3.1.51 \(\int \frac {x^{3/2}}{(a x+b x^3)^{9/2}} \, dx\)

Optimal. Leaf size=159 \[ \frac {9 b \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {a x+b x^3}}\right )}{2 a^{11/2}}-\frac {9 \sqrt {a x+b x^3}}{2 a^5 x^{5/2}}+\frac {3}{a^4 x^{3/2} \sqrt {a x+b x^3}}+\frac {3}{5 a^3 \sqrt {x} \left (a x+b x^3\right )^{3/2}}+\frac {9 \sqrt {x}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac {x^{3/2}}{7 a \left (a x+b x^3\right )^{7/2}} \]

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Rubi [A]  time = 0.24, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2023, 2025, 2029, 206} \begin {gather*} \frac {9 \sqrt {x}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac {3}{5 a^3 \sqrt {x} \left (a x+b x^3\right )^{3/2}}+\frac {3}{a^4 x^{3/2} \sqrt {a x+b x^3}}-\frac {9 \sqrt {a x+b x^3}}{2 a^5 x^{5/2}}+\frac {9 b \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {a x+b x^3}}\right )}{2 a^{11/2}}+\frac {x^{3/2}}{7 a \left (a x+b x^3\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(3/2)/(a*x + b*x^3)^(9/2),x]

[Out]

x^(3/2)/(7*a*(a*x + b*x^3)^(7/2)) + (9*Sqrt[x])/(35*a^2*(a*x + b*x^3)^(5/2)) + 3/(5*a^3*Sqrt[x]*(a*x + b*x^3)^
(3/2)) + 3/(a^4*x^(3/2)*Sqrt[a*x + b*x^3]) - (9*Sqrt[a*x + b*x^3])/(2*a^5*x^(5/2)) + (9*b*ArcTanh[(Sqrt[a]*Sqr
t[x])/Sqrt[a*x + b*x^3]])/(2*a^(11/2))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2023

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(c^(j - 1)*(c*x)^(m - j
+ 1)*(a*x^j + b*x^n)^(p + 1))/(a*(n - j)*(p + 1)), x] + Dist[(c^j*(m + n*p + n - j + 1))/(a*(n - j)*(p + 1)),
Int[(c*x)^(m - j)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] &
& (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[p, -1]

Rule 2025

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*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]) && LtQ[m + j*p + 1, 0]

Rule 2029

Int[(x_)^(m_.)/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[-2/(n - j), Subst[Int[1/(1 - a*x^2
), x], x, x^(j/2)/Sqrt[a*x^j + b*x^n]], x] /; FreeQ[{a, b, j, n}, x] && EqQ[m, j/2 - 1] && NeQ[n, j]

Rubi steps

\begin {align*} \int \frac {x^{3/2}}{\left (a x+b x^3\right )^{9/2}} \, dx &=\frac {x^{3/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac {9 \int \frac {\sqrt {x}}{\left (a x+b x^3\right )^{7/2}} \, dx}{7 a}\\ &=\frac {x^{3/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac {9 \sqrt {x}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac {9 \int \frac {1}{\sqrt {x} \left (a x+b x^3\right )^{5/2}} \, dx}{5 a^2}\\ &=\frac {x^{3/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac {9 \sqrt {x}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac {3}{5 a^3 \sqrt {x} \left (a x+b x^3\right )^{3/2}}+\frac {3 \int \frac {1}{x^{3/2} \left (a x+b x^3\right )^{3/2}} \, dx}{a^3}\\ &=\frac {x^{3/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac {9 \sqrt {x}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac {3}{5 a^3 \sqrt {x} \left (a x+b x^3\right )^{3/2}}+\frac {3}{a^4 x^{3/2} \sqrt {a x+b x^3}}+\frac {9 \int \frac {1}{x^{5/2} \sqrt {a x+b x^3}} \, dx}{a^4}\\ &=\frac {x^{3/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac {9 \sqrt {x}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac {3}{5 a^3 \sqrt {x} \left (a x+b x^3\right )^{3/2}}+\frac {3}{a^4 x^{3/2} \sqrt {a x+b x^3}}-\frac {9 \sqrt {a x+b x^3}}{2 a^5 x^{5/2}}-\frac {(9 b) \int \frac {1}{\sqrt {x} \sqrt {a x+b x^3}} \, dx}{2 a^5}\\ &=\frac {x^{3/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac {9 \sqrt {x}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac {3}{5 a^3 \sqrt {x} \left (a x+b x^3\right )^{3/2}}+\frac {3}{a^4 x^{3/2} \sqrt {a x+b x^3}}-\frac {9 \sqrt {a x+b x^3}}{2 a^5 x^{5/2}}+\frac {(9 b) \operatorname {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a x+b x^3}}\right )}{2 a^5}\\ &=\frac {x^{3/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac {9 \sqrt {x}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac {3}{5 a^3 \sqrt {x} \left (a x+b x^3\right )^{3/2}}+\frac {3}{a^4 x^{3/2} \sqrt {a x+b x^3}}-\frac {9 \sqrt {a x+b x^3}}{2 a^5 x^{5/2}}+\frac {9 b \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {a x+b x^3}}\right )}{2 a^{11/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 44, normalized size = 0.28 \begin {gather*} -\frac {b x^{7/2} \, _2F_1\left (-\frac {7}{2},2;-\frac {5}{2};\frac {b x^2}{a}+1\right )}{7 a^2 \left (x \left (a+b x^2\right )\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)/(a*x + b*x^3)^(9/2),x]

