3.59 \(\int \sin ^3(a+b x) \sqrt{d \tan (a+b x)} \, dx\)

Optimal. Leaf size=105 \[ -\frac{d \sin ^3(a+b x)}{3 b \sqrt{d \tan (a+b x)}}-\frac{5 d \sin (a+b x)}{6 b \sqrt{d \tan (a+b x)}}+\frac{5 \sqrt{\sin (2 a+2 b x)} \csc (a+b x) F\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (a+b x)}}{12 b} \]

[Out]

(-5*d*Sin[a + b*x])/(6*b*Sqrt[d*Tan[a + b*x]]) - (d*Sin[a + b*x]^3)/(3*b*Sqrt[d*Tan[a + b*x]]) + (5*Csc[a + b*
x]*EllipticF[a - Pi/4 + b*x, 2]*Sqrt[Sin[2*a + 2*b*x]]*Sqrt[d*Tan[a + b*x]])/(12*b)

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Rubi [A]  time = 0.133208, antiderivative size = 105, 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, 2573, 2641} \[ -\frac{d \sin ^3(a+b x)}{3 b \sqrt{d \tan (a+b x)}}-\frac{5 d \sin (a+b x)}{6 b \sqrt{d \tan (a+b x)}}+\frac{5 \sqrt{\sin (2 a+2 b x)} \csc (a+b x) F\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (a+b x)}}{12 b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^3*Sqrt[d*Tan[a + b*x]],x]

[Out]

(-5*d*Sin[a + b*x])/(6*b*Sqrt[d*Tan[a + b*x]]) - (d*Sin[a + b*x]^3)/(3*b*Sqrt[d*Tan[a + b*x]]) + (5*Csc[a + b*
x]*EllipticF[a - Pi/4 + b*x, 2]*Sqrt[Sin[2*a + 2*b*x]]*Sqrt[d*Tan[a + b*x]])/(12*b)

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) \sqrt{d \tan (a+b x)} \, dx &=-\frac{d \sin ^3(a+b x)}{3 b \sqrt{d \tan (a+b x)}}+\frac{5}{6} \int \sin (a+b x) \sqrt{d \tan (a+b x)} \, dx\\ &=-\frac{5 d \sin (a+b x)}{6 b \sqrt{d \tan (a+b x)}}-\frac{d \sin ^3(a+b x)}{3 b \sqrt{d \tan (a+b x)}}+\frac{5}{12} \int \csc (a+b x) \sqrt{d \tan (a+b x)} \, dx\\ &=-\frac{5 d \sin (a+b x)}{6 b \sqrt{d \tan (a+b x)}}-\frac{d \sin ^3(a+b x)}{3 b \sqrt{d \tan (a+b x)}}+\frac{\left (5 \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}{12 \sqrt{\sin (a+b x)}}\\ &=-\frac{5 d \sin (a+b x)}{6 b \sqrt{d \tan (a+b x)}}-\frac{d \sin ^3(a+b x)}{3 b \sqrt{d \tan (a+b x)}}+\frac{1}{12} \left (5 \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 \sin (a+b x)}{6 b \sqrt{d \tan (a+b x)}}-\frac{d \sin ^3(a+b x)}{3 b \sqrt{d \tan (a+b x)}}+\frac{5 \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)}}{12 b}\\ \end{align*}

Mathematica [C]  time = 1.82994, size = 139, normalized size = 1.32 \[ -\frac{\cos (2 (a+b x)) \sec (a+b x) \sqrt{d \tan (a+b x)} \left ((\cos (2 (a+b x))-6) \sqrt{\tan (a+b x)} \sqrt{\sec ^2(a+b x)}-5 \sqrt [4]{-1} \sec ^2(a+b x) F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-1} \sqrt{\tan (a+b x)}\right )\right |-1\right )\right )}{6 b \sqrt{\tan (a+b x)} \left (\tan ^2(a+b x)-1\right ) \sqrt{\sec ^2(a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^3*Sqrt[d*Tan[a + b*x]],x]

[Out]

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

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Maple [A]  time = 0.207, size = 216, normalized size = 2.1 \begin{align*} -{\frac{\sqrt{2} \left ( \cos \left ( bx+a \right ) -1 \right ) \left ( \cos \left ( bx+a \right ) +1 \right ) ^{2}}{12\,b \left ( \sin \left ( bx+a \right ) \right ) ^{4}} \left ( 5\,\sin \left ( bx+a \right ) \sqrt{{\frac{\cos \left ( bx+a \right ) -1}{\sin \left ( bx+a \right ) }}}\sqrt{{\frac{1-\cos \left ( bx+a \right ) +\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 ) }}}{\it EllipticF} \left ( \sqrt{{\frac{1-\cos \left ( bx+a \right ) +\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}+2\, \left ( \cos \left ( bx+a \right ) \right ) ^{3}\sqrt{2}+7\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}\sqrt{2}-7\,\cos \left ( bx+a \right ) \sqrt{2} \right ) \sqrt{{\frac{\sin \left ( bx+a \right ) d}{\cos \left ( bx+a \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)^3*(d*tan(b*x+a))^(1/2),x)

[Out]

-1/12/b*2^(1/2)*(cos(b*x+a)-1)*(5*sin(b*x+a)*((cos(b*x+a)-1)/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)*EllipticF(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1
/2),1/2*2^(1/2))-2*cos(b*x+a)^4*2^(1/2)+2*cos(b*x+a)^3*2^(1/2)+7*cos(b*x+a)^2*2^(1/2)-7*cos(b*x+a)*2^(1/2))*(c
os(b*x+a)+1)^2*(d*sin(b*x+a)/cos(b*x+a))^(1/2)/sin(b*x+a)^4

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{d \tan \left (b x + a\right )} \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))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(d*tan(b*x + a))*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 (\cos \left (b x + a\right )^{2} - 1\right )} \sqrt{d \tan \left (b x + a\right )} \sin \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))^(1/2),x, algorithm="fricas")

[Out]

integral(-(cos(b*x + a)^2 - 1)*sqrt(d*tan(b*x + a))*sin(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))**(1/2),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))^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError