3.1.40 \(\int \frac {x^{25/2}}{(a x+b x^3)^{9/2}} \, dx\)

Optimal. Leaf size=130 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a x+b x^3}}\right )}{b^{9/2}}-\frac {x^{3/2}}{b^4 \sqrt {a x+b x^3}}-\frac {x^{9/2}}{3 b^3 \left (a x+b x^3\right )^{3/2}}-\frac {x^{15/2}}{5 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {x^{21/2}}{7 b \left (a x+b x^3\right )^{7/2}} \]

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Rubi [A]  time = 0.21, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2022, 2029, 206} \begin {gather*} -\frac {x^{15/2}}{5 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {x^{9/2}}{3 b^3 \left (a x+b x^3\right )^{3/2}}-\frac {x^{3/2}}{b^4 \sqrt {a x+b x^3}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a x+b x^3}}\right )}{b^{9/2}}-\frac {x^{21/2}}{7 b \left (a x+b x^3\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(25/2)/(a*x + b*x^3)^(9/2),x]

[Out]

-x^(21/2)/(7*b*(a*x + b*x^3)^(7/2)) - x^(15/2)/(5*b^2*(a*x + b*x^3)^(5/2)) - x^(9/2)/(3*b^3*(a*x + b*x^3)^(3/2
)) - x^(3/2)/(b^4*Sqrt[a*x + b*x^3]) + ArcTanh[(Sqrt[b]*x^(3/2))/Sqrt[a*x + b*x^3]]/b^(9/2)

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 2022

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*(n - j)*(p + 1)), x] - Dist[(c^n*(m + j*p - n + j + 1))/(b*(n - j)*(p + 1)), I
nt[(c*x)^(m - n)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (I
ntegersQ[j, n] || GtQ[c, 0]) && LtQ[p, -1] && GtQ[m + j*p + 1, n - j]

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^{25/2}}{\left (a x+b x^3\right )^{9/2}} \, dx &=-\frac {x^{21/2}}{7 b \left (a x+b x^3\right )^{7/2}}+\frac {\int \frac {x^{19/2}}{\left (a x+b x^3\right )^{7/2}} \, dx}{b}\\ &=-\frac {x^{21/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {x^{15/2}}{5 b^2 \left (a x+b x^3\right )^{5/2}}+\frac {\int \frac {x^{13/2}}{\left (a x+b x^3\right )^{5/2}} \, dx}{b^2}\\ &=-\frac {x^{21/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {x^{15/2}}{5 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {x^{9/2}}{3 b^3 \left (a x+b x^3\right )^{3/2}}+\frac {\int \frac {x^{7/2}}{\left (a x+b x^3\right )^{3/2}} \, dx}{b^3}\\ &=-\frac {x^{21/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {x^{15/2}}{5 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {x^{9/2}}{3 b^3 \left (a x+b x^3\right )^{3/2}}-\frac {x^{3/2}}{b^4 \sqrt {a x+b x^3}}+\frac {\int \frac {\sqrt {x}}{\sqrt {a x+b x^3}} \, dx}{b^4}\\ &=-\frac {x^{21/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {x^{15/2}}{5 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {x^{9/2}}{3 b^3 \left (a x+b x^3\right )^{3/2}}-\frac {x^{3/2}}{b^4 \sqrt {a x+b x^3}}+\frac {\operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x^{3/2}}{\sqrt {a x+b x^3}}\right )}{b^4}\\ &=-\frac {x^{21/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {x^{15/2}}{5 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {x^{9/2}}{3 b^3 \left (a x+b x^3\right )^{3/2}}-\frac {x^{3/2}}{b^4 \sqrt {a x+b x^3}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a x+b x^3}}\right )}{b^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 120, normalized size = 0.92 \begin {gather*} \frac {\sqrt {x} \left (105 \sqrt {a} \left (a+b x^2\right )^3 \sqrt {\frac {b x^2}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )-\sqrt {b} x \left (105 a^3+350 a^2 b x^2+406 a b^2 x^4+176 b^3 x^6\right )\right )}{105 b^{9/2} \left (a+b x^2\right )^3 \sqrt {x \left (a+b x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(25/2)/(a*x + b*x^3)^(9/2),x]