[Out]

-1/7*(b*x^(7/2)*Hypergeometric2F1[-7/2, 2, -5/2, 1 + (b*x^2)/a])/(a^2*(x*(a + b*x^2))^(7/2))

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IntegrateAlgebraic [A]  time = 1.95, size = 113, normalized size = 0.71 \begin {gather*} \frac {9 b \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {a x+b x^3}}\right )}{2 a^{11/2}}+\frac {\sqrt {a x+b x^3} \left (-35 a^4-528 a^3 b x^2-1218 a^2 b^2 x^4-1050 a b^3 x^6-315 b^4 x^8\right )}{70 a^5 x^{5/2} \left (a+b x^2\right )^4} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(3/2)/(a*x + b*x^3)^(9/2),x]

[Out]

(Sqrt[a*x + b*x^3]*(-35*a^4 - 528*a^3*b*x^2 - 1218*a^2*b^2*x^4 - 1050*a*b^3*x^6 - 315*b^4*x^8))/(70*a^5*x^(5/2
)*(a + b*x^2)^4) + (9*b*ArcTanh[(Sqrt[a]*Sqrt[x])/Sqrt[a*x + b*x^3]])/(2*a^(11/2))

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fricas [A]  time = 0.43, size = 396, normalized size = 2.49 \begin {gather*} \left [\frac {315 \, {\left (b^{5} x^{11} + 4 \, a b^{4} x^{9} + 6 \, a^{2} b^{3} x^{7} + 4 \, a^{3} b^{2} x^{5} + a^{4} b x^{3}\right )} \sqrt {a} \log \left (\frac {b x^{3} + 2 \, a x + 2 \, \sqrt {b x^{3} + a x} \sqrt {a} \sqrt {x}}{x^{3}}\right ) - 2 \, {\left (315 \, a b^{4} x^{8} + 1050 \, a^{2} b^{3} x^{6} + 1218 \, a^{3} b^{2} x^{4} + 528 \, a^{4} b x^{2} + 35 \, a^{5}\right )} \sqrt {b x^{3} + a x} \sqrt {x}}{140 \, {\left (a^{6} b^{4} x^{11} + 4 \, a^{7} b^{3} x^{9} + 6 \, a^{8} b^{2} x^{7} + 4 \, a^{9} b x^{5} + a^{10} x^{3}\right )}}, -\frac {315 \, {\left (b^{5} x^{11} + 4 \, a b^{4} x^{9} + 6 \, a^{2} b^{3} x^{7} + 4 \, a^{3} b^{2} x^{5} + a^{4} b x^{3}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{3} + a x} \sqrt {-a}}{a \sqrt {x}}\right ) + {\left (315 \, a b^{4} x^{8} + 1050 \, a^{2} b^{3} x^{6} + 1218 \, a^{3} b^{2} x^{4} + 528 \, a^{4} b x^{2} + 35 \, a^{5}\right )} \sqrt {b x^{3} + a x} \sqrt {x}}{70 \, {\left (a^{6} b^{4} x^{11} + 4 \, a^{7} b^{3} x^{9} + 6 \, a^{8} b^{2} x^{7} + 4 \, a^{9} b x^{5} + a^{10} x^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)/(b*x^3+a*x)^(9/2),x, algorithm="fricas")

[Out]

