\(\int x (c x^2)^{5/2} (a+b x) \, dx\) [774]

   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 = 16, antiderivative size = 41 \[ \int x \left (c x^2\right )^{5/2} (a+b x) \, dx=\frac {1}{7} a c^2 x^6 \sqrt {c x^2}+\frac {1}{8} b c^2 x^7 \sqrt {c x^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {15, 45} \[ \int x \left (c x^2\right )^{5/2} (a+b x) \, dx=\frac {1}{7} a c^2 x^6 \sqrt {c x^2}+\frac {1}{8} b c^2 x^7 \sqrt {c x^2} \]

[In]

Int[x*(c*x^2)^(5/2)*(a + b*x),x]

[Out]

(a*c^2*x^6*Sqrt[c*x^2])/7 + (b*c^2*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^2 \sqrt {c x^2}\right ) \int x^6 (a+b x) \, dx}{x} \\ & = \frac {\left (c^2 \sqrt {c x^2}\right ) \int \left (a x^6+b x^7\right ) \, dx}{x} \\ & = \frac {1}{7} a c^2 x^6 \sqrt {c x^2}+\frac {1}{8} b c^2 x^7 \sqrt {c x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.59 \[ \int x \left (c x^2\right )^{5/2} (a+b x) \, dx=\frac {1}{56} x^2 \left (c x^2\right )^{5/2} (8 a+7 b x) \]

[In]

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

[Out]

(x^2*(c*x^2)^(5/2)*(8*a + 7*b*x))/56

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.51

method result size
gosper \(\frac {x^{2} \left (7 b x +8 a \right ) \left (c \,x^{2}\right )^{\frac {5}{2}}}{56}\) \(21\)
default \(\frac {x^{2} \left (7 b x +8 a \right ) \left (c \,x^{2}\right )^{\frac {5}{2}}}{56}\) \(21\)
risch \(\frac {a \,c^{2} x^{6} \sqrt {c \,x^{2}}}{7}+\frac {b \,c^{2} x^{7} \sqrt {c \,x^{2}}}{8}\) \(34\)
trager \(\frac {c^{2} \left (7 b \,x^{7}+8 a \,x^{6}+7 b \,x^{6}+8 a \,x^{5}+7 b \,x^{5}+8 a \,x^{4}+7 b \,x^{4}+8 a \,x^{3}+7 b \,x^{3}+8 a \,x^{2}+7 b \,x^{2}+8 a x +7 b x +8 a +7 b \right ) \left (-1+x \right ) \sqrt {c \,x^{2}}}{56 x}\) \(100\)

[In]

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

[Out]

1/56*x^2*(7*b*x+8*a)*(c*x^2)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.68 \[ \int x \left (c x^2\right )^{5/2} (a+b x) \, dx=\frac {1}{56} \, {\left (7 \, b c^{2} x^{7} + 8 \, a c^{2} x^{6}\right )} \sqrt {c x^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.80 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.71 \[ \int x \left (c x^2\right )^{5/2} (a+b x) \, dx=\frac {a x^{2} \left (c x^{2}\right )^{\frac {5}{2}}}{7} + \frac {b x^{3} \left (c x^{2}\right )^{\frac {5}{2}}}{8} \]

[In]

integrate(x*(c*x**2)**(5/2)*(b*x+a),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.68 \[ \int x \left (c x^2\right )^{5/2} (a+b x) \, dx=\frac {\left (c x^{2}\right )^{\frac {7}{2}} b x}{8 \, c} + \frac {\left (c x^{2}\right )^{\frac {7}{2}} a}{7 \, c} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.68 \[ \int x \left (c x^2\right )^{5/2} (a+b x) \, dx=\frac {1}{56} \, {\left (7 \, b c^{2} x^{8} \mathrm {sgn}\left (x\right ) + 8 \, a c^{2} x^{7} \mathrm {sgn}\left (x\right )\right )} \sqrt {c} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int x \left (c x^2\right )^{5/2} (a+b x) \, dx=\int x\,{\left (c\,x^2\right )}^{5/2}\,\left (a+b\,x\right ) \,d x \]

[In]

int(x*(c*x^2)^(5/2)*(a + b*x),x)

[Out]

int(x*(c*x^2)^(5/2)*(a + b*x), x)