3.326 \(\int \frac{1}{(e \cos (c+d x))^{4/3} \sqrt{a+a \sin (c+d x)}} \, dx\)

Optimal. Leaf size=75 \[ \frac{3 (\sin (c+d x)+1)^{2/3} \, _2F_1\left (-\frac{1}{6},\frac{5}{3};\frac{5}{6};\frac{1}{2} (1-\sin (c+d x))\right )}{2^{2/3} d e \sqrt{a \sin (c+d x)+a} \sqrt [3]{e \cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0945724, antiderivative size = 75, 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} \[ \frac{3 (\sin (c+d x)+1)^{2/3} \, _2F_1\left (-\frac{1}{6},\frac{5}{3};\frac{5}{6};\frac{1}{2} (1-\sin (c+d x))\right )}{2^{2/3} d e \sqrt{a \sin (c+d x)+a} \sqrt [3]{e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/((e*Cos[c + d*x])^(4/3)*Sqrt[a + a*Sin[c + d*x]]),x]

[Out]

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

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

Mathematica [A]  time = 0.0869293, size = 75, normalized size = 1. \[ \frac{3 (\sin (c+d x)+1)^{2/3} \, _2F_1\left (-\frac{1}{6},\frac{5}{3};\frac{5}{6};\frac{1}{2} (1-\sin (c+d x))\right )}{2^{2/3} d e \sqrt{a (\sin (c+d x)+1)} \sqrt [3]{e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((e*Cos[c + d*x])^(4/3)*Sqrt[a + a*Sin[c + d*x]]),x]

[Out]

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

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Maple [F]  time = 0.11, size = 0, normalized size = 0. \begin{align*} \int{ \left ( e\cos \left ( dx+c \right ) \right ) ^{-{\frac{4}{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(1/(e*cos(d*x+c))^(4/3)/(a+a*sin(d*x+c))^(1/2),x)

[Out]

int(1/(e*cos(d*x+c))^(4/3)/(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{1}{\left (e \cos \left (d x + c\right )\right )^{\frac{4}{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(1/(e*cos(d*x+c))^(4/3)/(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(1/((e*cos(d*x + c))^(4/3)*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}} \sqrt{a \sin \left (d x + c\right ) + a}}{a e^{2} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + a e^{2} \cos \left (d x + c\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cos(d*x+c))^(4/3)/(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral((e*cos(d*x + c))^(2/3)*sqrt(a*sin(d*x + c) + a)/(a*e^2*cos(d*x + c)^2*sin(d*x + c) + a*e^2*cos(d*x +
c)^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(1/(e*cos(d*x+c))**(4/3)/(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{1}{\left (e \cos \left (d x + c\right )\right )^{\frac{4}{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(1/(e*cos(d*x+c))^(4/3)/(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(1/((e*cos(d*x + c))^(4/3)*sqrt(a*sin(d*x + c) + a)), x)