3.90 \(\int \frac {1}{\sqrt {x} (a x+b x^3)^{9/2}} \, dx\)

Optimal. Leaf size=189 \[ -\frac {99 b^2 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {a x+b x^3}}\right )}{8 a^{13/2}}+\frac {99 b \sqrt {a x+b x^3}}{8 a^6 x^{5/2}}-\frac {33 \sqrt {a x+b x^3}}{4 a^5 x^{9/2}}+\frac {33}{5 a^4 x^{7/2} \sqrt {a x+b x^3}}+\frac {33}{35 a^3 x^{5/2} \left (a x+b x^3\right )^{3/2}}+\frac {11}{35 a^2 x^{3/2} \left (a x+b x^3\right )^{5/2}}+\frac {1}{7 a \sqrt {x} \left (a x+b x^3\right )^{7/2}} \]

[Out]

11/35/a^2/x^(3/2)/(b*x^3+a*x)^(5/2)+33/35/a^3/x^(5/2)/(b*x^3+a*x)^(3/2)-99/8*b^2*arctanh(a^(1/2)*x^(1/2)/(b*x^
3+a*x)^(1/2))/a^(13/2)+1/7/a/(b*x^3+a*x)^(7/2)/x^(1/2)+33/5/a^4/x^(7/2)/(b*x^3+a*x)^(1/2)-33/4*(b*x^3+a*x)^(1/
2)/a^5/x^(9/2)+99/8*b*(b*x^3+a*x)^(1/2)/a^6/x^(5/2)

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Rubi [A]  time = 0.29, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2023, 2025, 2029, 206} \[ -\frac {99 b^2 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {a x+b x^3}}\right )}{8 a^{13/2}}+\frac {99 b \sqrt {a x+b x^3}}{8 a^6 x^{5/2}}-\frac {33 \sqrt {a x+b x^3}}{4 a^5 x^{9/2}}+\frac {33}{5 a^4 x^{7/2} \sqrt {a x+b x^3}}+\frac {33}{35 a^3 x^{5/2} \left (a x+b x^3\right )^{3/2}}+\frac {11}{35 a^2 x^{3/2} \left (a x+b x^3\right )^{5/2}}+\frac {1}{7 a \sqrt {x} \left (a x+b x^3\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[x]*(a*x + b*x^3)^(9/2)),x]

[Out]

1/(7*a*Sqrt[x]*(a*x + b*x^3)^(7/2)) + 11/(35*a^2*x^(3/2)*(a*x + b*x^3)^(5/2)) + 33/(35*a^3*x^(5/2)*(a*x + b*x^
3)^(3/2)) + 33/(5*a^4*x^(7/2)*Sqrt[a*x + b*x^3]) - (33*Sqrt[a*x + b*x^3])/(4*a^5*x^(9/2)) + (99*b*Sqrt[a*x + b
*x^3])/(8*a^6*x^(5/2)) - (99*b^2*ArcTanh[(Sqrt[a]*Sqrt[x])/Sqrt[a*x + b*x^3]])/(8*a^(13/2))

