3.325 \(\int \frac {1}{\sqrt [3]{e \cos (c+d x)} \sqrt {a+a \sin (c+d x)}} \, dx\)

Optimal. Leaf size=77 \[ -\frac {3 \sqrt [6]{\sin (c+d x)+1} (e \cos (c+d x))^{2/3} \, _2F_1\left (\frac {1}{3},\frac {7}{6};\frac {4}{3};\frac {1}{2} (1-\sin (c+d x))\right )}{2 \sqrt [6]{2} d e \sqrt {a \sin (c+d x)+a}} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 77, 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 = {2689, 70, 69} \[ -\frac {3 \sqrt [6]{\sin (c+d x)+1} (e \cos (c+d x))^{2/3} \, _2F_1\left (\frac {1}{3},\frac {7}{6};\frac {4}{3};\frac {1}{2} (1-\sin (c+d x))\right )}{2 \sqrt [6]{2} d e \sqrt {a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]))

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 77, normalized size = 1.00 \[ -\frac {3 \sqrt [6]{\sin (c+d x)+1} (e \cos (c+d x))^{2/3} \, _2F_1\left (\frac {1}{3},\frac {7}{6};\frac {4}{3};\frac {1}{2} (1-\sin (c+d x))\right )}{2 \sqrt [6]{2} d e \sqrt {a (\sin (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (e \cos \left (d x + c\right )\right )^{\frac {2}{3}} \sqrt {a \sin \left (d x + c\right ) + a}}{a e \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a e \cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [F]  time = 0.21, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (e \cos \left (d x +c \right )\right )^{\frac {1}{3}} \sqrt {a +a \sin \left (d x +c \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (e \cos \left (d x + c\right )\right )^{\frac {1}{3}} \sqrt {a \sin \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )} \sqrt [3]{e \cos {\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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