3.1.81 \(\int \frac {x^{5/2}}{(b \sqrt {x}+a x)^{3/2}} \, dx\)

Optimal. Leaf size=171 \[ \frac {315 b^4 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {a x+b \sqrt {x}}}\right )}{32 a^{11/2}}-\frac {315 b^3 \sqrt {a x+b \sqrt {x}}}{32 a^5}+\frac {105 b^2 \sqrt {x} \sqrt {a x+b \sqrt {x}}}{16 a^4}-\frac {21 b x \sqrt {a x+b \sqrt {x}}}{4 a^3}+\frac {9 x^{3/2} \sqrt {a x+b \sqrt {x}}}{2 a^2}-\frac {4 x^{5/2}}{a \sqrt {a x+b \sqrt {x}}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.15, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2018, 668, 670, 640, 620, 206} \begin {gather*} -\frac {315 b^3 \sqrt {a x+b \sqrt {x}}}{32 a^5}+\frac {105 b^2 \sqrt {x} \sqrt {a x+b \sqrt {x}}}{16 a^4}+\frac {315 b^4 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {a x+b \sqrt {x}}}\right )}{32 a^{11/2}}+\frac {9 x^{3/2} \sqrt {a x+b \sqrt {x}}}{2 a^2}-\frac {21 b x \sqrt {a x+b \sqrt {x}}}{4 a^3}-\frac {4 x^{5/2}}{a \sqrt {a x+b \sqrt {x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/(b*Sqrt[x] + a*x)^(3/2),x]

[Out]

(-4*x^(5/2))/(a*Sqrt[b*Sqrt[x] + a*x]) - (315*b^3*Sqrt[b*Sqrt[x] + a*x])/(32*a^5) + (105*b^2*Sqrt[x]*Sqrt[b*Sq
rt[x] + a*x])/(16*a^4) - (21*b*x*Sqrt[b*Sqrt[x] + a*x])/(4*a^3) + (9*x^(3/2)*Sqrt[b*Sqrt[x] + a*x])/(2*a^2) +
(315*b^4*ArcTanh[(Sqrt[a]*Sqrt[x])/Sqrt[b*Sqrt[x] + a*x]])/(32*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 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^{5/2}}{\left (b \sqrt {x}+a x\right )^{3/2}} \, dx &=2 \operatorname {Subst}\left (\int \frac {x^6}{\left (b x+a x^2\right )^{3/2}} \, dx,x,\sqrt {x}\right )\\ &=-\frac {4 x^{5/2}}{a \sqrt {b \sqrt {x}+a x}}+\frac {18 \operatorname {Subst}\left (\int \frac {x^4}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{a}\\ &=-\frac {4 x^{5/2}}{a \sqrt {b \sqrt {x}+a x}}+\frac {9 x^{3/2} \sqrt {b \sqrt {x}+a x}}{2 a^2}-\frac {(63 b) \operatorname {Subst}\left (\int \frac {x^3}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{4 a^2}\\ &=-\frac {4 x^{5/2}}{a \sqrt {b \sqrt {x}+a x}}-\frac {21 b x \sqrt {b \sqrt {x}+a x}}{4 a^3}+\frac {9 x^{3/2} \sqrt {b \sqrt {x}+a x}}{2 a^2}+\frac {\left (105 b^2\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{8 a^3}\\ &=-\frac {4 x^{5/2}}{a \sqrt {b \sqrt {x}+a x}}+\frac {105 b^2 \sqrt {x} \sqrt {b \sqrt {x}+a x}}{16 a^4}-\frac {21 b x \sqrt {b \sqrt {x}+a x}}{4 a^3}+\frac {9 x^{3/2} \sqrt {b \sqrt {x}+a x}}{2 a^2}-\frac {\left (315 b^3\right ) \operatorname {Subst}\left (\int \frac {x}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{32 a^4}\\ &=-\frac {4 x^{5/2}}{a \sqrt {b \sqrt {x}+a x}}-\frac {315 b^3 \sqrt {b \sqrt {x}+a x}}{32 a^5}+\frac {105 b^2 \sqrt {x} \sqrt {b \sqrt {x}+a x}}{16 a^4}-\frac {21 b x \sqrt {b \sqrt {x}+a x}}{4 a^3}+\frac {9 x^{3/2} \sqrt {b \sqrt {x}+a x}}{2 a^2}+\frac {\left (315 b^4\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{64 a^5}\\ &=-\frac {4 x^{5/2}}{a \sqrt {b \sqrt {x}+a x}}-\frac {315 b^3 \sqrt {b \sqrt {x}+a x}}{32 a^5}+\frac {105 b^2 \sqrt {x} \sqrt {b \sqrt {x}+a x}}{16 a^4}-\frac {21 b x \sqrt {b \sqrt {x}+a x}}{4 a^3}+\frac {9 x^{3/2} \sqrt {b \sqrt {x}+a x}}{2 a^2}+\frac {\left (315 b^4\right ) \operatorname {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {b \sqrt {x}+a x}}\right )}{32 a^5}\\ &=-\frac {4 x^{5/2}}{a \sqrt {b \sqrt {x}+a x}}-\frac {315 b^3 \sqrt {b \sqrt {x}+a x}}{32 a^5}+\frac {105 b^2 \sqrt {x} \sqrt {b \sqrt {x}+a x}}{16 a^4}-\frac {21 b x \sqrt {b \sqrt {x}+a x}}{4 a^3}+\frac {9 x^{3/2} \sqrt {b \sqrt {x}+a x}}{2 a^2}+\frac {315 b^4 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b \sqrt {x}+a x}}\right )}{32 a^{11/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.08, size = 62, normalized size = 0.36 \begin {gather*} \frac {4 x^3 \sqrt {\frac {a \sqrt {x}}{b}+1} \, _2F_1\left (\frac {3}{2},\frac {11}{2};\frac {13}{2};-\frac {a \sqrt {x}}{b}\right )}{11 b \sqrt {a x+b \sqrt {x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)/(b*Sqrt[x] + a*x)^(3/2),x]

