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

Optimal. Leaf size=125 \[ -\frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a x^2+b x^3}}\right )}{8 b^{7/2}}+\frac {5 a^2 \sqrt {a x^2+b x^3}}{8 b^3 \sqrt {x}}-\frac {5 a \sqrt {x} \sqrt {a x^2+b x^3}}{12 b^2}+\frac {x^{3/2} \sqrt {a x^2+b x^3}}{3 b} \]

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Rubi [A]  time = 0.17, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2024, 2029, 206} \begin {gather*} \frac {5 a^2 \sqrt {a x^2+b x^3}}{8 b^3 \sqrt {x}}-\frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a x^2+b x^3}}\right )}{8 b^{7/2}}-\frac {5 a \sqrt {x} \sqrt {a x^2+b x^3}}{12 b^2}+\frac {x^{3/2} \sqrt {a x^2+b x^3}}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(7/2)/Sqrt[a*x^2 + b*x^3],x]

[Out]

(5*a^2*Sqrt[a*x^2 + b*x^3])/(8*b^3*Sqrt[x]) - (5*a*Sqrt[x]*Sqrt[a*x^2 + b*x^3])/(12*b^2) + (x^(3/2)*Sqrt[a*x^2
+ b*x^3])/(3*b) - (5*a^3*ArcTanh[(Sqrt[b]*x^(3/2))/Sqrt[a*x^2 + b*x^3]])/(8*b^(7/2))

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 2024

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n +
1)*(a*x^j + b*x^n)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^(n - j)*(m + j*p - n + j + 1))/(b*(m + n*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]) && GtQ[m + j*p + 1 - n + j, 0] && NeQ[m + n*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 {x^{7/2}}{\sqrt {a x^2+b x^3}} \, dx &=\frac {x^{3/2} \sqrt {a x^2+b x^3}}{3 b}-\frac {(5 a) \int \frac {x^{5/2}}{\sqrt {a x^2+b x^3}} \, dx}{6 b}\\ &=-\frac {5 a \sqrt {x} \sqrt {a x^2+b x^3}}{12 b^2}+\frac {x^{3/2} \sqrt {a x^2+b x^3}}{3 b}+\frac {\left (5 a^2\right ) \int \frac {x^{3/2}}{\sqrt {a x^2+b x^3}} \, dx}{8 b^2}\\ &=\frac {5 a^2 \sqrt {a x^2+b x^3}}{8 b^3 \sqrt {x}}-\frac {5 a \sqrt {x} \sqrt {a x^2+b x^3}}{12 b^2}+\frac {x^{3/2} \sqrt {a x^2+b x^3}}{3 b}-\frac {\left (5 a^3\right ) \int \frac {\sqrt {x}}{\sqrt {a x^2+b x^3}} \, dx}{16 b^3}\\ &=\frac {5 a^2 \sqrt {a x^2+b x^3}}{8 b^3 \sqrt {x}}-\frac {5 a \sqrt {x} \sqrt {a x^2+b x^3}}{12 b^2}+\frac {x^{3/2} \sqrt {a x^2+b x^3}}{3 b}-\frac {\left (5 a^3\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x^{3/2}}{\sqrt {a x^2+b x^3}}\right )}{8 b^3}\\ &=\frac {5 a^2 \sqrt {a x^2+b x^3}}{8 b^3 \sqrt {x}}-\frac {5 a \sqrt {x} \sqrt {a x^2+b x^3}}{12 b^2}+\frac {x^{3/2} \sqrt {a x^2+b x^3}}{3 b}-\frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a x^2+b x^3}}\right )}{8 b^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 104, normalized size = 0.83 \begin {gather*} \frac {\sqrt {x^2 (a+b x)} \left (\sqrt {b} \sqrt {x} \sqrt {\frac {b x}{a}+1} \left (15 a^2-10 a b x+8 b^2 x^2\right )-15 a^{5/2} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )\right )}{24 b^{7/2} x \sqrt {\frac {b x}{a}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x^2*(a + b*x)]*(Sqrt[b]*Sqrt[x]*Sqrt[1 + (b*x)/a]*(15*a^2 - 10*a*b*x + 8*b^2*x^2) - 15*a^(5/2)*ArcSinh[(
Sqrt[b]*Sqrt[x])/Sqrt[a]]))/(24*b^(7/2)*x*Sqrt[1 + (b*x)/a])

