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

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

[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))

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Rubi [A]  time = 0.169449, antiderivative size = 125, normalized size of antiderivative = 1., 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} \[ \frac{5 a^2 \sqrt{a x^2+b x^3}}{8 b^3 \sqrt{x}}-\frac{5 a^3 \tanh ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a x^2+b x^3}}\right )}{8 b^{7/2}}-\frac{5 a \sqrt{x} \sqrt{a x^2+b x^3}}{12 b^2}+\frac{x^{3/2} \sqrt{a x^2+b x^3}}{3 b} \]

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 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]

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])

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*}

Mathematica [A]  time = 0.108556, size = 104, normalized size = 0.83 \[ \frac{\sqrt{x^2 (a+b x)} \left (\sqrt{b} \sqrt{x} \sqrt{\frac{b x}{a}+1} \left (15 a^2-10 a b x+8 b^2 x^2\right )-15 a^{5/2} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )\right )}{24 b^{7/2} x \sqrt{\frac{b x}{a}+1}} \]

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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Maple [A]  time = 0.006, size = 103, normalized size = 0.8 \begin{align*} -{\frac{1}{48}\sqrt{x} \left ( -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\,\sqrt{x \left ( bx+a \right ) }\ln \left ( 1/2\,{\frac{2\,\sqrt{b{x}^{2}+ax}\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ){a}^{3}b \right ){\frac{1}{\sqrt{b{x}^{3}+a{x}^{2}}}}{b}^{-{\frac{9}{2}}}} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{\frac{7}{2}}}{\sqrt{b x^{3} + a x^{2}}}\,{d x} \end{align*}

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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Fricas [A]  time = 0.792716, size = 428, normalized size = 3.42 \begin{align*} \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{align*}

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

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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Giac [A]  time = 1.30755, size = 86, normalized size = 0.69 \begin{align*} \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{align*}

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)