3.83 $$\int x^{5/2} (b x+c x^2)^{3/2} \, dx$$

Optimal. Leaf size=136 $\frac{256 b^4 \left (b x+c x^2\right )^{5/2}}{15015 c^5 x^{5/2}}-\frac{128 b^3 \left (b x+c x^2\right )^{5/2}}{3003 c^4 x^{3/2}}+\frac{32 b^2 \left (b x+c x^2\right )^{5/2}}{429 c^3 \sqrt{x}}-\frac{16 b \sqrt{x} \left (b x+c x^2\right )^{5/2}}{143 c^2}+\frac{2 x^{3/2} \left (b x+c x^2\right )^{5/2}}{13 c}$

[Out]

(256*b^4*(b*x + c*x^2)^(5/2))/(15015*c^5*x^(5/2)) - (128*b^3*(b*x + c*x^2)^(5/2))/(3003*c^4*x^(3/2)) + (32*b^2
*(b*x + c*x^2)^(5/2))/(429*c^3*Sqrt[x]) - (16*b*Sqrt[x]*(b*x + c*x^2)^(5/2))/(143*c^2) + (2*x^(3/2)*(b*x + c*x
^2)^(5/2))/(13*c)

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Rubi [A]  time = 0.0560172, 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} $\frac{256 b^4 \left (b x+c x^2\right )^{5/2}}{15015 c^5 x^{5/2}}-\frac{128 b^3 \left (b x+c x^2\right )^{5/2}}{3003 c^4 x^{3/2}}+\frac{32 b^2 \left (b x+c x^2\right )^{5/2}}{429 c^3 \sqrt{x}}-\frac{16 b \sqrt{x} \left (b x+c x^2\right )^{5/2}}{143 c^2}+\frac{2 x^{3/2} \left (b x+c x^2\right )^{5/2}}{13 c}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^(5/2)*(b*x + c*x^2)^(3/2),x]

[Out]

(256*b^4*(b*x + c*x^2)^(5/2))/(15015*c^5*x^(5/2)) - (128*b^3*(b*x + c*x^2)^(5/2))/(3003*c^4*x^(3/2)) + (32*b^2
*(b*x + c*x^2)^(5/2))/(429*c^3*Sqrt[x]) - (16*b*Sqrt[x]*(b*x + c*x^2)^(5/2))/(143*c^2) + (2*x^(3/2)*(b*x + c*x
^2)^(5/2))/(13*c)

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 x^{5/2} \left (b x+c x^2\right )^{3/2} \, dx &=\frac{2 x^{3/2} \left (b x+c x^2\right )^{5/2}}{13 c}-\frac{(8 b) \int x^{3/2} \left (b x+c x^2\right )^{3/2} \, dx}{13 c}\\ &=-\frac{16 b \sqrt{x} \left (b x+c x^2\right )^{5/2}}{143 c^2}+\frac{2 x^{3/2} \left (b x+c x^2\right )^{5/2}}{13 c}+\frac{\left (48 b^2\right ) \int \sqrt{x} \left (b x+c x^2\right )^{3/2} \, dx}{143 c^2}\\ &=\frac{32 b^2 \left (b x+c x^2\right )^{5/2}}{429 c^3 \sqrt{x}}-\frac{16 b \sqrt{x} \left (b x+c x^2\right )^{5/2}}{143 c^2}+\frac{2 x^{3/2} \left (b x+c x^2\right )^{5/2}}{13 c}-\frac{\left (64 b^3\right ) \int \frac{\left (b x+c x^2\right )^{3/2}}{\sqrt{x}} \, dx}{429 c^3}\\ &=-\frac{128 b^3 \left (b x+c x^2\right )^{5/2}}{3003 c^4 x^{3/2}}+\frac{32 b^2 \left (b x+c x^2\right )^{5/2}}{429 c^3 \sqrt{x}}-\frac{16 b \sqrt{x} \left (b x+c x^2\right )^{5/2}}{143 c^2}+\frac{2 x^{3/2} \left (b x+c x^2\right )^{5/2}}{13 c}+\frac{\left (128 b^4\right ) \int \frac{\left (b x+c x^2\right )^{3/2}}{x^{3/2}} \, dx}{3003 c^4}\\ &=\frac{256 b^4 \left (b x+c x^2\right )^{5/2}}{15015 c^5 x^{5/2}}-\frac{128 b^3 \left (b x+c x^2\right )^{5/2}}{3003 c^4 x^{3/2}}+\frac{32 b^2 \left (b x+c x^2\right )^{5/2}}{429 c^3 \sqrt{x}}-\frac{16 b \sqrt{x} \left (b x+c x^2\right )^{5/2}}{143 c^2}+\frac{2 x^{3/2} \left (b x+c x^2\right )^{5/2}}{13 c}\\ \end{align*}

Mathematica [A]  time = 0.0381884, size = 64, normalized size = 0.47 $\frac{2 (x (b+c x))^{5/2} \left (560 b^2 c^2 x^2-320 b^3 c x+128 b^4-840 b c^3 x^3+1155 c^4 x^4\right )}{15015 c^5 x^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^(5/2)*(b*x + c*x^2)^(3/2),x]

[Out]

(2*(x*(b + c*x))^(5/2)*(128*b^4 - 320*b^3*c*x + 560*b^2*c^2*x^2 - 840*b*c^3*x^3 + 1155*c^4*x^4))/(15015*c^5*x^
(5/2))

