3.322 \(\int \frac{(e \cos (c+d x))^{5/3}}{\sqrt{a+a \sin (c+d x)}} \, dx\)

Optimal. Leaf size=78 \[ -\frac{3 a \sqrt [6]{\sin (c+d x)+1} (e \cos (c+d x))^{8/3} \, _2F_1\left (\frac{1}{6},\frac{4}{3};\frac{7}{3};\frac{1}{2} (1-\sin (c+d x))\right )}{4 \sqrt [6]{2} d e (a \sin (c+d x)+a)^{3/2}} \]

[Out]

(-3*a*(e*Cos[c + d*x])^(8/3)*Hypergeometric2F1[1/6, 4/3, 7/3, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x])^(1/6))/
(4*2^(1/6)*d*e*(a + a*Sin[c + d*x])^(3/2))

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Rubi [A]  time = 0.0973489, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2689, 70, 69} \[ -\frac{3 a \sqrt [6]{\sin (c+d x)+1} (e \cos (c+d x))^{8/3} \, _2F_1\left (\frac{1}{6},\frac{4}{3};\frac{7}{3};\frac{1}{2} (1-\sin (c+d x))\right )}{4 \sqrt [6]{2} d e (a \sin (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(5/3)/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-3*a*(e*Cos[c + d*x])^(8/3)*Hypergeometric2F1[1/6, 4/3, 7/3, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x])^(1/6))/
(4*2^(1/6)*d*e*(a + a*Sin[c + d*x])^(3/2))

Rule 2689

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[(a^2*
(g*Cos[e + f*x])^(p + 1))/(f*g*(a + b*Sin[e + f*x])^((p + 1)/2)*(a - b*Sin[e + f*x])^((p + 1)/2)), Subst[Int[(
a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&
EqQ[a^2 - b^2, 0] &&  !IntegerQ[m]

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*Simp[(b*c)/(b*c - a*d) + (b*d*x)/(b*c -
a*d), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

\begin{align*} \int \frac{(e \cos (c+d x))^{5/3}}{\sqrt{a+a \sin (c+d x)}} \, dx &=\frac{\left (a^2 (e \cos (c+d x))^{8/3}\right ) \operatorname{Subst}\left (\int \frac{\sqrt [3]{a-a x}}{\sqrt [6]{a+a x}} \, dx,x,\sin (c+d x)\right )}{d e (a-a \sin (c+d x))^{4/3} (a+a \sin (c+d x))^{4/3}}\\ &=\frac{\left (a^2 (e \cos (c+d x))^{8/3} \sqrt [6]{\frac{a+a \sin (c+d x)}{a}}\right ) \operatorname{Subst}\left (\int \frac{\sqrt [3]{a-a x}}{\sqrt [6]{\frac{1}{2}+\frac{x}{2}}} \, dx,x,\sin (c+d x)\right )}{\sqrt [6]{2} d e (a-a \sin (c+d x))^{4/3} (a+a \sin (c+d x))^{3/2}}\\ &=-\frac{3 a (e \cos (c+d x))^{8/3} \, _2F_1\left (\frac{1}{6},\frac{4}{3};\frac{7}{3};\frac{1}{2} (1-\sin (c+d x))\right ) \sqrt [6]{1+\sin (c+d x)}}{4 \sqrt [6]{2} d e (a+a \sin (c+d x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.108071, size = 77, normalized size = 0.99 \[ -\frac{3 (e \cos (c+d x))^{8/3} \, _2F_1\left (\frac{1}{6},\frac{4}{3};\frac{7}{3};\frac{1}{2} (1-\sin (c+d x))\right )}{4 \sqrt [6]{2} d e (\sin (c+d x)+1)^{5/6} \sqrt{a (\sin (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^(5/3)/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-3*(e*Cos[c + d*x])^(8/3)*Hypergeometric2F1[1/6, 4/3, 7/3, (1 - Sin[c + d*x])/2])/(4*2^(1/6)*d*e*(1 + Sin[c +
d*x])^(5/6)*Sqrt[a*(1 + Sin[c + d*x])])

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Maple [F]  time = 0.124, size = 0, normalized size = 0. \begin{align*} \int{ \left ( e\cos \left ( dx+c \right ) \right ) ^{{\frac{5}{3}}}{\frac{1}{\sqrt{a+a\sin \left ( dx+c \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(5/3)/(a+a*sin(d*x+c))^(1/2),x)

[Out]

int((e*cos(d*x+c))^(5/3)/(a+a*sin(d*x+c))^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \cos \left (d x + c\right )\right )^{\frac{5}{3}}}{\sqrt{a \sin \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(5/3)/(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^(5/3)/sqrt(a*sin(d*x + c) + a), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\left (e \cos \left (d x + c\right )\right )^{\frac{2}{3}} e \cos \left (d x + c\right )}{\sqrt{a \sin \left (d x + c\right ) + a}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(5/3)/(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral((e*cos(d*x + c))^(2/3)*e*cos(d*x + c)/sqrt(a*sin(d*x + c) + 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((e*cos(d*x+c))**(5/3)/(a+a*sin(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \cos \left (d x + c\right )\right )^{\frac{5}{3}}}{\sqrt{a \sin \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(5/3)/(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate((e*cos(d*x + c))^(5/3)/sqrt(a*sin(d*x + c) + a), x)