3.52 \(\int \frac{x^4}{(b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=97 \[ \frac{15 b^2 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{4 c^{7/2}}+\frac{5 x \sqrt{b x+c x^2}}{2 c^2}-\frac{15 b \sqrt{b x+c x^2}}{4 c^3}-\frac{2 x^3}{c \sqrt{b x+c x^2}} \]

[Out]

(-2*x^3)/(c*Sqrt[b*x + c*x^2]) - (15*b*Sqrt[b*x + c*x^2])/(4*c^3) + (5*x*Sqrt[b*x + c*x^2])/(2*c^2) + (15*b^2*
ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(4*c^(7/2))

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Rubi [A]  time = 0.0398188, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {668, 670, 640, 620, 206} \[ \frac{15 b^2 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{4 c^{7/2}}+\frac{5 x \sqrt{b x+c x^2}}{2 c^2}-\frac{15 b \sqrt{b x+c x^2}}{4 c^3}-\frac{2 x^3}{c \sqrt{b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^4/(b*x + c*x^2)^(3/2),x]

[Out]

(-2*x^3)/(c*Sqrt[b*x + c*x^2]) - (15*b*Sqrt[b*x + c*x^2])/(4*c^3) + (5*x*Sqrt[b*x + c*x^2])/(2*c^2) + (15*b^2*
ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(4*c^(7/2))

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 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 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 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^4}{\left (b x+c x^2\right )^{3/2}} \, dx &=-\frac{2 x^3}{c \sqrt{b x+c x^2}}+\frac{5 \int \frac{x^2}{\sqrt{b x+c x^2}} \, dx}{c}\\ &=-\frac{2 x^3}{c \sqrt{b x+c x^2}}+\frac{5 x \sqrt{b x+c x^2}}{2 c^2}-\frac{(15 b) \int \frac{x}{\sqrt{b x+c x^2}} \, dx}{4 c^2}\\ &=-\frac{2 x^3}{c \sqrt{b x+c x^2}}-\frac{15 b \sqrt{b x+c x^2}}{4 c^3}+\frac{5 x \sqrt{b x+c x^2}}{2 c^2}+\frac{\left (15 b^2\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{8 c^3}\\ &=-\frac{2 x^3}{c \sqrt{b x+c x^2}}-\frac{15 b \sqrt{b x+c x^2}}{4 c^3}+\frac{5 x \sqrt{b x+c x^2}}{2 c^2}+\frac{\left (15 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )}{4 c^3}\\ &=-\frac{2 x^3}{c \sqrt{b x+c x^2}}-\frac{15 b \sqrt{b x+c x^2}}{4 c^3}+\frac{5 x \sqrt{b x+c x^2}}{2 c^2}+\frac{15 b^2 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{4 c^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0123995, size = 50, normalized size = 0.52 \[ \frac{2 x^4 \sqrt{\frac{c x}{b}+1} \, _2F_1\left (\frac{3}{2},\frac{7}{2};\frac{9}{2};-\frac{c x}{b}\right )}{7 b \sqrt{x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/(b*x + c*x^2)^(3/2),x]

[Out]

(2*x^4*Sqrt[1 + (c*x)/b]*Hypergeometric2F1[3/2, 7/2, 9/2, -((c*x)/b)])/(7*b*Sqrt[x*(b + c*x)])

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Maple [A]  time = 0.063, size = 93, normalized size = 1. \begin{align*}{\frac{{x}^{3}}{2\,c}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}-{\frac{5\,b{x}^{2}}{4\,{c}^{2}}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}-{\frac{15\,{b}^{2}x}{4\,{c}^{3}}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}+{\frac{15\,{b}^{2}}{8}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(c*x^2+b*x)^(3/2),x)

[Out]

1/2*x^3/c/(c*x^2+b*x)^(1/2)-5/4*b/c^2*x^2/(c*x^2+b*x)^(1/2)-15/4*b^2/c^3/(c*x^2+b*x)^(1/2)*x+15/8*b^2/c^(7/2)*
ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.08516, size = 410, normalized size = 4.23 \begin{align*} \left [\frac{15 \,{\left (b^{2} c x + b^{3}\right )} \sqrt{c} \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) + 2 \,{\left (2 \, c^{3} x^{2} - 5 \, b c^{2} x - 15 \, b^{2} c\right )} \sqrt{c x^{2} + b x}}{8 \,{\left (c^{5} x + b c^{4}\right )}}, -\frac{15 \,{\left (b^{2} c x + b^{3}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) -{\left (2 \, c^{3} x^{2} - 5 \, b c^{2} x - 15 \, b^{2} c\right )} \sqrt{c x^{2} + b x}}{4 \,{\left (c^{5} x + b c^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

[1/8*(15*(b^2*c*x + b^3)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(2*c^3*x^2 - 5*b*c^2*x - 15*
b^2*c)*sqrt(c*x^2 + b*x))/(c^5*x + b*c^4), -1/4*(15*(b^2*c*x + b^3)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)
/(c*x)) - (2*c^3*x^2 - 5*b*c^2*x - 15*b^2*c)*sqrt(c*x^2 + b*x))/(c^5*x + b*c^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\left (x \left (b + c x\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(c*x**2+b*x)**(3/2),x)

[Out]

Integral(x**4/(x*(b + c*x))**(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(c*x^2+b*x)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError