3.1100 \(\int \frac{(e x)^{3/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx\)

Optimal. Leaf size=139 \[ \frac{\sqrt{a} (e x)^{3/2} \left (\frac{a}{b x^2}+1\right )^{3/4} (6 b c-5 a d) F\left (\left .\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{6 b^{3/2} \left (a+b x^2\right )^{3/4}}+\frac{e \sqrt{e x} \sqrt [4]{a+b x^2} (6 b c-5 a d)}{6 b^2}+\frac{d (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 b e} \]

[Out]

((6*b*c - 5*a*d)*e*Sqrt[e*x]*(a + b*x^2)^(1/4))/(6*b^2) + (d*(e*x)^(5/2)*(a + b*
x^2)^(1/4))/(3*b*e) + (Sqrt[a]*(6*b*c - 5*a*d)*(1 + a/(b*x^2))^(3/4)*(e*x)^(3/2)
*EllipticF[ArcCot[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(6*b^(3/2)*(a + b*x^2)^(3/4))

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Rubi [A]  time = 0.319791, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269 \[ \frac{\sqrt{a} (e x)^{3/2} \left (\frac{a}{b x^2}+1\right )^{3/4} (6 b c-5 a d) F\left (\left .\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{6 b^{3/2} \left (a+b x^2\right )^{3/4}}+\frac{e \sqrt{e x} \sqrt [4]{a+b x^2} (6 b c-5 a d)}{6 b^2}+\frac{d (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 b e} \]

Antiderivative was successfully verified.

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

[Out]

((6*b*c - 5*a*d)*e*Sqrt[e*x]*(a + b*x^2)^(1/4))/(6*b^2) + (d*(e*x)^(5/2)*(a + b*
x^2)^(1/4))/(3*b*e) + (Sqrt[a]*(6*b*c - 5*a*d)*(1 + a/(b*x^2))^(3/4)*(e*x)^(3/2)
*EllipticF[ArcCot[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(6*b^(3/2)*(a + b*x^2)^(3/4))

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Rubi in Sympy [A]  time = 33.1938, size = 124, normalized size = 0.89 \[ - \frac{\sqrt{a} \left (e x\right )^{\frac{3}{2}} \left (5 a d - 6 b c\right ) \left (\frac{a}{b x^{2}} + 1\right )^{\frac{3}{4}} F\left (\frac{\operatorname{atan}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{2}\middle | 2\right )}{6 b^{\frac{3}{2}} \left (a + b x^{2}\right )^{\frac{3}{4}}} + \frac{d \left (e x\right )^{\frac{5}{2}} \sqrt [4]{a + b x^{2}}}{3 b e} - \frac{e \sqrt{e x} \sqrt [4]{a + b x^{2}} \left (5 a d - 6 b c\right )}{6 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((e*x)**(3/2)*(d*x**2+c)/(b*x**2+a)**(3/4),x)

[Out]

-sqrt(a)*(e*x)**(3/2)*(5*a*d - 6*b*c)*(a/(b*x**2) + 1)**(3/4)*elliptic_f(atan(sq
rt(a)/(sqrt(b)*x))/2, 2)/(6*b**(3/2)*(a + b*x**2)**(3/4)) + d*(e*x)**(5/2)*(a +
b*x**2)**(1/4)/(3*b*e) - e*sqrt(e*x)*(a + b*x**2)**(1/4)*(5*a*d - 6*b*c)/(6*b**2
)

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Mathematica [C]  time = 0.115883, size = 97, normalized size = 0.7 \[ \frac{e \sqrt{e x} \left (a \left (\frac{b x^2}{a}+1\right )^{3/4} (5 a d-6 b c) \, _2F_1\left (\frac{1}{4},\frac{3}{4};\frac{5}{4};-\frac{b x^2}{a}\right )-\left (a+b x^2\right ) \left (5 a d-2 b \left (3 c+d x^2\right )\right )\right )}{6 b^2 \left (a+b x^2\right )^{3/4}} \]

Antiderivative was successfully verified.

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

[Out]

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

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Maple [F]  time = 0.044, size = 0, normalized size = 0. \[ \int{(d{x}^{2}+c) \left ( ex \right ) ^{{\frac{3}{2}}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{4}}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((e*x)^(3/2)*(d*x^2+c)/(b*x^2+a)^(3/4),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Sympy [A]  time = 59.3142, size = 94, normalized size = 0.68 \[ \frac{c e^{\frac{3}{2}} x^{\frac{5}{2}} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{4}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac{3}{4}} \Gamma \left (\frac{9}{4}\right )} + \frac{d e^{\frac{3}{2}} x^{\frac{9}{2}} \Gamma \left (\frac{9}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{4}, \frac{9}{4} \\ \frac{13}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac{3}{4}} \Gamma \left (\frac{13}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((e*x)**(3/2)*(d*x**2+c)/(b*x**2+a)**(3/4),x)

[Out]

c*e**(3/2)*x**(5/2)*gamma(5/4)*hyper((3/4, 5/4), (9/4,), b*x**2*exp_polar(I*pi)/
a)/(2*a**(3/4)*gamma(9/4)) + d*e**(3/2)*x**(9/2)*gamma(9/4)*hyper((3/4, 9/4), (1
3/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(3/4)*gamma(13/4))

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (d x^{2} + c\right )} \left (e x\right )^{\frac{3}{2}}}{{\left (b x^{2} + a\right )}^{\frac{3}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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