\(\int x^{3/2} (b x+c x^2)^{3/2} \, dx\) [84]

Optimal result

Integrand size = 19, antiderivative size = 108 \[ \int x^{3/2} \left (b x+c x^2\right )^{3/2} \, dx=-\frac {32 b^3 \left (b x+c x^2\right )^{5/2}}{1155 c^4 x^{5/2}}+\frac {16 b^2 \left (b x+c x^2\right )^{5/2}}{231 c^3 x^{3/2}}-\frac {4 b \left (b x+c x^2\right )^{5/2}}{33 c^2 \sqrt {x}}+\frac {2 \sqrt {x} \left (b x+c x^2\right )^{5/2}}{11 c} \]

[Out]

-32/1155*b^3*(c*x^2+b*x)^(5/2)/c^4/x^(5/2)+16/231*b^2*(c*x^2+b*x)^(5/2)/c^3/x^(3/2)-4/33*b*(c*x^2+b*x)^(5/2)/c
^2/x^(1/2)+2/11*(c*x^2+b*x)^(5/2)*x^(1/2)/c

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {670, 662} \[ \int x^{3/2} \left (b x+c x^2\right )^{3/2} \, dx=-\frac {32 b^3 \left (b x+c x^2\right )^{5/2}}{1155 c^4 x^{5/2}}+\frac {16 b^2 \left (b x+c x^2\right )^{5/2}}{231 c^3 x^{3/2}}-\frac {4 b \left (b x+c x^2\right )^{5/2}}{33 c^2 \sqrt {x}}+\frac {2 \sqrt {x} \left (b x+c x^2\right )^{5/2}}{11 c} \]

[In]

Int[x^(3/2)*(b*x + c*x^2)^(3/2),x]

[Out]

(-32*b^3*(b*x + c*x^2)^(5/2))/(1155*c^4*x^(5/2)) + (16*b^2*(b*x + c*x^2)^(5/2))/(231*c^3*x^(3/2)) - (4*b*(b*x
+ c*x^2)^(5/2))/(33*c^2*Sqrt[x]) + (2*Sqrt[x]*(b*x + c*x^2)^(5/2))/(11*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*} \text {integral}& = \frac {2 \sqrt {x} \left (b x+c x^2\right )^{5/2}}{11 c}-\frac {(6 b) \int \sqrt {x} \left (b x+c x^2\right )^{3/2} \, dx}{11 c} \\ & = -\frac {4 b \left (b x+c x^2\right )^{5/2}}{33 c^2 \sqrt {x}}+\frac {2 \sqrt {x} \left (b x+c x^2\right )^{5/2}}{11 c}+\frac {\left (8 b^2\right ) \int \frac {\left (b x+c x^2\right )^{3/2}}{\sqrt {x}} \, dx}{33 c^2} \\ & = \frac {16 b^2 \left (b x+c x^2\right )^{5/2}}{231 c^3 x^{3/2}}-\frac {4 b \left (b x+c x^2\right )^{5/2}}{33 c^2 \sqrt {x}}+\frac {2 \sqrt {x} \left (b x+c x^2\right )^{5/2}}{11 c}-\frac {\left (16 b^3\right ) \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{3/2}} \, dx}{231 c^3} \\ & = -\frac {32 b^3 \left (b x+c x^2\right )^{5/2}}{1155 c^4 x^{5/2}}+\frac {16 b^2 \left (b x+c x^2\right )^{5/2}}{231 c^3 x^{3/2}}-\frac {4 b \left (b x+c x^2\right )^{5/2}}{33 c^2 \sqrt {x}}+\frac {2 \sqrt {x} \left (b x+c x^2\right )^{5/2}}{11 c} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.49 \[ \int x^{3/2} \left (b x+c x^2\right )^{3/2} \, dx=\frac {2 (x (b+c x))^{5/2} \left (-16 b^3+40 b^2 c x-70 b c^2 x^2+105 c^3 x^3\right )}{1155 c^4 x^{5/2}} \]

[In]

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

[Out]

(2*(x*(b + c*x))^(5/2)*(-16*b^3 + 40*b^2*c*x - 70*b*c^2*x^2 + 105*c^3*x^3))/(1155*c^4*x^(5/2))

Maple [A] (verified)

