3.295 \(\int \frac{x^{13/2}}{\sqrt{a x^2+b x^5}} \, dx\)
Optimal. Leaf size=265 \[ \frac{7 a^{5/3} x^{3/2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x\right )^2}} \text{EllipticF}\left (\cos ^{-1}\left (\frac{\sqrt [3]{a}+\left (1-\sqrt{3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x}\right ),\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{40 \sqrt [4]{3} b^2 \sqrt{\frac{\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt{a x^2+b x^5}}-\frac{7 a \sqrt{a x^2+b x^5}}{20 b^2 \sqrt{x}}+\frac{x^{5/2} \sqrt{a x^2+b x^5}}{5 b} \]
[Out]
(-7*a*Sqrt[a*x^2 + b*x^5])/(20*b^2*Sqrt[x]) + (x^(5/2)*Sqrt[a*x^2 + b*x^5])/(5*b) + (7*a^(5/3)*x^(3/2)*(a^(1/3
) + b^(1/3)*x)*Sqrt[(a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2)/(a^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)^2]*Ellipti
cF[ArcCos[(a^(1/3) + (1 - Sqrt[3])*b^(1/3)*x)/(a^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)], (2 + Sqrt[3])/4])/(40*3^(1
/4)*b^2*Sqrt[(b^(1/3)*x*(a^(1/3) + b^(1/3)*x))/(a^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)^2]*Sqrt[a*x^2 + b*x^5])
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Rubi [A] time = 0.281867, antiderivative size = 265, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used =
{2024, 2032, 329, 225} \[ \frac{7 a^{5/3} x^{3/2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x\right )^2}} F\left (\cos ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}{\left (1+\sqrt{3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{40 \sqrt [4]{3} b^2 \sqrt{\frac{\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt{a x^2+b x^5}}-\frac{7 a \sqrt{a x^2+b x^5}}{20 b^2 \sqrt{x}}+\frac{x^{5/2} \sqrt{a x^2+b x^5}}{5 b} \]
Antiderivative was successfully verified.
[In]
Int[x^(13/2)/Sqrt[a*x^2 + b*x^5],x]
[Out]
(-7*a*Sqrt[a*x^2 + b*x^5])/(20*b^2*Sqrt[x]) + (x^(5/2)*Sqrt[a*x^2 + b*x^5])/(5*b) + (7*a^(5/3)*x^(3/2)*(a^(1/3
) + b^(1/3)*x)*Sqrt[(a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2)/(a^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)^2]*Ellipti
cF[ArcCos[(a^(1/3) + (1 - Sqrt[3])*b^(1/3)*x)/(a^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)], (2 + Sqrt[3])/4])/(40*3^(1
/4)*b^2*Sqrt[(b^(1/3)*x*(a^(1/3) + b^(1/3)*x))/(a^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)^2]*Sqrt[a*x^2 + b*x^5])
Rule 2024
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n +
1)*(a*x^j + b*x^n)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^(n - j)*(m + j*p - n + j + 1))/(b*(m + n*p + 1)
), Int[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] && !IntegerQ[p] && LtQ[0, j
, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[m + j*p + 1 - n + j, 0] && NeQ[m + n*p + 1, 0]
Rule 2032
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[m]*(a*x^j + b*x^n)^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]
Rule 329
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]
Rule 225
Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(x*(s
+ r*x^2)*Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/(s + (1 + Sqrt[3])*r*x^2)^2]*EllipticF[ArcCos[(s + (1 - Sqrt[3])*r*x^2
