3.719 \(\int \frac {1}{(d+e x)^{3/2} \sqrt [4]{a+c x^2}} \, dx\)

Optimal. Leaf size=200 \[ -\frac {2 \left (\sqrt {-a}-\sqrt {c} x\right ) \sqrt [4]{-\frac {\left (\sqrt {-a}+\sqrt {c} x\right ) \left (\sqrt {-a} e+\sqrt {c} d\right )}{\left (\sqrt {-a}-\sqrt {c} x\right ) \left (\sqrt {c} d-\sqrt {-a} e\right )}} \, _2F_1\left (-\frac {1}{2},\frac {1}{4};\frac {1}{2};\frac {2 \sqrt {-a} \sqrt {c} (d+e x)}{\left (\sqrt {c} d-\sqrt {-a} e\right ) \left (\sqrt {-a}-\sqrt {c} x\right )}\right )}{\sqrt [4]{a+c x^2} \sqrt {d+e x} \left (\sqrt {-a} e+\sqrt {c} d\right )} \]

[Out]

-2*hypergeom([-1/2, 1/4],[1/2],2*(e*x+d)*(-a)^(1/2)*c^(1/2)/(-e*(-a)^(1/2)+d*c^(1/2))/((-a)^(1/2)-x*c^(1/2)))*
((-a)^(1/2)-x*c^(1/2))*(-(e*(-a)^(1/2)+d*c^(1/2))*((-a)^(1/2)+x*c^(1/2))/(-e*(-a)^(1/2)+d*c^(1/2))/((-a)^(1/2)
-x*c^(1/2)))^(1/4)/(c*x^2+a)^(1/4)/(e*(-a)^(1/2)+d*c^(1/2))/(e*x+d)^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {727} \[ -\frac {2 \left (\sqrt {-a}-\sqrt {c} x\right ) \sqrt [4]{-\frac {\left (\sqrt {-a}+\sqrt {c} x\right ) \left (\sqrt {-a} e+\sqrt {c} d\right )}{\left (\sqrt {-a}-\sqrt {c} x\right ) \left (\sqrt {c} d-\sqrt {-a} e\right )}} \, _2F_1\left (-\frac {1}{2},\frac {1}{4};\frac {1}{2};\frac {2 \sqrt {-a} \sqrt {c} (d+e x)}{\left (\sqrt {c} d-\sqrt {-a} e\right ) \left (\sqrt {-a}-\sqrt {c} x\right )}\right )}{\sqrt [4]{a+c x^2} \sqrt {d+e x} \left (\sqrt {-a} e+\sqrt {c} d\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)^(3/2)*(a + c*x^2)^(1/4)),x]

[Out]

(-2*(Sqrt[-a] - Sqrt[c]*x)*(-(((Sqrt[c]*d + Sqrt[-a]*e)*(Sqrt[-a] + Sqrt[c]*x))/((Sqrt[c]*d - Sqrt[-a]*e)*(Sqr
t[-a] - Sqrt[c]*x))))^(1/4)*Hypergeometric2F1[-1/2, 1/4, 1/2, (2*Sqrt[-a]*Sqrt[c]*(d + e*x))/((Sqrt[c]*d - Sqr
t[-a]*e)*(Sqrt[-a] - Sqrt[c]*x))])/((Sqrt[c]*d + Sqrt[-a]*e)*Sqrt[d + e*x]*(a + c*x^2)^(1/4))

Rule 727

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((Rt[-(a*c), 2] - c*x)*(d + e*x)^(m
+ 1)*(a + c*x^2)^p*Hypergeometric2F1[m + 1, -p, m + 2, (2*c*Rt[-(a*c), 2]*(d + e*x))/((c*d - e*Rt[-(a*c), 2])
*(Rt[-(a*c), 2] - c*x))])/((m + 1)*(c*d + e*Rt[-(a*c), 2])*(((c*d + e*Rt[-(a*c), 2])*(Rt[-(a*c), 2] + c*x))/((
c*d - e*Rt[-(a*c), 2])*(-Rt[-(a*c), 2] + c*x)))^p), x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2, 0
] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.51, size = 108, normalized size = 0.54 \[ \frac {\left (a+c x^2\right )^{3/4} (c d x-a e) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};-\frac {(a e-c d x)^2}{a c (d+e x)^2}\right )}{a c (d+e x)^{5/2} \left (\frac {\left (a+c x^2\right ) \left (a e^2+c d^2\right )}{a c (d+e x)^2}\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)^(3/2)*(a + c*x^2)^(1/4)),x]

[Out]

((-(a*e) + c*d*x)*(a + c*x^2)^(3/4)*Hypergeometric2F1[1/4, 1/2, 3/2, -((a*e - c*d*x)^2/(a*c*(d + e*x)^2))])/(a
*c*(d + e*x)^(5/2)*(((c*d^2 + a*e^2)*(a + c*x^2))/(a*c*(d + e*x)^2))^(3/4))

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fricas [F]  time = 1.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c x^{2} + a\right )}^{\frac {3}{4}} \sqrt {e x + d}}{c e^{2} x^{4} + 2 \, c d e x^{3} + 2 \, a d e x + a d^{2} + {\left (c d^{2} + a e^{2}\right )} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c*x^2 + a)^(3/4)*sqrt(e*x + d)/(c*e^2*x^4 + 2*c*d*e*x^3 + 2*a*d*e*x + a*d^2 + (c*d^2 + a*e^2)*x^2),
x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (c x^{2} + a\right )}^{\frac {1}{4}} {\left (e x + d\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((c*x^2 + a)^(1/4)*(e*x + d)^(3/2)), x)

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maple [F]  time = 0.90, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (e x +d \right )^{\frac {3}{2}} \left (c \,x^{2}+a \right )^{\frac {1}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^(3/2)/(c*x^2+a)^(1/4),x)

[Out]

int(1/(e*x+d)^(3/2)/(c*x^2+a)^(1/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (c x^{2} + a\right )}^{\frac {1}{4}} {\left (e x + d\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((c*x^2 + a)^(1/4)*(e*x + d)^(3/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (c\,x^2+a\right )}^{1/4}\,{\left (d+e\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + c*x^2)^(1/4)*(d + e*x)^(3/2)),x)

[Out]

int(1/((a + c*x^2)^(1/4)*(d + e*x)^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt [4]{a + c x^{2}} \left (d + e x\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)**(3/2)/(c*x**2+a)**(1/4),x)

[Out]

Integral(1/((a + c*x**2)**(1/4)*(d + e*x)**(3/2)), x)

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