[1/140*(315*(b^5*x^11 + 4*a*b^4*x^9 + 6*a^2*b^3*x^7 + 4*a^3*b^2*x^5 + a^4*b*x^3)*sqrt(a)*log((b*x^3 + 2*a*x +
2*sqrt(b*x^3 + a*x)*sqrt(a)*sqrt(x))/x^3) - 2*(315*a*b^4*x^8 + 1050*a^2*b^3*x^6 + 1218*a^3*b^2*x^4 + 528*a^4*b
*x^2 + 35*a^5)*sqrt(b*x^3 + a*x)*sqrt(x))/(a^6*b^4*x^11 + 4*a^7*b^3*x^9 + 6*a^8*b^2*x^7 + 4*a^9*b*x^5 + a^10*x
^3), -1/70*(315*(b^5*x^11 + 4*a*b^4*x^9 + 6*a^2*b^3*x^7 + 4*a^3*b^2*x^5 + a^4*b*x^3)*sqrt(-a)*arctan(sqrt(b*x^
3 + a*x)*sqrt(-a)/(a*sqrt(x))) + (315*a*b^4*x^8 + 1050*a^2*b^3*x^6 + 1218*a^3*b^2*x^4 + 528*a^4*b*x^2 + 35*a^5
)*sqrt(b*x^3 + a*x)*sqrt(x))/(a^6*b^4*x^11 + 4*a^7*b^3*x^9 + 6*a^8*b^2*x^7 + 4*a^9*b*x^5 + a^10*x^3)]

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giac [A]  time = 0.25, size = 104, normalized size = 0.65 \begin {gather*} -\frac {9 \, b \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{2 \, \sqrt {-a} a^{5}} - \frac {\sqrt {b x^{2} + a}}{2 \, a^{5} x^{2}} - \frac {140 \, {\left (b x^{2} + a\right )}^{3} b + 35 \, {\left (b x^{2} + a\right )}^{2} a b + 14 \, {\left (b x^{2} + a\right )} a^{2} b + 5 \, a^{3} b}{35 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} a^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)/(b*x^3+a*x)^(9/2),x, algorithm="giac")

[Out]

-9/2*b*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^5) - 1/2*sqrt(b*x^2 + a)/(a^5*x^2) - 1/35*(140*(b*x^2 + a)
^3*b + 35*(b*x^2 + a)^2*a*b + 14*(b*x^2 + a)*a^2*b + 5*a^3*b)/((b*x^2 + a)^(7/2)*a^5)

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maple [A]  time = 0.06, size = 234, normalized size = 1.47 \begin {gather*} \frac {\sqrt {\left (b \,x^{2}+a \right ) x}\, \left (315 \sqrt {b \,x^{2}+a}\, b^{4} x^{8} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )-315 \sqrt {a}\, b^{4} x^{8}+945 \sqrt {b \,x^{2}+a}\, a \,b^{3} x^{6} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )-1050 a^{\frac {3}{2}} b^{3} x^{6}+945 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} x^{4} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )-1218 a^{\frac {5}{2}} b^{2} x^{4}+315 \sqrt {b \,x^{2}+a}\, a^{3} b \,x^{2} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )-528 a^{\frac {7}{2}} b \,x^{2}-35 a^{\frac {9}{2}}\right )}{70 \left (b \,x^{2}+a \right )^{4} a^{\frac {11}{2}} x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)/(b*x^3+a*x)^(9/2),x)

[Out]

1/70*((b*x^2+a)*x)^(1/2)/a^(11/2)*(315*ln(2*(a+(b*x^2+a)^(1/2)*a^(1/2))/x)*x^8*b^4*(b*x^2+a)^(1/2)-315*a^(1/2)
*x^8*b^4+945*ln(2*(a+(b*x^2+a)^(1/2)*a^(1/2))/x)*x^6*a*b^3*(b*x^2+a)^(1/2)-1050*a^(3/2)*x^6*b^3+945*ln(2*(a+(b
*x^2+a)^(1/2)*a^(1/2))/x)*x^4*a^2*b^2*(b*x^2+a)^(1/2)-1218*a^(5/2)*x^4*b^2+315*ln(2*(a+(b*x^2+a)^(1/2)*a^(1/2)
)/x)*x^2*a^3*b*(b*x^2+a)^(1/2)-528*a^(7/2)*x^2*b-35*a^(9/2))/x^(5/2)/(b*x^2+a)^4

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{\frac {3}{2}}}{{\left (b x^{3} + a x\right )}^{\frac {9}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)/(b*x^3+a*x)^(9/2),x, algorithm="maxima")

[Out]

integrate(x^(3/2)/(b*x^3 + a*x)^(9/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{3/2}}{{\left (b\,x^3+a\,x\right )}^{9/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)/(a*x + b*x^3)^(9/2),x)

[Out]

int(x^(3/2)/(a*x + b*x^3)^(9/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{\frac {3}{2}}}{\left (x \left (a + b x^{2}\right )\right )^{\frac {9}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(3/2)/(b*x**3+a*x)**(9/2),x)

[Out]

Integral(x**(3/2)/(x*(a + b*x**2))**(9/2), x)

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