)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4])/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[(r*x^2*(s + r*x^2))/(s + (1
+ Sqrt[3])*r*x^2)^2]), x]] /; FreeQ[{a, b}, x]
Rubi steps
\begin{align*} \int \frac{x^{13/2}}{\sqrt{a x^2+b x^5}} \, dx &=\frac{x^{5/2} \sqrt{a x^2+b x^5}}{5 b}-\frac{(7 a) \int \frac{x^{7/2}}{\sqrt{a x^2+b x^5}} \, dx}{10 b}\\ &=-\frac{7 a \sqrt{a x^2+b x^5}}{20 b^2 \sqrt{x}}+\frac{x^{5/2} \sqrt{a x^2+b x^5}}{5 b}+\frac{\left (7 a^2\right ) \int \frac{\sqrt{x}}{\sqrt{a x^2+b x^5}} \, dx}{40 b^2}\\ &=-\frac{7 a \sqrt{a x^2+b x^5}}{20 b^2 \sqrt{x}}+\frac{x^{5/2} \sqrt{a x^2+b x^5}}{5 b}+\frac{\left (7 a^2 x \sqrt{a+b x^3}\right ) \int \frac{1}{\sqrt{x} \sqrt{a+b x^3}} \, dx}{40 b^2 \sqrt{a x^2+b x^5}}\\ &=-\frac{7 a \sqrt{a x^2+b x^5}}{20 b^2 \sqrt{x}}+\frac{x^{5/2} \sqrt{a x^2+b x^5}}{5 b}+\frac{\left (7 a^2 x \sqrt{a+b x^3}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^6}} \, dx,x,\sqrt{x}\right )}{20 b^2 \sqrt{a x^2+b x^5}}\\ &=-\frac{7 a \sqrt{a x^2+b x^5}}{20 b^2 \sqrt{x}}+\frac{x^{5/2} \sqrt{a x^2+b x^5}}{5 b}+\frac{7 a^{5/3} x^{3/2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x\right )^2}} F\left (\cos ^{-1}\left (\frac{\sqrt [3]{a}+\left (1-\sqrt{3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{40 \sqrt [4]{3} b^2 \sqrt{\frac{\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt{a x^2+b x^5}}\\ \end{align*}
Mathematica [C] time = 0.0323308, size = 86, normalized size = 0.32 \[ \frac{x^{3/2} \left (7 a^2 \sqrt{\frac{b x^3}{a}+1} \, _2F_1\left (\frac{1}{6},\frac{1}{2};\frac{7}{6};-\frac{b x^3}{a}\right )-7 a^2-3 a b x^3+4 b^2 x^6\right )}{20 b^2 \sqrt{x^2 \left (a+b x^3\right )}} \]
Antiderivative was successfully verified.
[In]
Integrate[x^(13/2)/Sqrt[a*x^2 + b*x^5],x]
[Out]
(x^(3/2)*(-7*a^2 - 3*a*b*x^3 + 4*b^2*x^6 + 7*a^2*Sqrt[1 + (b*x^3)/a]*Hypergeometric2F1[1/6, 1/2, 7/6, -((b*x^3
)/a)]))/(20*b^2*Sqrt[x^2*(a + b*x^3)])
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Maple [C] time = 0.533, size = 2017, normalized size = 7.6 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^(13/2)/(b*x^5+a*x^2)^(1/2),x)
[Out]
-1/20/(b*x^5+a*x^2)^(1/2)*x^(3/2)*(b*x^3+a)/b^3/(-b^2*a)^(1/3)*(14*I*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+
(-b^2*a)^(1/3)))^(1/2)*((I*3^(1/2)*(-b^2*a)^(1/3)+2*b*x+(-b^2*a)^(1/3))/(1+I*3^(1/2))/(-b*x+(-b^2*a)^(1/3)))^(
1/2)*((I*3^(1/2)*(-b^2*a)^(1/3)-2*b*x-(-b^2*a)^(1/3))/(-1+I*3^(1/2))/(-b*x+(-b^2*a)^(1/3)))^(1/2)*EllipticF((-
(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-b^2*a)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3
^(1/2)-3))^(1/2))*3^(1/2)*x^2*a^2*b^2-28*I*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-b^2*a)^(1/3)))^(1/2)*((I
*3^(1/2)*(-b^2*a)^(1/3)+2*b*x+(-b^2*a)^(1/3))/(1+I*3^(1/2))/(-b*x+(-b^2*a)^(1/3)))^(1/2)*((I*3^(1/2)*(-b^2*a)^
(1/3)-2*b*x-(-b^2*a)^(1/3))/(-1+I*3^(1/2))/(-b*x+(-b^2*a)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(-1+I*3^
(1/2))/(-b*x+(-b^2*a)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-b^2*a)
^(1/3)*3^(1/2)*x*a^2*b+14*I*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-b^2*a)^(1/3)))^(1/2)*((I*3^(1/2)*(-b^2*
a)^(1/3)+2*b*x+(-b^2*a)^(1/3))/(1+I*3^(1/2))/(-b*x+(-b^2*a)^(1/3)))^(1/2)*((I*3^(1/2)*(-b^2*a)^(1/3)-2*b*x-(-b