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 {1}{\sqrt {x} \left (a x+b x^3\right )^{9/2}} \, dx &=\frac {1}{7 a \sqrt {x} \left (a x+b x^3\right )^{7/2}}+\frac {11 \int \frac {1}{x^{3/2} \left (a x+b x^3\right )^{7/2}} \, dx}{7 a}\\ &=\frac {1}{7 a \sqrt {x} \left (a x+b x^3\right )^{7/2}}+\frac {11}{35 a^2 x^{3/2} \left (a x+b x^3\right )^{5/2}}+\frac {99 \int \frac {1}{x^{5/2} \left (a x+b x^3\right )^{5/2}} \, dx}{35 a^2}\\ &=\frac {1}{7 a \sqrt {x} \left (a x+b x^3\right )^{7/2}}+\frac {11}{35 a^2 x^{3/2} \left (a x+b x^3\right )^{5/2}}+\frac {33}{35 a^3 x^{5/2} \left (a x+b x^3\right )^{3/2}}+\frac {33 \int \frac {1}{x^{7/2} \left (a x+b x^3\right )^{3/2}} \, dx}{5 a^3}\\ &=\frac {1}{7 a \sqrt {x} \left (a x+b x^3\right )^{7/2}}+\frac {11}{35 a^2 x^{3/2} \left (a x+b x^3\right )^{5/2}}+\frac {33}{35 a^3 x^{5/2} \left (a x+b x^3\right )^{3/2}}+\frac {33}{5 a^4 x^{7/2} \sqrt {a x+b x^3}}+\frac {33 \int \frac {1}{x^{9/2} \sqrt {a x+b x^3}} \, dx}{a^4}\\ &=\frac {1}{7 a \sqrt {x} \left (a x+b x^3\right )^{7/2}}+\frac {11}{35 a^2 x^{3/2} \left (a x+b x^3\right )^{5/2}}+\frac {33}{35 a^3 x^{5/2} \left (a x+b x^3\right )^{3/2}}+\frac {33}{5 a^4 x^{7/2} \sqrt {a x+b x^3}}-\frac {33 \sqrt {a x+b x^3}}{4 a^5 x^{9/2}}-\frac {(99 b) \int \frac {1}{x^{5/2} \sqrt {a x+b x^3}} \, dx}{4 a^5}\\ &=\frac {1}{7 a \sqrt {x} \left (a x+b x^3\right )^{7/2}}+\frac {11}{35 a^2 x^{3/2} \left (a x+b x^3\right )^{5/2}}+\frac {33}{35 a^3 x^{5/2} \left (a x+b x^3\right )^{3/2}}+\frac {33}{5 a^4 x^{7/2} \sqrt {a x+b x^3}}-\frac {33 \sqrt {a x+b x^3}}{4 a^5 x^{9/2}}+\frac {99 b \sqrt {a x+b x^3}}{8 a^6 x^{5/2}}+\frac {\left (99 b^2\right ) \int \frac {1}{\sqrt {x} \sqrt {a x+b x^3}} \, dx}{8 a^6}\\ &=\frac {1}{7 a \sqrt {x} \left (a x+b x^3\right )^{7/2}}+\frac {11}{35 a^2 x^{3/2} \left (a x+b x^3\right )^{5/2}}+\frac {33}{35 a^3 x^{5/2} \left (a x+b x^3\right )^{3/2}}+\frac {33}{5 a^4 x^{7/2} \sqrt {a x+b x^3}}-\frac {33 \sqrt {a x+b x^3}}{4 a^5 x^{9/2}}+\frac {99 b \sqrt {a x+b x^3}}{8 a^6 x^{5/2}}-\frac {\left (99 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a x+b x^3}}\right )}{8 a^6}\\ &=\frac {1}{7 a \sqrt {x} \left (a x+b x^3\right )^{7/2}}+\frac {11}{35 a^2 x^{3/2} \left (a x+b x^3\right )^{5/2}}+\frac {33}{35 a^3 x^{5/2} \left (a x+b x^3\right )^{3/2}}+\frac {33}{5 a^4 x^{7/2} \sqrt {a x+b x^3}}-\frac {33 \sqrt {a x+b x^3}}{4 a^5 x^{9/2}}+\frac {99 b \sqrt {a x+b x^3}}{8 a^6 x^{5/2}}-\frac {99 b^2 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {a x+b x^3}}\right )}{8 a^{13/2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 46, normalized size = 0.24 \[ \frac {b^2 x^{7/2} \, _2F_1\left (-\frac {7}{2},3;-\frac {5}{2};\frac {b x^2}{a}+1\right )}{7 a^3 \left (x \left (a+b x^2\right )\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[x]*(a*x + b*x^3)^(9/2)),x]

[Out]

