3.1.83 \(\int x^{5/2} (b x+c x^2)^{3/2} \, dx\) [83]

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]

256/15015*b^4*(c*x^2+b*x)^(5/2)/c^5/x^(5/2)-128/3003*b^3*(c*x^2+b*x)^(5/2)/c^4/x^(3/2)+2/13*x^(3/2)*(c*x^2+b*x
)^(5/2)/c+32/429*b^2*(c*x^2+b*x)^(5/2)/c^3/x^(1/2)-16/143*b*(c*x^2+b*x)^(5/2)*x^(1/2)/c^2

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Rubi [A]
time = 0.04, antiderivative size = 136, normalized size of antiderivative = 1.00, 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 = {670, 662} \begin {gather*} \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 {gather*}

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 662

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 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[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 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*}

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Mathematica [A]
time = 0.03, size = 64, normalized size = 0.47 \begin {gather*} \frac {2 (x (b+c x))^{5/2} \left (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\right )}{15015 c^5 x^{5/2}} \end {gather*}

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 = 0.43, size = 66, normalized size = 0.49

method	result	size
gosper	\(\frac {2 \left (c x +b \right ) \left (1155 c^{4} x^{4}-840 b \,c^{3} x^{3}+560 b^{2} c^{2} x^{2}-320 b^{3} c x +128 b^{4}\right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{15015 c^{5} x^{\frac {3}{2}}}\)	\(66\)
default	\(\frac {2 \sqrt {x \left (c x +b \right )}\, \left (c x +b \right )^{2} \left (1155 c^{4} x^{4}-840 b \,c^{3} x^{3}+560 b^{2} c^{2} x^{2}-320 b^{3} c x +128 b^{4}\right )}{15015 \sqrt {x}\, c^{5}}\)	\(66\)
risch	\(\frac {2 \left (c x +b \right ) \sqrt {x}\, \left (1155 x^{6} c^{6}+1470 b \,c^{5} x^{5}+35 b^{2} c^{4} x^{4}-40 b^{3} c^{3} x^{3}+48 b^{4} x^{2} c^{2}-64 b^{5} c x +128 b^{6}\right )}{15015 \sqrt {x \left (c x +b \right )}\, c^{5}}\)	\(86\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15015/x^(1/2)*(x*(c*x+b))^(1/2)*(c*x+b)^2*(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
^5

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Maxima [A]
time = 0.28, size = 147, normalized size = 1.08 \begin {gather*} \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 {gather*}

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 = 1.79, size = 82, normalized size = 0.60 \begin {gather*} \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 {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{\frac {5}{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**(5/2)*(c*x**2+b*x)**(3/2),x)

[Out]

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

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Giac [A]
time = 1.83, size = 158, normalized size = 1.16 \begin {gather*} \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 {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^{5/2}\,{\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^(5/2)*(b*x + c*x^2)^(3/2),x)

[Out]

int(x^(5/2)*(b*x + c*x^2)^(3/2), x)

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