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Maple [A]  time = 0.047, size = 66, normalized size = 0.5 \begin{align*}{\frac{ \left ( 2\,cx+2\,b \right ) \left ( 1155\,{x}^{4}{c}^{4}-840\,b{x}^{3}{c}^{3}+560\,{b}^{2}{x}^{2}{c}^{2}-320\,{b}^{3}xc+128\,{b}^{4} \right ) }{15015\,{c}^{5}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}{x}^{-{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(c*x^2+b*x)^(3/2),x)

[Out]

2/15015*(c*x+b)*(1155*c^4*x^4-840*b*c^3*x^3+560*b^2*c^2*x^2-320*b^3*c*x+128*b^4)*(c*x^2+b*x)^(3/2)/c^5/x^(3/2)

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Maxima [A]  time = 1.06191, size = 198, normalized size = 1.46 \begin{align*} \frac{2 \,{\left (5 \,{\left (693 \, c^{6} x^{6} + 63 \, b c^{5} x^{5} - 70 \, b^{2} c^{4} x^{4} + 80 \, b^{3} c^{3} x^{3} - 96 \, b^{4} c^{2} x^{2} + 128 \, b^{5} c x - 256 \, b^{6}\right )} x^{5} + 13 \,{\left (315 \, b c^{5} x^{6} + 35 \, b^{2} c^{4} x^{5} - 40 \, b^{3} c^{3} x^{4} + 48 \, b^{4} c^{2} x^{3} - 64 \, b^{5} c x^{2} + 128 \, b^{6} x\right )} x^{4}\right )} \sqrt{c x + b}}{45045 \, c^{5} x^{5}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

2/45045*(5*(693*c^6*x^6 + 63*b*c^5*x^5 - 70*b^2*c^4*x^4 + 80*b^3*c^3*x^3 - 96*b^4*c^2*x^2 + 128*b^5*c*x - 256*
b^6)*x^5 + 13*(315*b*c^5*x^6 + 35*b^2*c^4*x^5 - 40*b^3*c^3*x^4 + 48*b^4*c^2*x^3 - 64*b^5*c*x^2 + 128*b^6*x)*x^
4)*sqrt(c*x + b)/(c^5*x^5)

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Fricas [A]  time = 2.01196, size = 198, normalized size = 1.46 \begin{align*} \frac{2 \,{\left (1155 \, c^{6} x^{6} + 1470 \, b c^{5} x^{5} + 35 \, b^{2} c^{4} x^{4} - 40 \, b^{3} c^{3} x^{3} + 48 \, b^{4} c^{2} x^{2} - 64 \, b^{5} c x + 128 \, b^{6}\right )} \sqrt{c x^{2} + b x}}{15015 \, c^{5} \sqrt{x}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

2/15015*(1155*c^6*x^6 + 1470*b*c^5*x^5 + 35*b^2*c^4*x^4 - 40*b^3*c^3*x^3 + 48*b^4*c^2*x^2 - 64*b^5*c*x + 128*b
^6)*sqrt(c*x^2 + b*x)/(c^5*sqrt(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**(5/2)*(c*x**2+b*x)**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.22036, size = 213, normalized size = 1.57 \begin{align*} \frac{2}{9009} \, c{\left (\frac{256 \, b^{\frac{13}{2}}}{c^{6}} + \frac{693 \,{\left (c x + b\right )}^{\frac{13}{2}} - 4095 \,{\left (c x + b\right )}^{\frac{11}{2}} b + 10010 \,{\left (c x + b\right )}^{\frac{9}{2}} b^{2} - 12870 \,{\left (c x + b\right )}^{\frac{7}{2}} b^{3} + 9009 \,{\left (c x + b\right )}^{\frac{5}{2}} b^{4} - 3003 \,{\left (c x + b\right )}^{\frac{3}{2}} b^{5}}{c^{6}}\right )} - \frac{2}{3465} \, b{\left (\frac{128 \, b^{\frac{11}{2}}}{c^{5}} - \frac{315 \,{\left (c x + b\right )}^{\frac{11}{2}} - 1540 \,{\left (c x + b\right )}^{\frac{9}{2}} b + 2970 \,{\left (c x + b\right )}^{\frac{7}{2}} b^{2} - 2772 \,{\left (c x + b\right )}^{\frac{5}{2}} b^{3} + 1155 \,{\left (c x + b\right )}^{\frac{3}{2}} b^{4}}{c^{5}}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(c*x^2+b*x)^(3/2),x, algorithm="giac")

[Out]

2/9009*c*(256*b^(13/2)/c^6 + (693*(c*x + b)^(13/2) - 4095*(c*x + b)^(11/2)*b + 10010*(c*x + b)^(9/2)*b^2 - 128
70*(c*x + b)^(7/2)*b^3 + 9009*(c*x + b)^(5/2)*b^4 - 3003*(c*x + b)^(3/2)*b^5)/c^6) - 2/3465*b*(128*b^(11/2)/c^
5 - (315*(c*x + b)^(11/2) - 1540*(c*x + b)^(9/2)*b + 2970*(c*x + b)^(7/2)*b^2 - 2772*(c*x + b)^(5/2)*b^3 + 115
5*(c*x + b)^(3/2)*b^4)/c^5)