(b^2*x^(7/2)*Hypergeometric2F1[-7/2, 3, -5/2, 1 + (b*x^2)/a])/(7*a^3*(x*(a + b*x^2))^(7/2))

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fricas [A]  time = 0.69, size = 422, normalized size = 2.23 \[ \left [\frac {3465 \, {\left (b^{6} x^{13} + 4 \, a b^{5} x^{11} + 6 \, a^{2} b^{4} x^{9} + 4 \, a^{3} b^{3} x^{7} + a^{4} b^{2} x^{5}\right )} \sqrt {a} \log \left (\frac {b x^{3} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x} \sqrt {a} \sqrt {x}}{x^{3}}\right ) + 2 \, {\left (3465 \, a b^{5} x^{10} + 11550 \, a^{2} b^{4} x^{8} + 13398 \, a^{3} b^{3} x^{6} + 5808 \, a^{4} b^{2} x^{4} + 385 \, a^{5} b x^{2} - 70 \, a^{6}\right )} \sqrt {b x^{3} + a x} \sqrt {x}}{560 \, {\left (a^{7} b^{4} x^{13} + 4 \, a^{8} b^{3} x^{11} + 6 \, a^{9} b^{2} x^{9} + 4 \, a^{10} b x^{7} + a^{11} x^{5}\right )}}, \frac {3465 \, {\left (b^{6} x^{13} + 4 \, a b^{5} x^{11} + 6 \, a^{2} b^{4} x^{9} + 4 \, a^{3} b^{3} x^{7} + a^{4} b^{2} x^{5}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{3} + a x} \sqrt {-a}}{a \sqrt {x}}\right ) + {\left (3465 \, a b^{5} x^{10} + 11550 \, a^{2} b^{4} x^{8} + 13398 \, a^{3} b^{3} x^{6} + 5808 \, a^{4} b^{2} x^{4} + 385 \, a^{5} b x^{2} - 70 \, a^{6}\right )} \sqrt {b x^{3} + a x} \sqrt {x}}{280 \, {\left (a^{7} b^{4} x^{13} + 4 \, a^{8} b^{3} x^{11} + 6 \, a^{9} b^{2} x^{9} + 4 \, a^{10} b x^{7} + a^{11} x^{5}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(1/2)/(b*x^3+a*x)^(9/2),x, algorithm="fricas")

[Out]

[1/560*(3465*(b^6*x^13 + 4*a*b^5*x^11 + 6*a^2*b^4*x^9 + 4*a^3*b^3*x^7 + a^4*b^2*x^5)*sqrt(a)*log((b*x^3 + 2*a*
x - 2*sqrt(b*x^3 + a*x)*sqrt(a)*sqrt(x))/x^3) + 2*(3465*a*b^5*x^10 + 11550*a^2*b^4*x^8 + 13398*a^3*b^3*x^6 + 5
808*a^4*b^2*x^4 + 385*a^5*b*x^2 - 70*a^6)*sqrt(b*x^3 + a*x)*sqrt(x))/(a^7*b^4*x^13 + 4*a^8*b^3*x^11 + 6*a^9*b^
2*x^9 + 4*a^10*b*x^7 + a^11*x^5), 1/280*(3465*(b^6*x^13 + 4*a*b^5*x^11 + 6*a^2*b^4*x^9 + 4*a^3*b^3*x^7 + a^4*b
^2*x^5)*sqrt(-a)*arctan(sqrt(b*x^3 + a*x)*sqrt(-a)/(a*sqrt(x))) + (3465*a*b^5*x^10 + 11550*a^2*b^4*x^8 + 13398
*a^3*b^3*x^6 + 5808*a^4*b^2*x^4 + 385*a^5*b*x^2 - 70*a^6)*sqrt(b*x^3 + a*x)*sqrt(x))/(a^7*b^4*x^13 + 4*a^8*b^3
*x^11 + 6*a^9*b^2*x^9 + 4*a^10*b*x^7 + a^11*x^5)]