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IntegrateAlgebraic [A]  time = 0.18, size = 106, normalized size = 0.85 \begin {gather*} \frac {5 a^3 \log \left (\sqrt {a x^2+b x^3}-\sqrt {b} x^{3/2}\right )}{8 b^{7/2}}-\frac {5 a^3 \log \left (\sqrt {x}\right )}{4 b^{7/2}}+\frac {\left (15 a^2-10 a b x+8 b^2 x^2\right ) \sqrt {a x^2+b x^3}}{24 b^3 \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((15*a^2 - 10*a*b*x + 8*b^2*x^2)*Sqrt[a*x^2 + b*x^3])/(24*b^3*Sqrt[x]) - (5*a^3*Log[Sqrt[x]])/(4*b^(7/2)) + (5
*a^3*Log[-(Sqrt[b]*x^(3/2)) + Sqrt[a*x^2 + b*x^3]])/(8*b^(7/2))

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fricas [A]  time = 0.42, size = 180, normalized size = 1.44 \begin {gather*} \left [\frac {15 \, a^{3} \sqrt {b} x \log \left (\frac {2 \, b x^{2} + a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {b} \sqrt {x}}{x}\right ) + 2 \, {\left (8 \, b^{3} x^{2} - 10 \, a b^{2} x + 15 \, a^{2} b\right )} \sqrt {b x^{3} + a x^{2}} \sqrt {x}}{48 \, b^{4} x}, \frac {15 \, a^{3} \sqrt {-b} x \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-b}}{b x^{\frac {3}{2}}}\right ) + {\left (8 \, b^{3} x^{2} - 10 \, a b^{2} x + 15 \, a^{2} b\right )} \sqrt {b x^{3} + a x^{2}} \sqrt {x}}{24 \, b^{4} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)/(b*x^3+a*x^2)^(1/2),x, algorithm="fricas")

[Out]

[1/48*(15*a^3*sqrt(b)*x*log((2*b*x^2 + a*x - 2*sqrt(b*x^3 + a*x^2)*sqrt(b)*sqrt(x))/x) + 2*(8*b^3*x^2 - 10*a*b
^2*x + 15*a^2*b)*sqrt(b*x^3 + a*x^2)*sqrt(x))/(b^4*x), 1/24*(15*a^3*sqrt(-b)*x*arctan(sqrt(b*x^3 + a*x^2)*sqrt
(-b)/(b*x^(3/2))) + (8*b^3*x^2 - 10*a*b^2*x + 15*a^2*b)*sqrt(b*x^3 + a*x^2)*sqrt(x))/(b^4*x)]

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giac [A]  time = 0.20, size = 64, normalized size = 0.51 \begin {gather*} \frac {1}{24} \, \sqrt {b x + a} {\left (2 \, x {\left (\frac {4 \, x}{b} - \frac {5 \, a}{b^{2}}\right )} + \frac {15 \, a^{2}}{b^{3}}\right )} \sqrt {x} + \frac {5 \, a^{3} \log \left ({\left | -\sqrt {b} \sqrt {x} + \sqrt {b x + a} \right |}\right )}{8 \, b^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)/(b*x^3+a*x^2)^(1/2),x, algorithm="giac")

[Out]

1/24*sqrt(b*x + a)*(2*x*(4*x/b - 5*a/b^2) + 15*a^2/b^3)*sqrt(x) + 5/8*a^3*log(abs(-sqrt(b)*sqrt(x) + sqrt(b*x
+ a)))/b^(7/2)

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maple [A]  time = 0.07, size = 103, normalized size = 0.82 \begin {gather*} -\frac {\left (-16 b^{\frac {9}{2}} x^{4}+4 a \,b^{\frac {7}{2}} x^{3}-10 a^{2} b^{\frac {5}{2}} x^{2}-30 a^{3} b^{\frac {3}{2}} x +15 \sqrt {\left (b x +a \right ) x}\, a^{3} b \ln \left (\frac {2 b x +a +2 \sqrt {b \,x^{2}+a x}\, \sqrt {b}}{2 \sqrt {b}}\right )\right ) \sqrt {x}}{48 \sqrt {b \,x^{3}+a \,x^{2}}\, b^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)/(b*x^3+a*x^2)^(1/2),x)

[Out]

-1/48*x^(1/2)*(-16*b^(9/2)*x^4+4*b^(7/2)*x^3*a-10*b^(5/2)*x^2*a^2-30*b^(3/2)*x*a^3+15*(x*(b*x+a))^(1/2)*ln(1/2
*(2*(b*x^2+a*x)^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*a^3*b)/(b*x^3+a*x^2)^(1/2)/b^(9/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{\frac {7}{2}}}{\sqrt {b x^{3} + a x^{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^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^(7/2)/sqrt(b*x^3 + a*x^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{7/2}}{\sqrt {b\,x^3+a\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)/(a*x^2 + b*x^3)^(1/2),x)

[Out]

int(x^(7/2)/(a*x^2 + b*x^3)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{\frac {7}{2}}}{\sqrt {x^{2} \left (a + b x\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(7/2)/(b*x**3+a*x**2)**(1/2),x)

[Out]

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

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