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

Optimal. Leaf size=197 \[ -\frac {693 b^5 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {a x+b \sqrt {x}}}\right )}{64 a^{13/2}}+\frac {693 b^4 \sqrt {a x+b \sqrt {x}}}{64 a^6}-\frac {231 b^3 \sqrt {x} \sqrt {a x+b \sqrt {x}}}{32 a^5}+\frac {231 b^2 x \sqrt {a x+b \sqrt {x}}}{40 a^4}-\frac {99 b x^{3/2} \sqrt {a x+b \sqrt {x}}}{20 a^3}+\frac {22 x^2 \sqrt {a x+b \sqrt {x}}}{5 a^2}-\frac {4 x^3}{a \sqrt {a x+b \sqrt {x}}} \]

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Rubi [A]  time = 0.18, antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {2018, 668, 670, 640, 620, 206} \begin {gather*} \frac {693 b^4 \sqrt {a x+b \sqrt {x}}}{64 a^6}-\frac {231 b^3 \sqrt {x} \sqrt {a x+b \sqrt {x}}}{32 a^5}+\frac {231 b^2 x \sqrt {a x+b \sqrt {x}}}{40 a^4}-\frac {693 b^5 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {a x+b \sqrt {x}}}\right )}{64 a^{13/2}}-\frac {99 b x^{3/2} \sqrt {a x+b \sqrt {x}}}{20 a^3}+\frac {22 x^2 \sqrt {a x+b \sqrt {x}}}{5 a^2}-\frac {4 x^3}{a \sqrt {a x+b \sqrt {x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/(b*Sqrt[x] + a*x)^(3/2),x]

[Out]

(-4*x^3)/(a*Sqrt[b*Sqrt[x] + a*x]) + (693*b^4*Sqrt[b*Sqrt[x] + a*x])/(64*a^6) - (231*b^3*Sqrt[x]*Sqrt[b*Sqrt[x
] + a*x])/(32*a^5) + (231*b^2*x*Sqrt[b*Sqrt[x] + a*x])/(40*a^4) - (99*b*x^(3/2)*Sqrt[b*Sqrt[x] + a*x])/(20*a^3
) + (22*x^2*Sqrt[b*Sqrt[x] + a*x])/(5*a^2) - (693*b^5*ArcTanh[(Sqrt[a]*Sqrt[x])/Sqrt[b*Sqrt[x] + a*x]])/(64*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 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 668

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(p + 1)), x] - Dist[(e^2*(m + p))/(c*(p + 1)), Int[(d + e*x)^(m - 2)*(a + b*x +
c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &
& LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]

Rule 670

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[((m + p)*(2*c*d - b*e))/(c*(m + 2*p + 1)), Int[(d + e
*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 -
b*d*e + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rule 2018

