\(\int x^2 (c x^2)^{5/2} (a+b x) \, dx\) [773]

Optimal result

Integrand size = 18, antiderivative size = 41 \[ \int x^2 \left (c x^2\right )^{5/2} (a+b x) \, dx=\frac {1}{8} a c^2 x^7 \sqrt {c x^2}+\frac {1}{9} b c^2 x^8 \sqrt {c x^2} \]

[Out]

1/8*a*c^2*x^7*(c*x^2)^(1/2)+1/9*b*c^2*x^8*(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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {15, 45} \[ \int x^2 \left (c x^2\right )^{5/2} (a+b x) \, dx=\frac {1}{8} a c^2 x^7 \sqrt {c x^2}+\frac {1}{9} b c^2 x^8 \sqrt {c x^2} \]

[In]

Int[x^2*(c*x^2)^(5/2)*(a + b*x),x]

[Out]

(a*c^2*x^7*Sqrt[c*x^2])/8 + (b*c^2*x^8*Sqrt[c*x^2])/9

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^7 (a+b x) \, dx}{x} \\ & = \frac {\left (c^2 \sqrt {c x^2}\right ) \int \left (a x^7+b x^8\right ) \, dx}{x} \\ & = \frac {1}{8} a c^2 x^7 \sqrt {c x^2}+\frac {1}{9} b c^2 x^8 \sqrt {c x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.59 \[ \int x^2 \left (c x^2\right )^{5/2} (a+b x) \, dx=\frac {1}{72} x^3 \left (c x^2\right )^{5/2} (9 a+8 b x) \]

[In]

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

[Out]

(x^3*(c*x^2)^(5/2)*(9*a + 8*b*x))/72

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.51

method	result	size
gosper	\(\frac {x^{3} \left (8 b x +9 a \right ) \left (c \,x^{2}\right )^{\frac {5}{2}}}{72}\)	\(21\)
default	\(\frac {x^{3} \left (8 b x +9 a \right ) \left (c \,x^{2}\right )^{\frac {5}{2}}}{72}\)	\(21\)
risch	\(\frac {a \,c^{2} x^{7} \sqrt {c \,x^{2}}}{8}+\frac {b \,c^{2} x^{8} \sqrt {c \,x^{2}}}{9}\)	\(34\)
trager	\(\frac {c^{2} \left (8 b \,x^{8}+9 a \,x^{7}+8 b \,x^{7}+9 a \,x^{6}+8 b \,x^{6}+9 a \,x^{5}+8 b \,x^{5}+9 a \,x^{4}+8 b \,x^{4}+9 a \,x^{3}+8 b \,x^{3}+9 a \,x^{2}+8 b \,x^{2}+9 a x +8 b x +9 a +8 b \right ) \left (-1+x \right ) \sqrt {c \,x^{2}}}{72 x}\)	\(112\)

[In]

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

[Out]

1/72*x^3*(8*b*x+9*a)*(c*x^2)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.68 \[ \int x^2 \left (c x^2\right )^{5/2} (a+b x) \, dx=\frac {1}{72} \, {\left (8 \, b c^{2} x^{8} + 9 \, a c^{2} x^{7}\right )} \sqrt {c x^{2}} \]

[In]

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

[Out]

1/72*(8*b*c^2*x^8 + 9*a*c^2*x^7)*sqrt(c*x^2)

Sympy [A] (verification not implemented)

Time = 0.74 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.71 \[ \int x^2 \left (c x^2\right )^{5/2} (a+b x) \, dx=\frac {a x^{3} \left (c x^{2}\right )^{\frac {5}{2}}}{8} + \frac {b x^{4} \left (c x^{2}\right )^{\frac {5}{2}}}{9} \]

[In]

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

[Out]

a*x**3*(c*x**2)**(5/2)/8 + b*x**4*(c*x**2)**(5/2)/9

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.76 \[ \int x^2 \left (c x^2\right )^{5/2} (a+b x) \, dx=\frac {\left (c x^{2}\right )^{\frac {7}{2}} b x^{2}}{9 \, c} + \frac {\left (c x^{2}\right )^{\frac {7}{2}} a x}{8 \, c} \]

[In]

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

[Out]

1/9*(c*x^2)^(7/2)*b*x^2/c + 1/8*(c*x^2)^(7/2)*a*x/c

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.68 \[ \int x^2 \left (c x^2\right )^{5/2} (a+b x) \, dx=\frac {1}{72} \, {\left (8 \, b c^{2} x^{9} \mathrm {sgn}\left (x\right ) + 9 \, a c^{2} x^{8} \mathrm {sgn}\left (x\right )\right )} \sqrt {c} \]

[In]

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

[Out]

1/72*(8*b*c^2*x^9*sgn(x) + 9*a*c^2*x^8*sgn(x))*sqrt(c)

Mupad [F(-1)]

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

[In]

int(x^2*(c*x^2)^(5/2)*(a + b*x),x)

[Out]

int(x^2*(c*x^2)^(5/2)*(a + b*x), x)