\(\int (c x^2)^{3/2} (a+b x)^2 \, dx\) [815]

Optimal result

Integrand size = 17, antiderivative size = 60 \[ \int \left (c x^2\right )^{3/2} (a+b x)^2 \, dx=\frac {1}{4} a^2 c x^3 \sqrt {c x^2}+\frac {2}{5} a b c x^4 \sqrt {c x^2}+\frac {1}{6} b^2 c x^5 \sqrt {c x^2} \]

[Out]

1/4*a^2*c*x^3*(c*x^2)^(1/2)+2/5*a*b*c*x^4*(c*x^2)^(1/2)+1/6*b^2*c*x^5*(c*x^2)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {15, 45} \[ \int \left (c x^2\right )^{3/2} (a+b x)^2 \, dx=\frac {1}{4} a^2 c x^3 \sqrt {c x^2}+\frac {2}{5} a b c x^4 \sqrt {c x^2}+\frac {1}{6} b^2 c x^5 \sqrt {c x^2} \]

[In]

Int[(c*x^2)^(3/2)*(a + b*x)^2,x]

[Out]

(a^2*c*x^3*Sqrt[c*x^2])/4 + (2*a*b*c*x^4*Sqrt[c*x^2])/5 + (b^2*c*x^5*Sqrt[c*x^2])/6

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^3 (a+b x)^2 \, dx}{x} \\ & = \frac {\left (c \sqrt {c x^2}\right ) \int \left (a^2 x^3+2 a b x^4+b^2 x^5\right ) \, dx}{x} \\ & = \frac {1}{4} a^2 c x^3 \sqrt {c x^2}+\frac {2}{5} a b c x^4 \sqrt {c x^2}+\frac {1}{6} b^2 c x^5 \sqrt {c x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.55 \[ \int \left (c x^2\right )^{3/2} (a+b x)^2 \, dx=\frac {1}{60} x \left (c x^2\right )^{3/2} \left (15 a^2+24 a b x+10 b^2 x^2\right ) \]

[In]

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

[Out]

(x*(c*x^2)^(3/2)*(15*a^2 + 24*a*b*x + 10*b^2*x^2))/60

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.50

method	result	size
gosper	\(\frac {x \left (10 b^{2} x^{2}+24 a b x +15 a^{2}\right ) \left (c \,x^{2}\right )^{\frac {3}{2}}}{60}\)	\(30\)
default	\(\frac {x \left (10 b^{2} x^{2}+24 a b x +15 a^{2}\right ) \left (c \,x^{2}\right )^{\frac {3}{2}}}{60}\)	\(30\)
risch	\(\frac {a^{2} c \,x^{3} \sqrt {c \,x^{2}}}{4}+\frac {2 a b c \,x^{4} \sqrt {c \,x^{2}}}{5}+\frac {b^{2} c \,x^{5} \sqrt {c \,x^{2}}}{6}\)	\(49\)
trager	\(\frac {c \left (10 b^{2} x^{5}+24 a b \,x^{4}+10 b^{2} x^{4}+15 a^{2} x^{3}+24 a b \,x^{3}+10 b^{2} x^{3}+15 a^{2} x^{2}+24 a b \,x^{2}+10 b^{2} x^{2}+15 a^{2} x +24 a b x +10 b^{2} x +15 a^{2}+24 a b +10 b^{2}\right ) \left (-1+x \right ) \sqrt {c \,x^{2}}}{60 x}\)	\(118\)

[In]

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

[Out]

1/60*x*(10*b^2*x^2+24*a*b*x+15*a^2)*(c*x^2)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.60 \[ \int \left (c x^2\right )^{3/2} (a+b x)^2 \, dx=\frac {1}{60} \, {\left (10 \, b^{2} c x^{5} + 24 \, a b c x^{4} + 15 \, a^{2} c x^{3}\right )} \sqrt {c x^{2}} \]

[In]

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

[Out]

1/60*(10*b^2*c*x^5 + 24*a*b*c*x^4 + 15*a^2*c*x^3)*sqrt(c*x^2)

Sympy [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.82 \[ \int \left (c x^2\right )^{3/2} (a+b x)^2 \, dx=\frac {a^{2} x \left (c x^{2}\right )^{\frac {3}{2}}}{4} + \frac {2 a b x^{2} \left (c x^{2}\right )^{\frac {3}{2}}}{5} + \frac {b^{2} x^{3} \left (c x^{2}\right )^{\frac {3}{2}}}{6} \]

[In]

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

[Out]

a**2*x*(c*x**2)**(3/2)/4 + 2*a*b*x**2*(c*x**2)**(3/2)/5 + b**2*x**3*(c*x**2)**(3/2)/6

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.73 \[ \int \left (c x^2\right )^{3/2} (a+b x)^2 \, dx=\frac {1}{4} \, \left (c x^{2}\right )^{\frac {3}{2}} a^{2} x + \frac {\left (c x^{2}\right )^{\frac {5}{2}} b^{2} x}{6 \, c} + \frac {2 \, \left (c x^{2}\right )^{\frac {5}{2}} a b}{5 \, c} \]

[In]

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

[Out]

1/4*(c*x^2)^(3/2)*a^2*x + 1/6*(c*x^2)^(5/2)*b^2*x/c + 2/5*(c*x^2)^(5/2)*a*b/c

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.58 \[ \int \left (c x^2\right )^{3/2} (a+b x)^2 \, dx=\frac {1}{60} \, {\left (10 \, b^{2} x^{6} \mathrm {sgn}\left (x\right ) + 24 \, a b x^{5} \mathrm {sgn}\left (x\right ) + 15 \, a^{2} x^{4} \mathrm {sgn}\left (x\right )\right )} c^{\frac {3}{2}} \]

[In]

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

[Out]

1/60*(10*b^2*x^6*sgn(x) + 24*a*b*x^5*sgn(x) + 15*a^2*x^4*sgn(x))*c^(3/2)

Mupad [F(-1)]

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

[In]

int((c*x^2)^(3/2)*(a + b*x)^2,x)

[Out]

int((c*x^2)^(3/2)*(a + b*x)^2, x)