]^5)/(5*b*(d*Tan[a + b*x])^(3/2)) + (7*Ell
ipticE[a - Pi/4 + b*x, 2]*Sin[a + b*x])/(20*b*Sqrt[Sin[2*a + 2*b*x]]*Sqrt[d*Tan[a + b*x]])

Rule 2598

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*Ta
n[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 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 ^5(a+b x)}{\sqrt{d \tan (a+b x)}} \, dx &=-\frac{d \sin ^5(a+b x)}{5 b (d \tan (a+b x))^{3/2}}+\frac{7}{10} \int \frac{\sin ^3(a+b x)}{\sqrt{d \tan (a+b x)}} \, dx\\ &=-\frac{7 d \sin ^3(a+b x)}{30 b (d \tan (a+b x))^{3/2}}-\frac{d \sin ^5(a+b x)}{5 b (d \tan (a+b x))^{3/2}}+\frac{7}{20} \int \frac{\sin (a+b x)}{\sqrt{d \tan (a+b x)}} \, dx\\ &=-\frac{7 d \sin ^3(a+b x)}{30 b (d \tan (a+b x))^{3/2}}-\frac{d \sin ^5(a+b x)}{5 b (d \tan (a+b x))^{3/2}}+\frac{\left (7 \sqrt{\sin (a+b x)}\right ) \int \sqrt{\cos (a+b x)} \sqrt{\sin (a+b x)} \, dx}{20 \sqrt{\cos (a+b x)} \sqrt{d \tan (a+b x)}}\\ &=-\frac{7 d \sin ^3(a+b x)}{30 b (d \tan (a+b x))^{3/2}}-\frac{d \sin ^5(a+b x)}{5 b (d \tan (a+b x))^{3/2}}+\frac{(7 \sin (a+b x)) \int \sqrt{\sin (2 a+2 b x)} \, dx}{20 \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}}\\ &=-\frac{7 d \sin ^3(a+b x)}{30 b (d \tan (a+b x))^{3/2}}-\frac{d \sin ^5(a+b x)}{5 b (d \tan (a+b x))^{3/2}}+\frac{7 E\left (\left .a-\frac{\pi }{4}+b x\right |2\right ) \sin (a+b x)}{20 b \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}}\\ \end{align*}

Mathematica [C]  time = 0.840406, size = 86, normalized size = 0.8 \[ \frac{\sin (a+b x) \left (28 \tan (a+b x) \sqrt{\sec ^2(a+b x)} \, _2F_1\left (\frac{3}{4},\frac{3}{2};\frac{7}{4};-\tan ^2(a+b x)\right )-20 \sin (2 (a+b x))+3 \sin (4 (a+b x))\right )}{120 b \sqrt{d \tan (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^5/Sqrt[d*Tan[a + b*x]],x]

[Out]

(Sin[a + b*x]*(-20*Sin[2*(a + b*x)] + 3*Sin[4*(a + b*x)] + 28*Hypergeometric2F1[3/4, 3/2, 7/4, -Tan[a + b*x]^2
]*Sqrt[Sec[a + b*x]^2]*Tan[a + b*x]))/(120*b*Sqrt[d*Tan[a + b*x]])

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Maple [B]  time = 0.177, size = 558, normalized size = 5.2 \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)^5/(d*tan(b*x+a))^(1/2),x)

[Out]

-1/120/b*2^(1/2)*(cos(b*x+a)-1)^2*(12*cos(b*x+a)^6*2^(1/2)-38*cos(b*x+a)^4*2^(1/2)-21*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)+42*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)-21*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)+42*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)+47*cos(b*x+a)^2*2^(1/2)-21*cos(b*x+a)*2^(1/2))*(cos(b*x+a)+1)^2/cos(b*x+a)/sin(b*x+a)^4/(d*sin(b*
x+a)/cos(b*x+a))^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^5/(d*tan(b*x+a))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1\right )} \sqrt{d \tan \left (b x + a\right )} \sin \left (b x + a\right )}{d \tan \left (b x + a\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^5/(d*tan(b*x+a))^(1/2),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*tan(b*x + a)), 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)**5/(d*tan(b*x+a))**(1/2),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 )^{5}}{\sqrt{d \tan \left (b x + a\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^5/(d*tan(b*x+a))^(1/2),x, algorithm="giac")

[Out]

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