^2*a)^(1/3))/(-1+I*3^(1/2))/(-b*x+(-b^2*a)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-
b^2*a)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-b^2*a)^(2/3)*3^(1/2)*
a^2-14*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-b^2*a)^(1/3)))^(1/2)*((I*3^(1/2)*(-b^2*a)^(1/3)+2*b*x+(-b^2*
a)^(1/3))/(1+I*3^(1/2))/(-b*x+(-b^2*a)^(1/3)))^(1/2)*((I*3^(1/2)*(-b^2*a)^(1/3)-2*b*x-(-b^2*a)^(1/3))/(-1+I*3^
(1/2))/(-b*x+(-b^2*a)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-b^2*a)^(1/3)))^(1/2),
((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*x^2*a^2*b^2-4*I*(1/b^2*x*(-b*x+(-b^2*a)^(1/3
))*(I*3^(1/2)*(-b^2*a)^(1/3)+2*b*x+(-b^2*a)^(1/3))*(I*3^(1/2)*(-b^2*a)^(1/3)-2*b*x-(-b^2*a)^(1/3)))^(1/2)*((b*
x^3+a)*x)^(1/2)*(-b^2*a)^(1/3)*3^(1/2)*x^3*b^2+28*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-b^2*a)^(1/3)))^(1
/2)*((I*3^(1/2)*(-b^2*a)^(1/3)+2*b*x+(-b^2*a)^(1/3))/(1+I*3^(1/2))/(-b*x+(-b^2*a)^(1/3)))^(1/2)*((I*3^(1/2)*(-
b^2*a)^(1/3)-2*b*x-(-b^2*a)^(1/3))/(-1+I*3^(1/2))/(-b*x+(-b^2*a)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(
-1+I*3^(1/2))/(-b*x+(-b^2*a)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(
-b^2*a)^(1/3)*x*a^2*b-14*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-b^2*a)^(1/3)))^(1/2)*((I*3^(1/2)*(-b^2*a)^
(1/3)+2*b*x+(-b^2*a)^(1/3))/(1+I*3^(1/2))/(-b*x+(-b^2*a)^(1/3)))^(1/2)*((I*3^(1/2)*(-b^2*a)^(1/3)-2*b*x-(-b^2*
a)^(1/3))/(-1+I*3^(1/2))/(-b*x+(-b^2*a)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-b^2
*a)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-b^2*a)^(2/3)*a^2+12*x^3*
((b*x^3+a)*x)^(1/2)*b^2*(-b^2*a)^(1/3)*(1/b^2*x*(-b*x+(-b^2*a)^(1/3))*(I*3^(1/2)*(-b^2*a)^(1/3)+2*b*x+(-b^2*a)
^(1/3))*(I*3^(1/2)*(-b^2*a)^(1/3)-2*b*x-(-b^2*a)^(1/3)))^(1/2)+7*I*(1/b^2*x*(-b*x+(-b^2*a)^(1/3))*(I*3^(1/2)*(
-b^2*a)^(1/3)+2*b*x+(-b^2*a)^(1/3))*(I*3^(1/2)*(-b^2*a)^(1/3)-2*b*x-(-b^2*a)^(1/3)))^(1/2)*((b*x^3+a)*x)^(1/2)
*(-b^2*a)^(1/3)*3^(1/2)*a*b-21*a*((b*x^3+a)*x)^(1/2)*b*(-b^2*a)^(1/3)*(1/b^2*x*(-b*x+(-b^2*a)^(1/3))*(I*3^(1/2
)*(-b^2*a)^(1/3)+2*b*x+(-b^2*a)^(1/3))*(I*3^(1/2)*(-b^2*a)^(1/3)-2*b*x-(-b^2*a)^(1/3)))^(1/2))/((b*x^3+a)*x)^(
1/2)/(I*3^(1/2)-3)/(1/b^2*x*(-b*x+(-b^2*a)^(1/3))*(I*3^(1/2)*(-b^2*a)^(1/3)+2*b*x+(-b^2*a)^(1/3))*(I*3^(1/2)*(
-b^2*a)^(1/3)-2*b*x-(-b^2*a)^(1/3)))^(1/2)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{\frac{13}{2}}}{\sqrt{b x^{5} + a x^{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^(13/2)/(b*x^5+a*x^2)^(1/2),x, algorithm="maxima")
[Out]
integrate(x^(13/2)/sqrt(b*x^5 + a*x^2), x)
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b x^{5} + a x^{2}} x^{\frac{9}{2}}}{b x^{3} + a}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^(13/2)/(b*x^5+a*x^2)^(1/2),x, algorithm="fricas")
[Out]
integral(sqrt(b*x^5 + a*x^2)*x^(9/2)/(b*x^3 + a), 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(x**(13/2)/(b*x**5+a*x**2)**(1/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{\frac{13}{2}}}{\sqrt{b x^{5} + a x^{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^(13/2)/(b*x^5+a*x^2)^(1/2),x, algorithm="giac")
[Out]
integrate(x^(13/2)/sqrt(b*x^5 + a*x^2), x)