### 3.103 $$\int \frac{x^{11/2}}{(b x+c x^2)^{3/2}} \, dx$$

Optimal. Leaf size=136 $\frac{32 b^2 x^{5/2}}{35 c^3 \sqrt{b x+c x^2}}-\frac{128 b^3 x^{3/2}}{35 c^4 \sqrt{b x+c x^2}}-\frac{256 b^4 \sqrt{x}}{35 c^5 \sqrt{b x+c x^2}}-\frac{16 b x^{7/2}}{35 c^2 \sqrt{b x+c x^2}}+\frac{2 x^{9/2}}{7 c \sqrt{b x+c x^2}}$

[Out]

(-256*b^4*Sqrt[x])/(35*c^5*Sqrt[b*x + c*x^2]) - (128*b^3*x^(3/2))/(35*c^4*Sqrt[b*x + c*x^2]) + (32*b^2*x^(5/2)
)/(35*c^3*Sqrt[b*x + c*x^2]) - (16*b*x^(7/2))/(35*c^2*Sqrt[b*x + c*x^2]) + (2*x^(9/2))/(7*c*Sqrt[b*x + c*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.0567674, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 2, integrand size = 19, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.105, Rules used = {656, 648} $\frac{32 b^2 x^{5/2}}{35 c^3 \sqrt{b x+c x^2}}-\frac{128 b^3 x^{3/2}}{35 c^4 \sqrt{b x+c x^2}}-\frac{256 b^4 \sqrt{x}}{35 c^5 \sqrt{b x+c x^2}}-\frac{16 b x^{7/2}}{35 c^2 \sqrt{b x+c x^2}}+\frac{2 x^{9/2}}{7 c \sqrt{b x+c x^2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^(11/2)/(b*x + c*x^2)^(3/2),x]

[Out]

(-256*b^4*Sqrt[x])/(35*c^5*Sqrt[b*x + c*x^2]) - (128*b^3*x^(3/2))/(35*c^4*Sqrt[b*x + c*x^2]) + (32*b^2*x^(5/2)
)/(35*c^3*Sqrt[b*x + c*x^2]) - (16*b*x^(7/2))/(35*c^2*Sqrt[b*x + c*x^2]) + (2*x^(9/2))/(7*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^{11/2}}{\left (b x+c x^2\right )^{3/2}} \, dx &=\frac{2 x^{9/2}}{7 c \sqrt{b x+c x^2}}-\frac{(8 b) \int \frac{x^{9/2}}{\left (b x+c x^2\right )^{3/2}} \, dx}{7 c}\\ &=-\frac{16 b x^{7/2}}{35 c^2 \sqrt{b x+c x^2}}+\frac{2 x^{9/2}}{7 c \sqrt{b x+c x^2}}+\frac{\left (48 b^2\right ) \int \frac{x^{7/2}}{\left (b x+c x^2\right )^{3/2}} \, dx}{35 c^2}\\ &=\frac{32 b^2 x^{5/2}}{35 c^3 \sqrt{b x+c x^2}}-\frac{16 b x^{7/2}}{35 c^2 \sqrt{b x+c x^2}}+\frac{2 x^{9/2}}{7 c \sqrt{b x+c x^2}}-\frac{\left (64 b^3\right ) \int \frac{x^{5/2}}{\left (b x+c x^2\right )^{3/2}} \, dx}{35 c^3}\\ &=-\frac{128 b^3 x^{3/2}}{35 c^4 \sqrt{b x+c x^2}}+\frac{32 b^2 x^{5/2}}{35 c^3 \sqrt{b x+c x^2}}-\frac{16 b x^{7/2}}{35 c^2 \sqrt{b x+c x^2}}+\frac{2 x^{9/2}}{7 c \sqrt{b x+c x^2}}+\frac{\left (128 b^4\right ) \int \frac{x^{3/2}}{\left (b x+c x^2\right )^{3/2}} \, dx}{35 c^4}\\ &=-\frac{256 b^4 \sqrt{x}}{35 c^5 \sqrt{b x+c x^2}}-\frac{128 b^3 x^{3/2}}{35 c^4 \sqrt{b x+c x^2}}+\frac{32 b^2 x^{5/2}}{35 c^3 \sqrt{b x+c x^2}}-\frac{16 b x^{7/2}}{35 c^2 \sqrt{b x+c x^2}}+\frac{2 x^{9/2}}{7 c \sqrt{b x+c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0291625, size = 64, normalized size = 0.47 $\frac{2 \sqrt{x} \left (16 b^2 c^2 x^2-64 b^3 c x-128 b^4-8 b c^3 x^3+5 c^4 x^4\right )}{35 c^5 \sqrt{x (b+c x)}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^(11/2)/(b*x + c*x^2)^(3/2),x]

[Out]

