3.1.49 \(\int \frac {x^{7/2}}{(a x+b x^3)^{9/2}} \, dx\)

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

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Rubi [A]  time = 0.20, 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 = {2023, 2029, 206} \begin {gather*} \frac {x^{5/2}}{5 a^2 \left (a x+b x^3\right )^{5/2}}+\frac {x^{3/2}}{3 a^3 \left (a x+b x^3\right )^{3/2}}+\frac {\sqrt {x}}{a^4 \sqrt {a x+b x^3}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {a x+b x^3}}\right )}{a^{9/2}}+\frac {x^{7/2}}{7 a \left (a x+b x^3\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(7/2)/(a*x + b*x^3)^(9/2),x]

[Out]

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

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Mathematica [C]  time = 0.01, size = 43, normalized size = 0.33 \begin {gather*} \frac {x^{7/2} \, _2F_1\left (-\frac {7}{2},1;-\frac {5}{2};\frac {b x^2}{a}+1\right )}{7 a \left (x \left (a+b x^2\right )\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(7/2)/(a*x + b*x^3)^(9/2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 1.04, size = 99, normalized size = 0.76 \begin {gather*} \frac {\sqrt {a x+b x^3} \left (176 a^3+406 a^2 b x^2+350 a b^2 x^4+105 b^3 x^6\right )}{105 a^4 \sqrt {x} \left (a+b x^2\right )^4}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {a x+b x^3}}\right )}{a^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(7/2)/(a*x + b*x^3)^(9/2),x]

[Out]

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

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fricas [A]  time = 0.44, size = 360, normalized size = 2.77 \begin {gather*} \left [\frac {105 \, {\left (b^{4} x^{9} + 4 \, a b^{3} x^{7} + 6 \, a^{2} b^{2} x^{5} + 4 \, a^{3} b x^{3} + a^{4} x\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 (105 \, a b^{3} x^{6} + 350 \, a^{2} b^{2} x^{4} + 406 \, a^{3} b x^{2} + 176 \, a^{4}\right )} \sqrt {b x^{3} + a x} \sqrt {x}}{210 \, {\left (a^{5} b^{4} x^{9} + 4 \, a^{6} b^{3} x^{7} + 6 \, a^{7} b^{2} x^{5} + 4 \, a^{8} b x^{3} + a^{9} x\right )}}, \frac {105 \, {\left (b^{4} x^{9} + 4 \, a b^{3} x^{7} + 6 \, a^{2} b^{2} x^{5} + 4 \, a^{3} b x^{3} + a^{4} x\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{3} + a x} \sqrt {-a}}{a \sqrt {x}}\right ) + {\left (105 \, a b^{3} x^{6} + 350 \, a^{2} b^{2} x^{4} + 406 \, a^{3} b x^{2} + 176 \, a^{4}\right )} \sqrt {b x^{3} + a x} \sqrt {x}}{105 \, {\left (a^{5} b^{4} x^{9} + 4 \, a^{6} b^{3} x^{7} + 6 \, a^{7} b^{2} x^{5} + 4 \, a^{8} b x^{3} + a^{9} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)/(b*x^3+a*x)^(9/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.27, size = 114, normalized size = 0.88 \begin {gather*} \frac {\arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{4}} - \frac {105 \, \sqrt {a} \arctan \left (\frac {\sqrt {a}}{\sqrt {-a}}\right ) + 176 \, \sqrt {-a}}{105 \, \sqrt {-a} a^{\frac {9}{2}}} + \frac {105 \, {\left (b x^{2} + a\right )}^{3} + 35 \, {\left (b x^{2} + a\right )}^{2} a + 21 \, {\left (b x^{2} + a\right )} a^{2} + 15 \, a^{3}}{105 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)/(b*x^3+a*x)^(9/2),x, algorithm="giac")

[Out]

arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^4) - 1/105*(105*sqrt(a)*arctan(sqrt(a)/sqrt(-a)) + 176*sqrt(-a))/
(sqrt(-a)*a^(9/2)) + 1/105*(105*(b*x^2 + a)^3 + 35*(b*x^2 + a)^2*a + 21*(b*x^2 + a)*a^2 + 15*a^3)/((b*x^2 + a)
^(7/2)*a^4)

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maple [B]  time = 0.05, size = 217, normalized size = 1.67 \begin {gather*} -\frac {\sqrt {\left (b \,x^{2}+a \right ) x}\, \left (105 \sqrt {b \,x^{2}+a}\, b^{3} x^{6} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )-105 \sqrt {a}\, b^{3} x^{6}+315 \sqrt {b \,x^{2}+a}\, a \,b^{2} x^{4} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )-350 a^{\frac {3}{2}} b^{2} x^{4}+315 \sqrt {b \,x^{2}+a}\, a^{2} b \,x^{2} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )-406 a^{\frac {5}{2}} b \,x^{2}+105 \sqrt {b \,x^{2}+a}\, a^{3} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )-176 a^{\frac {7}{2}}\right )}{105 \left (b \,x^{2}+a \right )^{4} a^{\frac {9}{2}} \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)/(b*x^3+a*x)^(9/2),x)

[Out]

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

[Out]

integrate(x^(7/2)/(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^{7/2}}{{\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^(7/2)/(a*x + b*x^3)^(9/2),x)

[Out]

int(x^(7/2)/(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 {7}{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**(7/2)/(b*x**3+a*x)**(9/2),x)

[Out]

Integral(x**(7/2)/(x*(a + b*x**2))**(9/2), x)

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