Time = 2.01 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.51

method	result	size
gosper	\(-\frac {2 \left (c x +b \right ) \left (-105 c^{3} x^{3}+70 b \,c^{2} x^{2}-40 b^{2} c x +16 b^{3}\right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{1155 c^{4} x^{\frac {3}{2}}}\)	\(55\)
default	\(-\frac {2 \sqrt {x \left (c x +b \right )}\, \left (c x +b \right )^{2} \left (-105 c^{3} x^{3}+70 b \,c^{2} x^{2}-40 b^{2} c x +16 b^{3}\right )}{1155 \sqrt {x}\, c^{4}}\)	\(55\)
risch	\(-\frac {2 \left (c x +b \right ) \sqrt {x}\, \left (-105 c^{5} x^{5}-140 b \,x^{4} c^{4}-5 b^{2} c^{3} x^{3}+6 x^{2} b^{3} c^{2}-8 c x \,b^{4}+16 b^{5}\right )}{1155 \sqrt {x \left (c x +b \right )}\, c^{4}}\)	\(75\)

[In]

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

[Out]

-2/1155*(c*x+b)*(-105*c^3*x^3+70*b*c^2*x^2-40*b^2*c*x+16*b^3)*(c*x^2+b*x)^(3/2)/c^4/x^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.66 \[ \int x^{3/2} \left (b x+c x^2\right )^{3/2} \, dx=\frac {2 \, {\left (105 \, c^{5} x^{5} + 140 \, b c^{4} x^{4} + 5 \, b^{2} c^{3} x^{3} - 6 \, b^{3} c^{2} x^{2} + 8 \, b^{4} c x - 16 \, b^{5}\right )} \sqrt {c x^{2} + b x}}{1155 \, c^{4} \sqrt {x}} \]

[In]

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

[Out]

2/1155*(105*c^5*x^5 + 140*b*c^4*x^4 + 5*b^2*c^3*x^3 - 6*b^3*c^2*x^2 + 8*b^4*c*x - 16*b^5)*sqrt(c*x^2 + b*x)/(c
^4*sqrt(x))

Sympy [F]

\[ \int x^{3/2} \left (b x+c x^2\right )^{3/2} \, dx=\int x^{\frac {3}{2}} \left (x \left (b + c x\right )\right )^{\frac {3}{2}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.15 \[ \int x^{3/2} \left (b x+c x^2\right )^{3/2} \, dx=\frac {2 \, {\left ({\left (315 \, c^{5} x^{5} + 35 \, b c^{4} x^{4} - 40 \, b^{2} c^{3} x^{3} + 48 \, b^{3} c^{2} x^{2} - 64 \, b^{4} c x + 128 \, b^{5}\right )} x^{4} + 11 \, {\left (35 \, b c^{4} x^{5} + 5 \, b^{2} c^{3} x^{4} - 6 \, b^{3} c^{2} x^{3} + 8 \, b^{4} c x^{2} - 16 \, b^{5} x\right )} x^{3}\right )} \sqrt {c x + b}}{3465 \, c^{4} x^{4}} \]

[In]

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

[Out]

2/3465*((315*c^5*x^5 + 35*b*c^4*x^4 - 40*b^2*c^3*x^3 + 48*b^3*c^2*x^2 - 64*b^4*c*x + 128*b^5)*x^4 + 11*(35*b*c
^4*x^5 + 5*b^2*c^3*x^4 - 6*b^3*c^2*x^3 + 8*b^4*c*x^2 - 16*b^5*x)*x^3)*sqrt(c*x + b)/(c^4*x^4)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.24 \[ \int x^{3/2} \left (b x+c x^2\right )^{3/2} \, dx=-\frac {2}{3465} \, c {\left (\frac {128 \, b^{\frac {11}{2}}}{c^{5}} - \frac {315 \, {\left (c x + b\right )}^{\frac {11}{2}} - 1540 \, {\left (c x + b\right )}^{\frac {9}{2}} b + 2970 \, {\left (c x + b\right )}^{\frac {7}{2}} b^{2} - 2772 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{3} + 1155 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{4}}{c^{5}}\right )} + \frac {2}{315} \, b {\left (\frac {16 \, b^{\frac {9}{2}}}{c^{4}} + \frac {35 \, {\left (c x + b\right )}^{\frac {9}{2}} - 135 \, {\left (c x + b\right )}^{\frac {7}{2}} b + 189 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{2} - 105 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{3}}{c^{4}}\right )} \]

[In]

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

[Out]

-2/3465*c*(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 + 1155*(c*x + b)^(3/2)*b^4)/c^5) + 2/315*b*(16*b^(9/2)/c^4 + (35*(c*x + b)^(9/2) - 135*(c
*x + b)^(7/2)*b + 189*(c*x + b)^(5/2)*b^2 - 105*(c*x + b)^(3/2)*b^3)/c^4)

Mupad [F(-1)]

Timed out. \[ \int x^{3/2} \left (b x+c x^2\right )^{3/2} \, dx=\int x^{3/2}\,{\left (c\,x^2+b\,x\right )}^{3/2} \,d x \]

[In]

int(x^(3/2)*(b*x + c*x^2)^(3/2),x)

[Out]

int(x^(3/2)*(b*x + c*x^2)^(3/2), x)