[Out]

(Sqrt[x]*(-(Sqrt[b]*x*(105*a^3 + 350*a^2*b*x^2 + 406*a*b^2*x^4 + 176*b^3*x^6)) + 105*Sqrt[a]*(a + b*x^2)^3*Sqr
t[1 + (b*x^2)/a]*ArcSinh[(Sqrt[b]*x)/Sqrt[a]]))/(105*b^(9/2)*(a + b*x^2)^3*Sqrt[x*(a + b*x^2)])

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IntegrateAlgebraic [A]  time = 0.11, size = 110, normalized size = 0.85 \begin {gather*} \frac {x^{9/2} \left (a+b x^2\right )^{9/2} \left (\frac {-105 a^3 x-350 a^2 b x^3-406 a b^2 x^5-176 b^3 x^7}{105 b^4 \left (a+b x^2\right )^{7/2}}-\frac {\log \left (\sqrt {a+b x^2}-\sqrt {b} x\right )}{b^{9/2}}\right )}{\left (x \left (a+b x^2\right )\right )^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(25/2)/(a*x + b*x^3)^(9/2),x]

[Out]

(x^(9/2)*(a + b*x^2)^(9/2)*((-105*a^3*x - 350*a^2*b*x^3 - 406*a*b^2*x^5 - 176*b^3*x^7)/(105*b^4*(a + b*x^2)^(7
/2)) - Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]]/b^(9/2)))/(x*(a + b*x^2))^(9/2)

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fricas [A]  time = 0.43, size = 348, normalized size = 2.68 \begin {gather*} \left [\frac {105 \, {\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \sqrt {b} \log \left (2 \, b x^{2} + 2 \, \sqrt {b x^{3} + a x} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (176 \, b^{4} x^{6} + 406 \, a b^{3} x^{4} + 350 \, a^{2} b^{2} x^{2} + 105 \, a^{3} b\right )} \sqrt {b x^{3} + a x} \sqrt {x}}{210 \, {\left (b^{9} x^{8} + 4 \, a b^{8} x^{6} + 6 \, a^{2} b^{7} x^{4} + 4 \, a^{3} b^{6} x^{2} + a^{4} b^{5}\right )}}, -\frac {105 \, {\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{3} + a x} \sqrt {-b}}{b x^{\frac {3}{2}}}\right ) + {\left (176 \, b^{4} x^{6} + 406 \, a b^{3} x^{4} + 350 \, a^{2} b^{2} x^{2} + 105 \, a^{3} b\right )} \sqrt {b x^{3} + a x} \sqrt {x}}{105 \, {\left (b^{9} x^{8} + 4 \, a b^{8} x^{6} + 6 \, a^{2} b^{7} x^{4} + 4 \, a^{3} b^{6} x^{2} + a^{4} b^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(25/2)/(b*x^3+a*x)^(9/2),x, algorithm="fricas")

[Out]

