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

Optimal. Leaf size=146 \[ \frac {35 b^4 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {a x+b \sqrt {x}}}\right )}{32 a^{9/2}}-\frac {35 b^3 \sqrt {a x+b \sqrt {x}}}{32 a^4}+\frac {35 b^2 \sqrt {x} \sqrt {a x+b \sqrt {x}}}{48 a^3}-\frac {7 b x \sqrt {a x+b \sqrt {x}}}{12 a^2}+\frac {x^{3/2} \sqrt {a x+b \sqrt {x}}}{2 a} \]

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Rubi [A]  time = 0.12, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2018, 670, 640, 620, 206} \begin {gather*} -\frac {35 b^3 \sqrt {a x+b \sqrt {x}}}{32 a^4}+\frac {35 b^2 \sqrt {x} \sqrt {a x+b \sqrt {x}}}{48 a^3}+\frac {35 b^4 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {a x+b \sqrt {x}}}\right )}{32 a^{9/2}}-\frac {7 b x \sqrt {a x+b \sqrt {x}}}{12 a^2}+\frac {x^{3/2} \sqrt {a x+b \sqrt {x}}}{2 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(3/2)/Sqrt[b*Sqrt[x] + a*x],x]

[Out]

(-35*b^3*Sqrt[b*Sqrt[x] + a*x])/(32*a^4) + (35*b^2*Sqrt[x]*Sqrt[b*Sqrt[x] + a*x])/(48*a^3) - (7*b*x*Sqrt[b*Sqr
t[x] + a*x])/(12*a^2) + (x^(3/2)*Sqrt[b*Sqrt[x] + a*x])/(2*a) + (35*b^4*ArcTanh[(Sqrt[a]*Sqrt[x])/Sqrt[b*Sqrt[
x] + a*x]])/(32*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 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 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^{3/2}}{\sqrt {b \sqrt {x}+a x}} \, dx &=2 \operatorname {Subst}\left (\int \frac {x^4}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )\\ &=\frac {x^{3/2} \sqrt {b \sqrt {x}+a x}}{2 a}-\frac {(7 b) \operatorname {Subst}\left (\int \frac {x^3}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{4 a}\\ &=-\frac {7 b x \sqrt {b \sqrt {x}+a x}}{12 a^2}+\frac {x^{3/2} \sqrt {b \sqrt {x}+a x}}{2 a}+\frac {\left (35 b^2\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{24 a^2}\\ &=\frac {35 b^2 \sqrt {x} \sqrt {b \sqrt {x}+a x}}{48 a^3}-\frac {7 b x \sqrt {b \sqrt {x}+a x}}{12 a^2}+\frac {x^{3/2} \sqrt {b \sqrt {x}+a x}}{2 a}-\frac {\left (35 b^3\right ) \operatorname {Subst}\left (\int \frac {x}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{32 a^3}\\ &=-\frac {35 b^3 \sqrt {b \sqrt {x}+a x}}{32 a^4}+\frac {35 b^2 \sqrt {x} \sqrt {b \sqrt {x}+a x}}{48 a^3}-\frac {7 b x \sqrt {b \sqrt {x}+a x}}{12 a^2}+\frac {x^{3/2} \sqrt {b \sqrt {x}+a x}}{2 a}+\frac {\left (35 b^4\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{64 a^4}\\ &=-\frac {35 b^3 \sqrt {b \sqrt {x}+a x}}{32 a^4}+\frac {35 b^2 \sqrt {x} \sqrt {b \sqrt {x}+a x}}{48 a^3}-\frac {7 b x \sqrt {b \sqrt {x}+a x}}{12 a^2}+\frac {x^{3/2} \sqrt {b \sqrt {x}+a x}}{2 a}+\frac {\left (35 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^4}\\ &=-\frac {35 b^3 \sqrt {b \sqrt {x}+a x}}{32 a^4}+\frac {35 b^2 \sqrt {x} \sqrt {b \sqrt {x}+a x}}{48 a^3}-\frac {7 b x \sqrt {b \sqrt {x}+a x}}{12 a^2}+\frac {x^{3/2} \sqrt {b \sqrt {x}+a x}}{2 a}+\frac {35 b^4 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b \sqrt {x}+a x}}\right )}{32 a^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 142, normalized size = 0.97 \begin {gather*} -\frac {35 b^5 \left (\frac {a \sqrt {x}}{b}+1\right ) \left (-\frac {32 a^4 x^2}{35 b^4}+\frac {16 a^3 x^{3/2}}{15 b^3}-\frac {4 a^2 x}{3 b^2}+\frac {2 a \sqrt {x}}{b}-\frac {2 \sqrt {a} \sqrt [4]{x} \sinh ^{-1}\left (\frac {\sqrt {a} \sqrt [4]{x}}{\sqrt {b}}\right )}{\sqrt {b} \sqrt {\frac {a \sqrt {x}}{b}+1}}\right )}{64 a^5 \sqrt {\sqrt {x} \left (a \sqrt {x}+b\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-35*b^5*(1 + (a*Sqrt[x])/b)*((2*a*Sqrt[x])/b - (4*a^2*x)/(3*b^2) + (16*a^3*x^(3/2))/(15*b^3) - (32*a^4*x^2)/(
35*b^4) - (2*Sqrt[a]*x^(1/4)*ArcSinh[(Sqrt[a]*x^(1/4))/Sqrt[b]])/(Sqrt[b]*Sqrt[1 + (a*Sqrt[x])/b])))/(64*a^5*S
qrt[(b + a*Sqrt[x])*Sqrt[x]])

