\(\int x^3 (c x^2)^{3/2} (a+b x) \, dx\) [764]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 18, antiderivative size = 37 \[ \int x^3 \left (c x^2\right )^{3/2} (a+b x) \, dx=\frac {1}{7} a c x^6 \sqrt {c x^2}+\frac {1}{8} b c x^7 \sqrt {c x^2} \]

[Out]

1/7*a*c*x^6*(c*x^2)^(1/2)+1/8*b*c*x^7*(c*x^2)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {15, 45} \[ \int x^3 \left (c x^2\right )^{3/2} (a+b x) \, dx=\frac {1}{7} a c x^6 \sqrt {c x^2}+\frac {1}{8} b c x^7 \sqrt {c x^2} \]

[In]

Int[x^3*(c*x^2)^(3/2)*(a + b*x),x]

[Out]

(a*c*x^6*Sqrt[c*x^2])/7 + (b*c*x^7*Sqrt[c*x^2])/8

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[a^IntPart[m]*((a*x^n)^FracPart[m]/x^(n*FracPart[m])), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps \begin{align*} \text {integral}& = \frac {\left (c \sqrt {c x^2}\right ) \int x^6 (a+b x) \, dx}{x} \\ & = \frac {\left (c \sqrt {c x^2}\right ) \int \left (a x^6+b x^7\right ) \, dx}{x} \\ & = \frac {1}{7} a c x^6 \sqrt {c x^2}+\frac {1}{8} b c x^7 \sqrt {c x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.65 \[ \int x^3 \left (c x^2\right )^{3/2} (a+b x) \, dx=\frac {1}{56} x^4 \left (c x^2\right )^{3/2} (8 a+7 b x) \]

[In]

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

[Out]

(x^4*(c*x^2)^(3/2)*(8*a + 7*b*x))/56

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.57

method result size
gosper \(\frac {x^{4} \left (7 b x +8 a \right ) \left (c \,x^{2}\right )^{\frac {3}{2}}}{56}\) \(21\)
default \(\frac {x^{4} \left (7 b x +8 a \right ) \left (c \,x^{2}\right )^{\frac {3}{2}}}{56}\) \(21\)
risch \(\frac {a c \,x^{6} \sqrt {c \,x^{2}}}{7}+\frac {b c \,x^{7} \sqrt {c \,x^{2}}}{8}\) \(30\)
trager \(\frac {c \left (7 b \,x^{7}+8 a \,x^{6}+7 b \,x^{6}+8 a \,x^{5}+7 b \,x^{5}+8 a \,x^{4}+7 b \,x^{4}+8 a \,x^{3}+7 b \,x^{3}+8 a \,x^{2}+7 b \,x^{2}+8 a x +7 b x +8 a +7 b \right ) \left (-1+x \right ) \sqrt {c \,x^{2}}}{56 x}\) \(98\)

[In]

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

[Out]

1/56*x^4*(7*b*x+8*a)*(c*x^2)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.65 \[ \int x^3 \left (c x^2\right )^{3/2} (a+b x) \, dx=\frac {1}{56} \, {\left (7 \, b c x^{7} + 8 \, a c x^{6}\right )} \sqrt {c x^{2}} \]

[In]

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

[Out]

1/56*(7*b*c*x^7 + 8*a*c*x^6)*sqrt(c*x^2)

Sympy [A] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int x^3 \left (c x^2\right )^{3/2} (a+b x) \, dx=\frac {a x^{4} \left (c x^{2}\right )^{\frac {3}{2}}}{7} + \frac {b x^{5} \left (c x^{2}\right )^{\frac {3}{2}}}{8} \]

[In]

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

[Out]

a*x**4*(c*x**2)**(3/2)/7 + b*x**5*(c*x**2)**(3/2)/8

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.89 \[ \int x^3 \left (c x^2\right )^{3/2} (a+b x) \, dx=\frac {\left (c x^{2}\right )^{\frac {5}{2}} b x^{3}}{8 \, c} + \frac {\left (c x^{2}\right )^{\frac {5}{2}} a x^{2}}{7 \, c} \]

[In]

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

[Out]

1/8*(c*x^2)^(5/2)*b*x^3/c + 1/7*(c*x^2)^(5/2)*a*x^2/c

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.59 \[ \int x^3 \left (c x^2\right )^{3/2} (a+b x) \, dx=\frac {1}{56} \, {\left (7 \, b x^{8} \mathrm {sgn}\left (x\right ) + 8 \, a x^{7} \mathrm {sgn}\left (x\right )\right )} c^{\frac {3}{2}} \]

[In]

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

[Out]

1/56*(7*b*x^8*sgn(x) + 8*a*x^7*sgn(x))*c^(3/2)

Mupad [F(-1)]

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

[In]

int(x^3*(c*x^2)^(3/2)*(a + b*x),x)

[Out]

int(x^3*(c*x^2)^(3/2)*(a + b*x), x)