[Out]

(4*Sqrt[1 + (a*Sqrt[x])/b]*x^3*Hypergeometric2F1[3/2, 11/2, 13/2, -((a*Sqrt[x])/b)])/(11*b*Sqrt[b*Sqrt[x] + a*
x])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.45, size = 124, normalized size = 0.73 \begin {gather*} \frac {\sqrt {a x+b \sqrt {x}} \left (16 a^4 x^2-24 a^3 b x^{3/2}+42 a^2 b^2 x-105 a b^3 \sqrt {x}-315 b^4\right )}{32 a^5 \left (a \sqrt {x}+b\right )}-\frac {315 b^4 \log \left (-2 \sqrt {a} \sqrt {a x+b \sqrt {x}}+2 a \sqrt {x}+b\right )}{64 a^{11/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(5/2)/(b*Sqrt[x] + a*x)^(3/2),x]

[Out]

(Sqrt[b*Sqrt[x] + a*x]*(-315*b^4 - 105*a*b^3*Sqrt[x] + 42*a^2*b^2*x - 24*a^3*b*x^(3/2) + 16*a^4*x^2))/(32*a^5*
(b + a*Sqrt[x])) - (315*b^4*Log[b + 2*a*Sqrt[x] - 2*Sqrt[a]*Sqrt[b*Sqrt[x] + a*x]])/(64*a^(11/2))

________________________________________________________________________________________

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^(5/2)/(b*x^(1/2)+a*x)^(3/2),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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^(5/2)/(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

________________________________________________________________________________________

maple [B]  time = 0.06, size = 527, normalized size = 3.08 \begin {gather*} \frac {\sqrt {a x +b \sqrt {x}}\, \left (384 a^{3} b^{4} 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 )-69 a^{3} b^{4} x \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {a x +b \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )+768 a^{2} b^{5} \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 )-138 a^{2} b^{5} \sqrt {x}\, \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {a x +b \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )+276 \sqrt {a x +b \sqrt {x}}\, a^{\frac {9}{2}} b^{2} x^{\frac {3}{2}}+384 a \,b^{6} \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )-69 a \,b^{6} \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {a x +b \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )-768 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {7}{2}} b^{3} x +690 \sqrt {a x +b \sqrt {x}}\, a^{\frac {7}{2}} b^{3} x +32 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {11}{2}} x^{\frac {3}{2}}-1536 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {5}{2}} b^{4} \sqrt {x}+552 \sqrt {a x +b \sqrt {x}}\, a^{\frac {5}{2}} b^{4} \sqrt {x}-48 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {9}{2}} b x -768 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {3}{2}} b^{5}+138 \sqrt {a x +b \sqrt {x}}\, a^{\frac {3}{2}} b^{5}-192 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {7}{2}} b^{2} \sqrt {x}+256 \left (\left (a \sqrt {x}+b \right ) \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {5}{2}} b^{3}-112 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {5}{2}} b^{3}\right )}{64 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, \left (a \sqrt {x}+b \right )^{2} a^{\frac {13}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)/(a*x+b*x^(1/2))^(3/2),x)

[Out]

1/64*(a*x+b*x^(1/2))^(1/2)/a^(13/2)*(32*(a*x+b*x^(1/2))^(3/2)*a^(11/2)*x^(3/2)+276*x^(3/2)*(a*x+b*x^(1/2))^(1/
2)*a^(9/2)*b^2-48*(a*x+b*x^(1/2))^(3/2)*a^(9/2)*b*x-768*x*((a*x^(1/2)+b)*x^(1/2))^(1/2)*a^(7/2)*b^3+690*x*(a*x
+b*x^(1/2))^(1/2)*a^(7/2)*b^3+384*x*a^3*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^4-192*(a*x+b*x^(1/2))^(3/2)*a^(7/2)*b^2*x^(1/2)-69*x*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^3*b^4-1536*x^(1/2)*((a*x^(1/2)+b)*x^(1/2))^(1/2)*a^(5/2)*b^4+552*(a*x+b*x^(1/2))^(1/2)*a^(5/2)*b^
4*x^(1/2)+768*x^(1/2)*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))*b^5+256*((a*
x^(1/2)+b)*x^(1/2))^(3/2)*a^(5/2)*b^3-112*(a*x+b*x^(1/2))^(3/2)*a^(5/2)*b^3-138*x^(1/2)*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^2*b^5-768*((a*x^(1/2)+b)*x^(1/2))^(1/2)*a^(3/2)*b^5+138*(a*x+b*x^(
1/2))^(1/2)*a^(3/2)*b^5+384*a*b^6*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))-69*a
*b^6*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*x^(1/2)+b)*x^(1/2))^(1/2)/(a*x^(1/2)
+b)^2

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{\frac {5}{2}}}{{\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^(5/2)/(b*x^(1/2)+a*x)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^(5/2)/(a*x + b*sqrt(x))^(3/2), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{5/2}}{{\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^(5/2)/(a*x + b*x^(1/2))^(3/2),x)

[Out]

int(x^(5/2)/(a*x + b*x^(1/2))^(3/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{\frac {5}{2}}}{\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**(5/2)/(b*x**(1/2)+a*x)**(3/2),x)

[Out]

Integral(x**(5/2)/(a*x + b*sqrt(x))**(3/2), x)

________________________________________________________________________________________