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IntegrateAlgebraic [A]  time = 0.30, size = 108, normalized size = 0.74 \begin {gather*} \frac {\sqrt {a x+b \sqrt {x}} \left (48 a^3 x^{3/2}-56 a^2 b x+70 a b^2 \sqrt {x}-105 b^3\right )}{96 a^4}-\frac {35 b^4 \log \left (-2 a^{9/2} \sqrt {a x+b \sqrt {x}}+2 a^5 \sqrt {x}+a^4 b\right )}{64 a^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*Sqrt[x] + a*x]*(-105*b^3 + 70*a*b^2*Sqrt[x] - 56*a^2*b*x + 48*a^3*x^(3/2)))/(96*a^4) - (35*b^4*Log[a^4
*b + 2*a^5*Sqrt[x] - 2*a^(9/2)*Sqrt[b*Sqrt[x] + a*x]])/(64*a^(9/2))

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

[Out]

Timed out

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giac [A]  time = 0.41, size = 97, normalized size = 0.66 \begin {gather*} \frac {1}{96} \, \sqrt {a x + b \sqrt {x}} {\left (2 \, {\left (4 \, \sqrt {x} {\left (\frac {6 \, \sqrt {x}}{a} - \frac {7 \, b}{a^{2}}\right )} + \frac {35 \, b^{2}}{a^{3}}\right )} \sqrt {x} - \frac {105 \, b^{3}}{a^{4}}\right )} - \frac {35 \, b^{4} \log \left ({\left | -2 \, \sqrt {a} {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )} - b \right |}\right )}{64 \, a^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*sqrt(a*x + b*sqrt(x))*(2*(4*sqrt(x)*(6*sqrt(x)/a - 7*b/a^2) + 35*b^2/a^3)*sqrt(x) - 105*b^3/a^4) - 35/64*
b^4*log(abs(-2*sqrt(a)*(sqrt(a)*sqrt(x) - sqrt(a*x + b*sqrt(x))) - b))/a^(9/2)

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maple [A]  time = 0.05, size = 203, normalized size = 1.39 \begin {gather*} \frac {\sqrt {a x +b \sqrt {x}}\, \left (192 a \,b^{4} \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )-87 a \,b^{4} \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {a x +b \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )+348 \sqrt {a x +b \sqrt {x}}\, a^{\frac {5}{2}} b^{2} \sqrt {x}-384 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {3}{2}} b^{3}+174 \sqrt {a x +b \sqrt {x}}\, a^{\frac {3}{2}} b^{3}+96 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {7}{2}} \sqrt {x}-208 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {5}{2}} b \right )}{192 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {11}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(a*x+b*x^(1/2))^(1/2)*(96*x^(1/2)*(a*x+b*x^(1/2))^(3/2)*a^(7/2)+348*x^(1/2)*(a*x+b*x^(1/2))^(1/2)*a^(5/2
)*b^2-208*(a*x+b*x^(1/2))^(3/2)*a^(5/2)*b-384*((a*x^(1/2)+b)*x^(1/2))^(1/2)*a^(3/2)*b^3+174*(a*x+b*x^(1/2))^(1
/2)*a^(3/2)*b^3+192*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^4-87*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^4)/((a*x^(1/2)+b)*x^(1/2))^(1/2)/a^(11/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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