3.4.22 \(\int \frac {(e \cos (c+d x))^{5/3}}{\sqrt {a+a \sin (c+d x)}} \, dx\) [322]

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

[Out]

-3/8*a*(e*cos(d*x+c))^(8/3)*hypergeom([1/6, 4/3],[7/3],1/2-1/2*sin(d*x+c))*(1+sin(d*x+c))^(1/6)*2^(5/6)/d/e/(a
+a*sin(d*x+c))^(3/2)

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Rubi [A]
time = 0.07, antiderivative size = 78, normalized size of antiderivative = 1.00, 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 = {2768, 72, 71} \begin {gather*} -\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}} \end {gather*}

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 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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]))

Rule 72

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 2768

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]

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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.12, size = 77, normalized size = 0.99 \begin {gather*} -\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 (1+\sin (c+d x))^{5/6} \sqrt {a (1+\sin (c+d x))}} \end {gather*}

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 = 0.09, size = 0, normalized size = 0.00 \[\int \frac {\left (e \cos \left (d x +c \right )\right )^{\frac {5}{3}}}{\sqrt {a +a \sin \left (d x +c \right )}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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]

Exception raised: SystemError >> excessive stack use: stack is 6190 deep

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

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]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:Evaluation time:
236.93Unable to divide, perhaps due to rounding error%%%{2,[0,2,6,0]%%%}+%%%{6,[0,2,4,0]%%%}+%%%{6,[0,2,2,0]%%
%}+%%%{2,[0,2,

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{5/3}}{\sqrt {a+a\,\sin \left (c+d\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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