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

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

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Rubi [A]  time = 0.03, antiderivative size = 80, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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 {gather*}

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 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]

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]

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*}

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Mathematica [A]  time = 0.02, size = 41, normalized size = 0.51 \begin {gather*} \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)}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.48, size = 50, normalized size = 0.62 \begin {gather*} \frac {2 \sqrt {b x+c x^2} \left (-8 b^2-4 b c x+c^2 x^2\right )}{3 c^3 \sqrt {x} (b+c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.40, size = 50, normalized size = 0.62 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.18, size = 54, normalized size = 0.68 \begin {gather*} \frac {16 \, b^{\frac {3}{2}}}{3 \, c^{3}} - \frac {2 \, b^{2}}{\sqrt {c x + b} c^{3}} + \frac {2 \, {\left ({\left (c x + b\right )}^{\frac {3}{2}} c^{6} - 6 \, \sqrt {c x + b} b c^{6}\right )}}{3 \, c^{9}} \end {gather*}

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*b^2/(sqrt(c*x + b)*c^3) + 2/3*((c*x + b)^(3/2)*c^6 - 6*sqrt(c*x + b)*b*c^6)/c^9

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maple [A]  time = 0.04, size = 44, normalized size = 0.55 \begin {gather*} -\frac {2 \left (c x +b \right ) \left (-c^{2} x^{2}+4 b c x +8 b^{2}\right ) x^{\frac {3}{2}}}{3 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} c^{3}} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)/(b*x + c*x^2)^(3/2),x)

[Out]

int(x^(7/2)/(b*x + c*x^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{\frac {7}{2}}}{\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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