\(\int (c x^2)^{5/2} (a+b x) \, dx\) [775]

   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 = 15, antiderivative size = 41 \[ \int \left (c x^2\right )^{5/2} (a+b x) \, dx=\frac {1}{6} a c^2 x^5 \sqrt {c x^2}+\frac {1}{7} b c^2 x^6 \sqrt {c x^2} \]

[Out]

1/6*a*c^2*x^5*(c*x^2)^(1/2)+1/7*b*c^2*x^6*(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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {15, 45} \[ \int \left (c x^2\right )^{5/2} (a+b x) \, dx=\frac {1}{6} a c^2 x^5 \sqrt {c x^2}+\frac {1}{7} b c^2 x^6 \sqrt {c x^2} \]

[In]

Int[(c*x^2)^(5/2)*(a + b*x),x]

[Out]

(a*c^2*x^5*Sqrt[c*x^2])/6 + (b*c^2*x^6*Sqrt[c*x^2])/7

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^5 (a+b x) \, dx}{x} \\ & = \frac {\left (c^2 \sqrt {c x^2}\right ) \int \left (a x^5+b x^6\right ) \, dx}{x} \\ & = \frac {1}{6} a c^2 x^5 \sqrt {c x^2}+\frac {1}{7} b c^2 x^6 \sqrt {c x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.54 \[ \int \left (c x^2\right )^{5/2} (a+b x) \, dx=\frac {1}{42} x \left (c x^2\right )^{5/2} (7 a+6 b x) \]

[In]

Integrate[(c*x^2)^(5/2)*(a + b*x),x]

[Out]

(x*(c*x^2)^(5/2)*(7*a + 6*b*x))/42

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.46

method result size
gosper \(\frac {x \left (6 b x +7 a \right ) \left (c \,x^{2}\right )^{\frac {5}{2}}}{42}\) \(19\)
default \(\frac {x \left (6 b x +7 a \right ) \left (c \,x^{2}\right )^{\frac {5}{2}}}{42}\) \(19\)
risch \(\frac {a \,c^{2} x^{5} \sqrt {c \,x^{2}}}{6}+\frac {b \,c^{2} x^{6} \sqrt {c \,x^{2}}}{7}\) \(34\)
trager \(\frac {c^{2} \left (6 b \,x^{6}+7 a \,x^{5}+6 b \,x^{5}+7 a \,x^{4}+6 b \,x^{4}+7 a \,x^{3}+6 b \,x^{3}+7 a \,x^{2}+6 b \,x^{2}+7 a x +6 b x +7 a +6 b \right ) \left (-1+x \right ) \sqrt {c \,x^{2}}}{42 x}\) \(88\)

[In]

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

[Out]

1/42*x*(6*b*x+7*a)*(c*x^2)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.68 \[ \int \left (c x^2\right )^{5/2} (a+b x) \, dx=\frac {1}{42} \, {\left (6 \, b c^{2} x^{6} + 7 \, a c^{2} x^{5}\right )} \sqrt {c x^{2}} \]

[In]

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

[Out]

1/42*(6*b*c^2*x^6 + 7*a*c^2*x^5)*sqrt(c*x^2)

Sympy [A] (verification not implemented)

Time = 0.56 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.66 \[ \int \left (c x^2\right )^{5/2} (a+b x) \, dx=\frac {a x \left (c x^{2}\right )^{\frac {5}{2}}}{6} + \frac {b x^{2} \left (c x^{2}\right )^{\frac {5}{2}}}{7} \]

[In]

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

[Out]

a*x*(c*x**2)**(5/2)/6 + b*x**2*(c*x**2)**(5/2)/7

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.61 \[ \int \left (c x^2\right )^{5/2} (a+b x) \, dx=\frac {1}{6} \, \left (c x^{2}\right )^{\frac {5}{2}} a x + \frac {\left (c x^{2}\right )^{\frac {7}{2}} b}{7 \, c} \]

[In]

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

[Out]

1/6*(c*x^2)^(5/2)*a*x + 1/7*(c*x^2)^(7/2)*b/c

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.68 \[ \int \left (c x^2\right )^{5/2} (a+b x) \, dx=\frac {1}{42} \, {\left (6 \, b c^{2} x^{7} \mathrm {sgn}\left (x\right ) + 7 \, a c^{2} x^{6} \mathrm {sgn}\left (x\right )\right )} \sqrt {c} \]

[In]

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

[Out]

1/42*(6*b*c^2*x^7*sgn(x) + 7*a*c^2*x^6*sgn(x))*sqrt(c)

Mupad [F(-1)]

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

[In]

int((c*x^2)^(5/2)*(a + b*x),x)

[Out]

int((c*x^2)^(5/2)*(a + b*x), x)