(2*Sqrt[x]*(-128*b^4 - 64*b^3*c*x + 16*b^2*c^2*x^2 - 8*b*c^3*x^3 + 5*c^4*x^4))/(35*c^5*Sqrt[x*(b + c*x)])

________________________________________________________________________________________

Maple [A]  time = 0.046, size = 66, normalized size = 0.5 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( -5\,{x}^{4}{c}^{4}+8\,b{x}^{3}{c}^{3}-16\,{b}^{2}{x}^{2}{c}^{2}+64\,{b}^{3}xc+128\,{b}^{4} \right ) }{35\,{c}^{5}}{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^(11/2)/(c*x^2+b*x)^(3/2),x)

[Out]

-2/35*(c*x+b)*(-5*c^4*x^4+8*b*c^3*x^3-16*b^2*c^2*x^2+64*b^3*c*x+128*b^4)*x^(3/2)/c^5/(c*x^2+b*x)^(3/2)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{2 \,{\left (3 \,{\left (5 \, c^{5} x^{4} - b c^{4} x^{3} + 2 \, b^{2} c^{3} x^{2} - 8 \, b^{3} c^{2} x - 16 \, b^{4} c\right )} x^{4} - 2 \,{\left (3 \, b c^{4} x^{4} - 2 \, b^{2} c^{3} x^{3} + 11 \, b^{3} c^{2} x^{2} + 40 \, b^{4} c x + 24 \, b^{5}\right )} x^{3} + 14 \,{\left (b^{2} c^{3} x^{4} - 2 \, b^{3} c^{2} x^{3} - 7 \, b^{4} c x^{2} - 4 \, b^{5} x\right )} x^{2} - 70 \,{\left (b^{3} c^{2} x^{4} + 2 \, b^{4} c x^{3} + b^{5} x^{2}\right )} x\right )}}{105 \,{\left (c^{6} x^{4} + b c^{5} x^{3}\right )} \sqrt{c x + b}} + \int \frac{2 \,{\left (b^{4} c x + b^{5}\right )} x}{{\left (c^{6} x^{3} + 2 \, b c^{5} x^{2} + b^{2} c^{4} x\right )} \sqrt{c x + b}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(11/2)/(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

2/105*(3*(5*c^5*x^4 - b*c^4*x^3 + 2*b^2*c^3*x^2 - 8*b^3*c^2*x - 16*b^4*c)*x^4 - 2*(3*b*c^4*x^4 - 2*b^2*c^3*x^3
+ 11*b^3*c^2*x^2 + 40*b^4*c*x + 24*b^5)*x^3 + 14*(b^2*c^3*x^4 - 2*b^3*c^2*x^3 - 7*b^4*c*x^2 - 4*b^5*x)*x^2 -
70*(b^3*c^2*x^4 + 2*b^4*c*x^3 + b^5*x^2)*x)/((c^6*x^4 + b*c^5*x^3)*sqrt(c*x + b)) + integrate(2*(b^4*c*x + b^5
)*x/((c^6*x^3 + 2*b*c^5*x^2 + b^2*c^4*x)*sqrt(c*x + b)), x)

________________________________________________________________________________________

Fricas [A]  time = 1.98118, size = 159, normalized size = 1.17 \begin{align*} \frac{2 \,{\left (5 \, c^{4} x^{4} - 8 \, b c^{3} x^{3} + 16 \, b^{2} c^{2} x^{2} - 64 \, b^{3} c x - 128 \, b^{4}\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{35 \,{\left (c^{6} x^{2} + b c^{5} x\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(11/2)/(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

2/35*(5*c^4*x^4 - 8*b*c^3*x^3 + 16*b^2*c^2*x^2 - 64*b^3*c*x - 128*b^4)*sqrt(c*x^2 + b*x)*sqrt(x)/(c^6*x^2 + b*
c^5*x)

________________________________________________________________________________________

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**(11/2)/(c*x**2+b*x)**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.17735, size = 95, normalized size = 0.7 \begin{align*} \frac{256 \, b^{\frac{7}{2}}}{35 \, c^{5}} + \frac{2 \,{\left (5 \,{\left (c x + b\right )}^{\frac{7}{2}} - 28 \,{\left (c x + b\right )}^{\frac{5}{2}} b + 70 \,{\left (c x + b\right )}^{\frac{3}{2}} b^{2} - 140 \, \sqrt{c x + b} b^{3} - \frac{35 \, b^{4}}{\sqrt{c x + b}}\right )}}{35 \, c^{5}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(11/2)/(c*x^2+b*x)^(3/2),x, algorithm="giac")

[Out]

256/35*b^(7/2)/c^5 + 2/35*(5*(c*x + b)^(7/2) - 28*(c*x + b)^(5/2)*b + 70*(c*x + b)^(3/2)*b^2 - 140*sqrt(c*x +
b)*b^3 - 35*b^4/sqrt(c*x + b))/c^5