[1/210*(105*(b^4*x^8 + 4*a*b^3*x^6 + 6*a^2*b^2*x^4 + 4*a^3*b*x^2 + a^4)*sqrt(b)*log(2*b*x^2 + 2*sqrt(b*x^3 + a
*x)*sqrt(b)*sqrt(x) + a) - 2*(176*b^4*x^6 + 406*a*b^3*x^4 + 350*a^2*b^2*x^2 + 105*a^3*b)*sqrt(b*x^3 + a*x)*sqr
t(x))/(b^9*x^8 + 4*a*b^8*x^6 + 6*a^2*b^7*x^4 + 4*a^3*b^6*x^2 + a^4*b^5), -1/105*(105*(b^4*x^8 + 4*a*b^3*x^6 +
6*a^2*b^2*x^4 + 4*a^3*b*x^2 + a^4)*sqrt(-b)*arctan(sqrt(b*x^3 + a*x)*sqrt(-b)/(b*x^(3/2))) + (176*b^4*x^6 + 40
6*a*b^3*x^4 + 350*a^2*b^2*x^2 + 105*a^3*b)*sqrt(b*x^3 + a*x)*sqrt(x))/(b^9*x^8 + 4*a*b^8*x^6 + 6*a^2*b^7*x^4 +
4*a^3*b^6*x^2 + a^4*b^5)]

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giac [A]  time = 0.30, size = 86, normalized size = 0.66 \begin {gather*} -\frac {{\left (2 \, {\left (x^{2} {\left (\frac {88 \, x^{2}}{b} + \frac {203 \, a}{b^{2}}\right )} + \frac {175 \, a^{2}}{b^{3}}\right )} x^{2} + \frac {105 \, a^{3}}{b^{4}}\right )} x}{105 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}}} - \frac {\log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{b^{\frac {9}{2}}} + \frac {\log \left ({\left | a \right |}\right )}{2 \, b^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(25/2)/(b*x^3+a*x)^(9/2),x, algorithm="giac")

[Out]

-1/105*(2*(x^2*(88*x^2/b + 203*a/b^2) + 175*a^2/b^3)*x^2 + 105*a^3/b^4)*x/(b*x^2 + a)^(7/2) - log(abs(-sqrt(b)
*x + sqrt(b*x^2 + a)))/b^(9/2) + 1/2*log(abs(a))/b^(9/2)

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maple [A]  time = 0.08, size = 198, normalized size = 1.52 \begin {gather*} \frac {\sqrt {\left (b \,x^{2}+a \right ) x}\, \left (-176 b^{\frac {7}{2}} x^{7}+105 \sqrt {b \,x^{2}+a}\, b^{3} x^{6} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )-406 a \,b^{\frac {5}{2}} x^{5}+315 \sqrt {b \,x^{2}+a}\, a \,b^{2} x^{4} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )-350 a^{2} b^{\frac {3}{2}} x^{3}+315 \sqrt {b \,x^{2}+a}\, a^{2} b \,x^{2} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )-105 a^{3} \sqrt {b}\, x +105 \sqrt {b \,x^{2}+a}\, a^{3} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )\right )}{105 \left (b \,x^{2}+a \right )^{4} b^{\frac {9}{2}} \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(25/2)/(b*x^3+a*x)^(9/2),x)

[Out]

1/105*((b*x^2+a)*x)^(1/2)/b^(9/2)*(105*ln(b^(1/2)*x+(b*x^2+a)^(1/2))*x^6*b^3*(b*x^2+a)^(1/2)-176*x^7*b^(7/2)+3
15*ln(b^(1/2)*x+(b*x^2+a)^(1/2))*x^4*a*b^2*(b*x^2+a)^(1/2)-406*b^(5/2)*x^5*a+315*ln(b^(1/2)*x+(b*x^2+a)^(1/2))
*x^2*a^2*b*(b*x^2+a)^(1/2)-350*b^(3/2)*x^3*a^2+105*ln(b^(1/2)*x+(b*x^2+a)^(1/2))*a^3*(b*x^2+a)^(1/2)-105*b^(1/
2)*x*a^3)/x^(1/2)/(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 {25}{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^(25/2)/(b*x^3+a*x)^(9/2),x, algorithm="maxima")

[Out]

integrate(x^(25/2)/(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^{25/2}}{{\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^(25/2)/(a*x + b*x^3)^(9/2),x)

[Out]

int(x^(25/2)/(a*x + b*x^3)^(9/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(25/2)/(b*x**3+a*x)**(9/2),x)

[Out]

Timed out

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