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

Rubi steps

\begin {align*} \int \frac {x^3}{\left (b \sqrt {x}+a x\right )^{3/2}} \, dx &=2 \operatorname {Subst}\left (\int \frac {x^7}{\left (b x+a x^2\right )^{3/2}} \, dx,x,\sqrt {x}\right )\\ &=-\frac {4 x^3}{a \sqrt {b \sqrt {x}+a x}}+\frac {22 \operatorname {Subst}\left (\int \frac {x^5}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{a}\\ &=-\frac {4 x^3}{a \sqrt {b \sqrt {x}+a x}}+\frac {22 x^2 \sqrt {b \sqrt {x}+a x}}{5 a^2}-\frac {(99 b) \operatorname {Subst}\left (\int \frac {x^4}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{5 a^2}\\ &=-\frac {4 x^3}{a \sqrt {b \sqrt {x}+a x}}-\frac {99 b x^{3/2} \sqrt {b \sqrt {x}+a x}}{20 a^3}+\frac {22 x^2 \sqrt {b \sqrt {x}+a x}}{5 a^2}+\frac {\left (693 b^2\right ) \operatorname {Subst}\left (\int \frac {x^3}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{40 a^3}\\ &=-\frac {4 x^3}{a \sqrt {b \sqrt {x}+a x}}+\frac {231 b^2 x \sqrt {b \sqrt {x}+a x}}{40 a^4}-\frac {99 b x^{3/2} \sqrt {b \sqrt {x}+a x}}{20 a^3}+\frac {22 x^2 \sqrt {b \sqrt {x}+a x}}{5 a^2}-\frac {\left (231 b^3\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{16 a^4}\\ &=-\frac {4 x^3}{a \sqrt {b \sqrt {x}+a x}}-\frac {231 b^3 \sqrt {x} \sqrt {b \sqrt {x}+a x}}{32 a^5}+\frac {231 b^2 x \sqrt {b \sqrt {x}+a x}}{40 a^4}-\frac {99 b x^{3/2} \sqrt {b \sqrt {x}+a x}}{20 a^3}+\frac {22 x^2 \sqrt {b \sqrt {x}+a x}}{5 a^2}+\frac {\left (693 b^4\right ) \operatorname {Subst}\left (\int \frac {x}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{64 a^5}\\ &=-\frac {4 x^3}{a \sqrt {b \sqrt {x}+a x}}+\frac {693 b^4 \sqrt {b \sqrt {x}+a x}}{64 a^6}-\frac {231 b^3 \sqrt {x} \sqrt {b \sqrt {x}+a x}}{32 a^5}+\frac {231 b^2 x \sqrt {b \sqrt {x}+a x}}{40 a^4}-\frac {99 b x^{3/2} \sqrt {b \sqrt {x}+a x}}{20 a^3}+\frac {22 x^2 \sqrt {b \sqrt {x}+a x}}{5 a^2}-\frac {\left (693 b^5\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{128 a^6}\\ &=-\frac {4 x^3}{a \sqrt {b \sqrt {x}+a x}}+\frac {693 b^4 \sqrt {b \sqrt {x}+a x}}{64 a^6}-\frac {231 b^3 \sqrt {x} \sqrt {b \sqrt {x}+a x}}{32 a^5}+\frac {231 b^2 x \sqrt {b \sqrt {x}+a x}}{40 a^4}-\frac {99 b x^{3/2} \sqrt {b \sqrt {x}+a x}}{20 a^3}+\frac {22 x^2 \sqrt {b \sqrt {x}+a x}}{5 a^2}-\frac {\left (693 b^5\right ) \operatorname {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {b \sqrt {x}+a x}}\right )}{64 a^6}\\ &=-\frac {4 x^3}{a \sqrt {b \sqrt {x}+a x}}+\frac {693 b^4 \sqrt {b \sqrt {x}+a x}}{64 a^6}-\frac {231 b^3 \sqrt {x} \sqrt {b \sqrt {x}+a x}}{32 a^5}+\frac {231 b^2 x \sqrt {b \sqrt {x}+a x}}{40 a^4}-\frac {99 b x^{3/2} \sqrt {b \sqrt {x}+a x}}{20 a^3}+\frac {22 x^2 \sqrt {b \sqrt {x}+a x}}{5 a^2}-\frac {693 b^5 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b \sqrt {x}+a x}}\right )}{64 a^{13/2}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 64, normalized size = 0.32 \begin {gather*} \frac {4 x^{7/2} \sqrt {\frac {a \sqrt {x}}{b}+1} \, _2F_1\left (\frac {3}{2},\frac {13}{2};\frac {15}{2};-\frac {a \sqrt {x}}{b}\right )}{13 b \sqrt {a x+b \sqrt {x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[1 + (a*Sqrt[x])/b]*x^(7/2)*Hypergeometric2F1[3/2, 13/2, 15/2, -((a*Sqrt[x])/b)])/(13*b*Sqrt[b*Sqrt[x]
+ a*x])

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IntegrateAlgebraic [A]  time = 0.50, size = 137, normalized size = 0.70 \begin {gather*} \frac {693 b^5 \log \left (-2 \sqrt {a} \sqrt {a x+b \sqrt {x}}+2 a \sqrt {x}+b\right )}{128 a^{13/2}}+\frac {\sqrt {a x+b \sqrt {x}} \left (128 a^5 x^{5/2}-176 a^4 b x^2+264 a^3 b^2 x^{3/2}-462 a^2 b^3 x+1155 a b^4 \sqrt {x}+3465 b^5\right )}{320 a^6 \left (a \sqrt {x}+b\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*Sqrt[x] + a*x]*(3465*b^5 + 1155*a*b^4*Sqrt[x] - 462*a^2*b^3*x + 264*a^3*b^2*x^(3/2) - 176*a^4*b*x^2 +
128*a^5*x^(5/2)))/(320*a^6*(b + a*Sqrt[x])) + (693*b^5*Log[b + 2*a*Sqrt[x] - 2*Sqrt[a]*Sqrt[b*Sqrt[x] + a*x]])
/(128*a^(13/2))

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

[Out]

