3.69 \(\int \sin ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx\)

Optimal. Leaf size=110 \[ \frac{7 d^3 \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}-\frac{7 d^2 \sin (a+b x) E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{2 b \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}}+\frac{2 d \sin ^3(a+b x) \sqrt{d \tan (a+b x)}}{b} \]

[Out]

(7*d^3*Sin[a + b*x]^3)/(3*b*(d*Tan[a + b*x])^(3/2)) - (7*d^2*EllipticE[a - Pi/4 + b*x, 2]*Sin[a + b*x])/(2*b*S
qrt[Sin[2*a + 2*b*x]]*Sqrt[d*Tan[a + b*x]]) + (2*d*Sin[a + b*x]^3*Sqrt[d*Tan[a + b*x]])/b

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Rubi [A]  time = 0.141057, antiderivative size = 110, normalized size of antiderivative = 1., 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 = {2594, 2598, 2601, 2572, 2639} \[ \frac{7 d^3 \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}-\frac{7 d^2 \sin (a+b x) E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{2 b \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}}+\frac{2 d \sin ^3(a+b x) \sqrt{d \tan (a+b x)}}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^3*(d*Tan[a + b*x])^(3/2),x]

[Out]

(7*d^3*Sin[a + b*x]^3)/(3*b*(d*Tan[a + b*x])^(3/2)) - (7*d^2*EllipticE[a - Pi/4 + b*x, 2]*Sin[a + b*x])/(2*b*S
qrt[Sin[2*a + 2*b*x]]*Sqrt[d*Tan[a + b*x]]) + (2*d*Sin[a + b*x]^3*Sqrt[d*Tan[a + b*x]])/b

Rule 2594

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*(n - 1)), x] - Dist[(b^2*(m + n - 1))/(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] && GtQ[n, 1] && IntegersQ[2*m, 2*n] &&  !(GtQ[m,
1] &&  !IntegerQ[(m - 1)/2])

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 \sin ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx &=\frac{2 d \sin ^3(a+b x) \sqrt{d \tan (a+b x)}}{b}-\left (7 d^2\right ) \int \frac{\sin ^3(a+b x)}{\sqrt{d \tan (a+b x)}} \, dx\\ &=\frac{7 d^3 \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}+\frac{2 d \sin ^3(a+b x) \sqrt{d \tan (a+b x)}}{b}-\frac{1}{2} \left (7 d^2\right ) \int \frac{\sin (a+b x)}{\sqrt{d \tan (a+b x)}} \, dx\\ &=\frac{7 d^3 \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}+\frac{2 d \sin ^3(a+b x) \sqrt{d \tan (a+b x)}}{b}-\frac{\left (7 d^2 \sqrt{\sin (a+b x)}\right ) \int \sqrt{\cos (a+b x)} \sqrt{\sin (a+b x)} \, dx}{2 \sqrt{\cos (a+b x)} \sqrt{d \tan (a+b x)}}\\ &=\frac{7 d^3 \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}+\frac{2 d \sin ^3(a+b x) \sqrt{d \tan (a+b x)}}{b}-\frac{\left (7 d^2 \sin (a+b x)\right ) \int \sqrt{\sin (2 a+2 b x)} \, dx}{2 \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}}\\ &=\frac{7 d^3 \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}-\frac{7 d^2 E\left (\left .a-\frac{\pi }{4}+b x\right |2\right ) \sin (a+b x)}{2 b \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}}+\frac{2 d \sin ^3(a+b x) \sqrt{d \tan (a+b x)}}{b}\\ \end{align*}

Mathematica [C]  time = 0.580102, size = 90, normalized size = 0.82 \[ \frac{(d \tan (a+b x))^{3/2} \left (2 \cos (a+b x) (\cos (2 (a+b x))+13) \sqrt{\sec ^2(a+b x)}-28 \sec (a+b x) \, _2F_1\left (\frac{3}{4},\frac{3}{2};\frac{7}{4};-\tan ^2(a+b x)\right )\right )}{12 b \sqrt{\sec ^2(a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^3*(d*Tan[a + b*x])^(3/2),x]

[Out]

((-28*Hypergeometric2F1[3/4, 3/2, 7/4, -Tan[a + b*x]^2]*Sec[a + b*x] + 2*Cos[a + b*x]*(13 + Cos[2*(a + b*x)])*
Sqrt[Sec[a + b*x]^2])*(d*Tan[a + b*x])^(3/2))/(12*b*Sqrt[Sec[a + b*x]^2])

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Maple [B]  time = 0.155, size = 548, normalized size = 5. \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)^3*(d*tan(b*x+a))^(3/2),x)

[Out]

-1/12/b*2^(1/2)*(cos(b*x+a)-1)^2*(2*cos(b*x+a)^4*2^(1/2)+21*cos(b*x+a)*EllipticF((-(cos(b*x+a)-1-sin(b*x+a))/s
in(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)-11*cos(b*x+a)
^2*2^(1/2)+21*cos(b*x+a)*2^(1/2)-12*2^(1/2))*cos(b*x+a)*(cos(b*x+a)+1)^2*(d*sin(b*x+a)/cos(b*x+a))^(3/2)/sin(b
*x+a)^6

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \tan \left (b x + a\right )\right )^{\frac{3}{2}} \sin \left (b x + a\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^3*(d*tan(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

integrate((d*tan(b*x + a))^(3/2)*sin(b*x + a)^3, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (d \cos \left (b x + a\right )^{2} - d\right )} \sqrt{d \tan \left (b x + a\right )} \sin \left (b x + a\right ) \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)^3*(d*tan(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^3*(d*tan(b*x+a))^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError