3.78 \(\int \sin ^3(a+b x) (d \tan (a+b x))^{5/2} \, dx\)

Optimal. Leaf size=137 \[ \frac{d^3 \sin ^3(a+b x)}{b \sqrt{d \tan (a+b x)}}+\frac{5 d^3 \sin (a+b x)}{2 b \sqrt{d \tan (a+b x)}}-\frac{5 d^2 \sqrt{\sin (2 a+2 b x)} \csc (a+b x) F\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (a+b x)}}{4 b}+\frac{2 d \sin ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b} \]

[Out]

(5*d^3*Sin[a + b*x])/(2*b*Sqrt[d*Tan[a + b*x]]) + (d^3*Sin[a + b*x]^3)/(b*Sqrt[d*Tan[a + b*x]]) - (5*d^2*Csc[a
+ b*x]*EllipticF[a - Pi/4 + b*x, 2]*Sqrt[Sin[2*a + 2*b*x]]*Sqrt[d*Tan[a + b*x]])/(4*b) + (2*d*Sin[a + b*x]^3*
(d*Tan[a + b*x])^(3/2))/(3*b)

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Rubi [A]  time = 0.174526, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2594, 2598, 2601, 2573, 2641} \[ \frac{d^3 \sin ^3(a+b x)}{b \sqrt{d \tan (a+b x)}}+\frac{5 d^3 \sin (a+b x)}{2 b \sqrt{d \tan (a+b x)}}-\frac{5 d^2 \sqrt{\sin (2 a+2 b x)} \csc (a+b x) F\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (a+b x)}}{4 b}+\frac{2 d \sin ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^3*(d*Tan[a + b*x])^(5/2),x]

[Out]

(5*d^3*Sin[a + b*x])/(2*b*Sqrt[d*Tan[a + b*x]]) + (d^3*Sin[a + b*x]^3)/(b*Sqrt[d*Tan[a + b*x]]) - (5*d^2*Csc[a
+ b*x]*EllipticF[a - Pi/4 + b*x, 2]*Sqrt[Sin[2*a + 2*b*x]]*Sqrt[d*Tan[a + b*x]])/(4*b) + (2*d*Sin[a + b*x]^3*
(d*Tan[a + b*x])^(3/2))/(3*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 2573

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(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))^{5/2} \, dx &=\frac{2 d \sin ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b}-\left (3 d^2\right ) \int \sin ^3(a+b x) \sqrt{d \tan (a+b x)} \, dx\\ &=\frac{d^3 \sin ^3(a+b x)}{b \sqrt{d \tan (a+b x)}}+\frac{2 d \sin ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b}-\frac{1}{2} \left (5 d^2\right ) \int \sin (a+b x) \sqrt{d \tan (a+b x)} \, dx\\ &=\frac{5 d^3 \sin (a+b x)}{2 b \sqrt{d \tan (a+b x)}}+\frac{d^3 \sin ^3(a+b x)}{b \sqrt{d \tan (a+b x)}}+\frac{2 d \sin ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b}-\frac{1}{4} \left (5 d^2\right ) \int \csc (a+b x) \sqrt{d \tan (a+b x)} \, dx\\ &=\frac{5 d^3 \sin (a+b x)}{2 b \sqrt{d \tan (a+b x)}}+\frac{d^3 \sin ^3(a+b x)}{b \sqrt{d \tan (a+b x)}}+\frac{2 d \sin ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b}-\frac{\left (5 d^2 \sqrt{\cos (a+b x)} \sqrt{d \tan (a+b x)}\right ) \int \frac{1}{\sqrt{\cos (a+b x)} \sqrt{\sin (a+b x)}} \, dx}{4 \sqrt{\sin (a+b x)}}\\ &=\frac{5 d^3 \sin (a+b x)}{2 b \sqrt{d \tan (a+b x)}}+\frac{d^3 \sin ^3(a+b x)}{b \sqrt{d \tan (a+b x)}}+\frac{2 d \sin ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b}-\frac{1}{4} \left (5 d^2 \csc (a+b x) \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}\right ) \int \frac{1}{\sqrt{\sin (2 a+2 b x)}} \, dx\\ &=\frac{5 d^3 \sin (a+b x)}{2 b \sqrt{d \tan (a+b x)}}+\frac{d^3 \sin ^3(a+b x)}{b \sqrt{d \tan (a+b x)}}-\frac{5 d^2 \csc (a+b x) F\left (\left .a-\frac{\pi }{4}+b x\right |2\right ) \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}}{4 b}+\frac{2 d \sin ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b}\\ \end{align*}

