3.105 \(\int \frac{x^{7/2}}{(b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=80 \[ -\frac{16 b^2 \sqrt{x}}{3 c^3 \sqrt{b x+c x^2}}-\frac{8 b x^{3/2}}{3 c^2 \sqrt{b x+c x^2}}+\frac{2 x^{5/2}}{3 c \sqrt{b x+c x^2}} \]

[Out]

(-16*b^2*Sqrt[x])/(3*c^3*Sqrt[b*x + c*x^2]) - (8*b*x^(3/2))/(3*c^2*Sqrt[b*x + c*x^2]) + (2*x^(5/2))/(3*c*Sqrt[
b*x + c*x^2])

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Rubi [A]  time = 0.0275102, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {656, 648} \[ -\frac{16 b^2 \sqrt{x}}{3 c^3 \sqrt{b x+c x^2}}-\frac{8 b x^{3/2}}{3 c^2 \sqrt{b x+c x^2}}+\frac{2 x^{5/2}}{3 c \sqrt{b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^(7/2)/(b*x + c*x^2)^(3/2),x]

[Out]

(-16*b^2*Sqrt[x])/(3*c^3*Sqrt[b*x + c*x^2]) - (8*b*x^(3/2))/(3*c^2*Sqrt[b*x + c*x^2]) + (2*x^(5/2))/(3*c*Sqrt[
b*x + c*x^2])

Rule 656

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[(Simplify[m + p]*(2*c*d - b*e))/(c*(m + 2*p + 1)), In
t[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && E
qQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && IGtQ[Simplify[m + p], 0]

Rule 648

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] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c
*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p, 0]

Rubi steps

\begin{align*} \int \frac{x^{7/2}}{\left (b x+c x^2\right )^{3/2}} \, dx &=\frac{2 x^{5/2}}{3 c \sqrt{b x+c x^2}}-\frac{(4 b) \int \frac{x^{5/2}}{\left (b x+c x^2\right )^{3/2}} \, dx}{3 c}\\ &=-\frac{8 b x^{3/2}}{3 c^2 \sqrt{b x+c x^2}}+\frac{2 x^{5/2}}{3 c \sqrt{b x+c x^2}}+\frac{\left (8 b^2\right ) \int \frac{x^{3/2}}{\left (b x+c x^2\right )^{3/2}} \, dx}{3 c^2}\\ &=-\frac{16 b^2 \sqrt{x}}{3 c^3 \sqrt{b x+c x^2}}-\frac{8 b x^{3/2}}{3 c^2 \sqrt{b x+c x^2}}+\frac{2 x^{5/2}}{3 c \sqrt{b x+c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0184767, size = 41, normalized size = 0.51 \[ \frac{2 \sqrt{x} \left (-8 b^2-4 b c x+c^2 x^2\right )}{3 c^3 \sqrt{x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x]*(-8*b^2 - 4*b*c*x + c^2*x^2))/(3*c^3*Sqrt[x*(b + c*x)])

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Maple [A]  time = 0.047, size = 44, normalized size = 0.6 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( -{c}^{2}{x}^{2}+4\,bcx+8\,{b}^{2} \right ) }{3\,{c}^{3}}{x}^{{\frac{3}{2}}} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(c*x+b)*(-c^2*x^2+4*b*c*x+8*b^2)*x^(3/2)/c^3/(c*x^2+b*x)^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)/(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.99862, size = 107, normalized size = 1.34 \begin{align*} \frac{2 \,{\left (c^{2} x^{2} - 4 \, b c x - 8 \, b^{2}\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{3 \,{\left (c^{4} x^{2} + b c^{3} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(c^2*x^2 - 4*b*c*x - 8*b^2)*sqrt(c*x^2 + b*x)*sqrt(x)/(c^4*x^2 + b*c^3*x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.28931, size = 59, normalized size = 0.74 \begin{align*} \frac{16 \, b^{\frac{3}{2}}}{3 \, c^{3}} + \frac{2 \,{\left ({\left (c x + b\right )}^{\frac{3}{2}} - 6 \, \sqrt{c x + b} b - \frac{3 \, b^{2}}{\sqrt{c x + b}}\right )}}{3 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16/3*b^(3/2)/c^3 + 2/3*((c*x + b)^(3/2) - 6*sqrt(c*x + b)*b - 3*b^2/sqrt(c*x + b))/c^3