Timed out

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument ValueDone

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maple [B]  time = 0.06, size = 549, normalized size = 2.79 \begin {gather*} \frac {\sqrt {a x +b \sqrt {x}}\, \left (-4480 a^{3} b^{5} x \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )+1015 a^{3} b^{5} x \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {a x +b \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )-8960 a^{2} b^{6} \sqrt {x}\, \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )+2030 a^{2} b^{6} \sqrt {x}\, \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {a x +b \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )-4060 \sqrt {a x +b \sqrt {x}}\, a^{\frac {9}{2}} b^{3} x^{\frac {3}{2}}-4480 a \,b^{7} \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )+1015 a \,b^{7} \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {a x +b \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )+256 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {13}{2}} x^{2}-10150 \sqrt {a x +b \sqrt {x}}\, a^{\frac {7}{2}} b^{4} x +8960 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {7}{2}} b^{4} x -352 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {11}{2}} b \,x^{\frac {3}{2}}-8120 \sqrt {a x +b \sqrt {x}}\, a^{\frac {5}{2}} b^{5} \sqrt {x}+17920 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {5}{2}} b^{5} \sqrt {x}+528 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {9}{2}} b^{2} x -2030 \sqrt {a x +b \sqrt {x}}\, a^{\frac {3}{2}} b^{6}+8960 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {3}{2}} b^{6}+3136 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {7}{2}} b^{3} \sqrt {x}+2000 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {5}{2}} b^{4}-2560 \left (\left (a \sqrt {x}+b \right ) \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {5}{2}} b^{4}\right )}{640 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, \left (a \sqrt {x}+b \right )^{2} a^{\frac {15}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/640*(a*x+b*x^(1/2))^(1/2)*(256*(a*x+b*x^(1/2))^(3/2)*a^(13/2)*x^2-352*(a*x+b*x^(1/2))^(3/2)*a^(11/2)*x^(3/2)
*b-4060*(a*x+b*x^(1/2))^(1/2)*a^(9/2)*x^(3/2)*b^3+528*(a*x+b*x^(1/2))^(3/2)*a^(9/2)*x*b^2+3136*(a*x+b*x^(1/2))
^(3/2)*a^(7/2)*x^(1/2)*b^3-10150*(a*x+b*x^(1/2))^(1/2)*a^(7/2)*x*b^4+8960*a^(7/2)*x*((a*x^(1/2)+b)*x^(1/2))^(1
/2)*b^4+2000*(a*x+b*x^(1/2))^(3/2)*a^(5/2)*b^4-8120*(a*x+b*x^(1/2))^(1/2)*a^(5/2)*x^(1/2)*b^5+17920*a^(5/2)*x^
(1/2)*((a*x^(1/2)+b)*x^(1/2))^(1/2)*b^5-2560*a^(5/2)*((a*x^(1/2)+b)*x^(1/2))^(3/2)*b^4-2030*(a*x+b*x^(1/2))^(1
/2)*a^(3/2)*b^6+8960*a^(3/2)*((a*x^(1/2)+b)*x^(1/2))^(1/2)*b^6+2030*ln(1/2*(2*a*x^(1/2)+b+2*(a*x+b*x^(1/2))^(1
/2)*a^(1/2))/a^(1/2))*x^(1/2)*a^2*b^6+1015*ln(1/2*(2*a*x^(1/2)+b+2*(a*x+b*x^(1/2))^(1/2)*a^(1/2))/a^(1/2))*x*a
^3*b^5-8960*a^2*ln(1/2*(2*a*x^(1/2)+b+2*((a*x^(1/2)+b)*x^(1/2))^(1/2)*a^(1/2))/a^(1/2))*x^(1/2)*b^6-4480*a^3*l
n(1/2*(2*a*x^(1/2)+b+2*((a*x^(1/2)+b)*x^(1/2))^(1/2)*a^(1/2))/a^(1/2))*x*b^5+1015*ln(1/2*(2*a*x^(1/2)+b+2*(a*x
+b*x^(1/2))^(1/2)*a^(1/2))/a^(1/2))*a*b^7-4480*a*ln(1/2*(2*a*x^(1/2)+b+2*((a*x^(1/2)+b)*x^(1/2))^(1/2)*a^(1/2)
)/a^(1/2))*b^7)/a^(15/2)/((a*x^(1/2)+b)*x^(1/2))^(1/2)/(a*x^(1/2)+b)^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a*x + b*x^(1/2))^(3/2),x)

[Out]

int(x^3/(a*x + b*x^(1/2))^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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