Mathematica [C]  time = 3.24586, size = 153, normalized size = 1.12 \[ -\frac{\csc (a+b x) \sqrt{\sec ^2(a+b x)} (d \tan (a+b x))^{5/2} \left ((77 \cos (2 (a+b x))+22 \cos (4 (a+b x))-\cos (6 (a+b x))+22) \sqrt{\tan (a+b x)} \sqrt{\sec ^2(a+b x)}+120 \sqrt [4]{-1} \cos (2 (a+b x)) F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-1} \sqrt{\tan (a+b x)}\right )\right |-1\right )\right )}{48 b \tan ^{\frac{3}{2}}(a+b x) \left (\tan ^2(a+b x)-1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^3*(d*Tan[a + b*x])^(5/2),x]

[Out]

-(Csc[a + b*x]*Sqrt[Sec[a + b*x]^2]*(120*(-1)^(1/4)*Cos[2*(a + b*x)]*EllipticF[I*ArcSinh[(-1)^(1/4)*Sqrt[Tan[a
+ b*x]]], -1] + (22 + 77*Cos[2*(a + b*x)] + 22*Cos[4*(a + b*x)] - Cos[6*(a + b*x)])*Sqrt[Sec[a + b*x]^2]*Sqrt
[Tan[a + b*x]])*(d*Tan[a + b*x])^(5/2))/(48*b*Tan[a + b*x]^(3/2)*(-1 + Tan[a + b*x]^2))

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Maple [A]  time = 0.128, size = 248, normalized size = 1.8 \begin{align*} -{\frac{\sqrt{2} \left ( \cos \left ( bx+a \right ) -1 \right ) \cos \left ( bx+a \right ) \left ( \cos \left ( bx+a \right ) +1 \right ) ^{2}}{12\,b \left ( \sin \left ( bx+a \right ) \right ) ^{6}} \left ( 2\,\sqrt{2} \left ( \cos \left ( bx+a \right ) \right ) ^{5}-15\,\cos \left ( bx+a \right ) \sqrt{{\frac{\cos \left ( bx+a \right ) -1}{\sin \left ( bx+a \right ) }}}\sqrt{{\frac{\cos \left ( bx+a \right ) -1+\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}\sqrt{-{\frac{\cos \left ( bx+a \right ) -1-\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}\sin \left ( bx+a \right ){\it EllipticF} \left ( \sqrt{-{\frac{\cos \left ( bx+a \right ) -1-\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}},1/2\,\sqrt{2} \right ) -2\, \left ( \cos \left ( bx+a \right ) \right ) ^{4}\sqrt{2}-13\, \left ( \cos \left ( bx+a \right ) \right ) ^{3}\sqrt{2}+13\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}\sqrt{2}-4\,\cos \left ( bx+a \right ) \sqrt{2}+4\,\sqrt{2} \right ) \left ({\frac{\sin \left ( bx+a \right ) d}{\cos \left ( bx+a \right ) }} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)^3*(d*tan(b*x+a))^(5/2),x)

[Out]

-1/12/b*2^(1/2)*(cos(b*x+a)-1)*(2*2^(1/2)*cos(b*x+a)^5-15*cos(b*x+a)*((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)*sin(b*x+a)*EllipticF((-(co
s(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))-2*cos(b*x+a)^4*2^(1/2)-13*cos(b*x+a)^3*2^(1/2)+13*cos(b*
x+a)^2*2^(1/2)-4*cos(b*x+a)*2^(1/2)+4*2^(1/2))*cos(b*x+a)*(cos(b*x+a)+1)^2*(d*sin(b*x+a)/cos(b*x+a))^(5/2)/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{5}{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))^(5/2),x, algorithm="maxima")

[Out]

integrate((d*tan(b*x + a))^(5/2)*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^{2} \cos \left (b x + a\right )^{2} - d^{2}\right )} \sqrt{d \tan \left (b x + a\right )} \sin \left (b x + a\right ) \tan \left (b x + a\right )^{2}, 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))^(5/2),x, algorithm="fricas")

[Out]

integral(-(d^2*cos(b*x + a)^2 - d^2)*sqrt(d*tan(b*x + a))*sin(b*x + a)*tan(b*x + a)^2, 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))**(5/2),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))^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError