3.295 \(\int \frac {x^{13/2}}{\sqrt {a x^2+b x^5}} \, dx\)

Optimal. Leaf size=265 \[ \frac {7 a^{5/3} x^{3/2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} F\left (\cos ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}{\left (1+\sqrt {3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{40 \sqrt [4]{3} b^2 \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a x^2+b x^5}}-\frac {7 a \sqrt {a x^2+b x^5}}{20 b^2 \sqrt {x}}+\frac {x^{5/2} \sqrt {a x^2+b x^5}}{5 b} \]

[Out]

1/5*x^(5/2)*(b*x^5+a*x^2)^(1/2)/b-7/20*a*(b*x^5+a*x^2)^(1/2)/b^2/x^(1/2)+7/120*a^(5/3)*x^(3/2)*(a^(1/3)+b^(1/3
)*x)*((a^(1/3)+b^(1/3)*x*(1-3^(1/2)))^2/(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))^2)^(1/2)/(a^(1/3)+b^(1/3)*x*(1-3^(1/2)
))*(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))*EllipticF((1-(a^(1/3)+b^(1/3)*x*(1-3^(1/2)))^2/(a^(1/3)+b^(1/3)*x*(1+3^(1/2
)))^2)^(1/2),1/4*6^(1/2)+1/4*2^(1/2))*((a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))
^2)^(1/2)*3^(3/4)/b^2/(b*x^5+a*x^2)^(1/2)/(b^(1/3)*x*(a^(1/3)+b^(1/3)*x)/(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))^2)^(1
/2)

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Rubi [A]  time = 0.28, antiderivative size = 265, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, 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 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]

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 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]

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*}

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Mathematica [C]  time = 0.05, 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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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x^{5} + a x^{2}} x^{\frac {9}{2}}}{b x^{3} + a}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{\frac {13}{2}}}{\sqrt {b x^{5} + a x^{2}}}\,{d x} \]

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)

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maple [C]  time = 0.85, size = 2017, normalized size = 7.61 \[ \text {result too large to display} \]

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/(-a*b^2)^(1/3)*(14*I*(-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(
-a*b^2)^(1/3)))^(1/2)*((2*b*x+I*3^(1/2)*(-a*b^2)^(1/3)+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1
/2)*((-2*b*x+I*3^(1/2)*(-a*b^2)^(1/3)-(-a*b^2)^(1/3))/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(
I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1
/2)-3))^(1/2))*3^(1/2)*x^2*a^2*b^2-28*I*(-a*b^2)^(1/3)*(-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3))
)^(1/2)*((2*b*x+I*3^(1/2)*(-a*b^2)^(1/3)+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((-2*b*x+I
*3^(1/2)*(-a*b^2)^(1/3)-(-a*b^2)^(1/3))/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x
*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))
*3^(1/2)*x*a^2*b+14*I*(-a*b^2)^(2/3)*(-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((2*b*x+I*
3^(1/2)*(-a*b^2)^(1/3)+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((-2*b*x+I*3^(1/2)*(-a*b^2)^
(1/3)-(-a*b^2)^(1/3))/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(
-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*3^(1/2)*a^2-4*I*(
-a*b^2)^(1/3)*((b*x^3+a)*x)^(1/2)*(1/b^2*x*(-b*x+(-a*b^2)^(1/3))*(2*b*x+I*3^(1/2)*(-a*b^2)^(1/3)+(-a*b^2)^(1/3
))*(-2*b*x+I*3^(1/2)*(-a*b^2)^(1/3)-(-a*b^2)^(1/3)))^(1/2)*3^(1/2)*x^3*b^2-14*(-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1
)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((2*b*x+I*3^(1/2)*(-a*b^2)^(1/3)+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(
1/3)))^(1/2)*((-2*b*x+I*3^(1/2)*(-a*b^2)^(1/3)-(-a*b^2)^(1/3))/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*Elli
pticF((-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2)
)/(I*3^(1/2)-3))^(1/2))*x^2*a^2*b^2+28*(-a*b^2)^(1/3)*(-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))
^(1/2)*((2*b*x+I*3^(1/2)*(-a*b^2)^(1/3)+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((-2*b*x+I*
3^(1/2)*(-a*b^2)^(1/3)-(-a*b^2)^(1/3))/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*
b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*
x*a^2*b-14*(-a*b^2)^(2/3)*(-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((2*b*x+I*3^(1/2)*(-a
*b^2)^(1/3)+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((-2*b*x+I*3^(1/2)*(-a*b^2)^(1/3)-(-a*b
^2)^(1/3))/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^
2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*a^2+12*x^3*((b*x^3+a)*x)^(1/
2)*b^2*(-a*b^2)^(1/3)*(1/b^2*x*(-b*x+(-a*b^2)^(1/3))*(2*b*x+I*3^(1/2)*(-a*b^2)^(1/3)+(-a*b^2)^(1/3))*(-2*b*x+I
*3^(1/2)*(-a*b^2)^(1/3)-(-a*b^2)^(1/3)))^(1/2)+7*I*(-a*b^2)^(1/3)*((b*x^3+a)*x)^(1/2)*(1/b^2*x*(-b*x+(-a*b^2)^
(1/3))*(2*b*x+I*3^(1/2)*(-a*b^2)^(1/3)+(-a*b^2)^(1/3))*(-2*b*x+I*3^(1/2)*(-a*b^2)^(1/3)-(-a*b^2)^(1/3)))^(1/2)
*3^(1/2)*a*b-21*a*((b*x^3+a)*x)^(1/2)*b*(-a*b^2)^(1/3)*(1/b^2*x*(-b*x+(-a*b^2)^(1/3))*(2*b*x+I*3^(1/2)*(-a*b^2
)^(1/3)+(-a*b^2)^(1/3))*(-2*b*x+I*3^(1/2)*(-a*b^2)^(1/3)-(-a*b^2)^(1/3)))^(1/2))/((b*x^3+a)*x)^(1/2)/(I*3^(1/2
)-3)/(1/b^2*x*(-b*x+(-a*b^2)^(1/3))*(2*b*x+I*3^(1/2)*(-a*b^2)^(1/3)+(-a*b^2)^(1/3))*(-2*b*x+I*3^(1/2)*(-a*b^2)
^(1/3)-(-a*b^2)^(1/3)))^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{\frac {13}{2}}}{\sqrt {b x^{5} + a x^{2}}}\,{d x} \]

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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^{13/2}}{\sqrt {b\,x^5+a\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(13/2)/(a*x^2 + b*x^5)^(1/2),x)

[Out]

int(x^(13/2)/(a*x^2 + b*x^5)^(1/2), x)

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sympy [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(x**(13/2)/(b*x**5+a*x**2)**(1/2),x)

[Out]

Timed out

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