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giac [A]  time = 0.27, size = 138, normalized size = 0.73 \[ \frac {99 \, b^{2} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{8 \, \sqrt {-a} a^{6}} + \frac {350 \, {\left (b x^{2} + a\right )}^{3} b^{2} + 70 \, {\left (b x^{2} + a\right )}^{2} a b^{2} + 21 \, {\left (b x^{2} + a\right )} a^{2} b^{2} + 5 \, a^{3} b^{2}}{35 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} a^{6}} + \frac {19 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2} - 21 \, \sqrt {b x^{2} + a} a b^{2}}{8 \, a^{6} b^{2} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(1/2)/(b*x^3+a*x)^(9/2),x, algorithm="giac")

[Out]

99/8*b^2*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^6) + 1/35*(350*(b*x^2 + a)^3*b^2 + 70*(b*x^2 + a)^2*a*b^
2 + 21*(b*x^2 + a)*a^2*b^2 + 5*a^3*b^2)/((b*x^2 + a)^(7/2)*a^6) + 1/8*(19*(b*x^2 + a)^(3/2)*b^2 - 21*sqrt(b*x^
2 + a)*a*b^2)/(a^6*b^2*x^4)

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maple [A]  time = 0.06, size = 247, normalized size = 1.31 \[ -\frac {\sqrt {\left (b \,x^{2}+a \right ) x}\, \left (3465 \sqrt {b \,x^{2}+a}\, b^{5} x^{10} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )-3465 \sqrt {a}\, b^{5} x^{10}+10395 \sqrt {b \,x^{2}+a}\, a \,b^{4} x^{8} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )-11550 a^{\frac {3}{2}} b^{4} x^{8}+10395 \sqrt {b \,x^{2}+a}\, a^{2} b^{3} x^{6} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )-13398 a^{\frac {5}{2}} b^{3} x^{6}+3465 \sqrt {b \,x^{2}+a}\, a^{3} b^{2} x^{4} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )-5808 a^{\frac {7}{2}} b^{2} x^{4}-385 a^{\frac {9}{2}} b \,x^{2}+70 a^{\frac {11}{2}}\right )}{280 \left (b \,x^{2}+a \right )^{4} a^{\frac {13}{2}} x^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(1/2)/(b*x^3+a*x)^(9/2),x)

[Out]

-1/280*((b*x^2+a)*x)^(1/2)/a^(13/2)*(3465*ln(2*(a+(b*x^2+a)^(1/2)*a^(1/2))/x)*x^10*b^5*(b*x^2+a)^(1/2)-3465*a^
(1/2)*x^10*b^5+10395*ln(2*(a+(b*x^2+a)^(1/2)*a^(1/2))/x)*x^8*a*b^4*(b*x^2+a)^(1/2)-11550*a^(3/2)*x^8*b^4+10395
*ln(2*(a+(b*x^2+a)^(1/2)*a^(1/2))/x)*x^6*a^2*b^3*(b*x^2+a)^(1/2)-13398*a^(5/2)*x^6*b^3+3465*ln(2*(a+(b*x^2+a)^
(1/2)*a^(1/2))/x)*x^4*a^3*b^2*(b*x^2+a)^(1/2)-5808*a^(7/2)*x^4*b^2-385*a^(9/2)*x^2*b+70*a^(11/2))/x^(9/2)/(b*x
^2+a)^4

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{3} + a x\right )}^{\frac {9}{2}} \sqrt {x}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(1/2)/(b*x^3+a*x)^(9/2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^3 + a*x)^(9/2)*sqrt(x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(1/2)*(a*x + b*x^3)^(9/2)),x)

[Out]

int(1/(x^(1/2)*(a*x + b*x^3)^(9/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {x} \left (x \left (a + b x^{2}\right )\right )^{\frac {9}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(1/2)/(b*x**3+a*x)**(9/2),x)

[Out]

Integral(1/(sqrt(x)*(x*(a + b*x**2))**(9/2)), x)

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