3.9.17 \(\int \frac {(c x^2)^{3/2} (a+b x)^2}{x^2} \, dx\) [817]

Optimal. Leaf size=58 \[ \frac {1}{2} a^2 c x \sqrt {c x^2}+\frac {2}{3} a b c x^2 \sqrt {c x^2}+\frac {1}{4} b^2 c x^3 \sqrt {c x^2} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 45} \begin {gather*} \frac {1}{2} a^2 c x \sqrt {c x^2}+\frac {2}{3} a b c x^2 \sqrt {c x^2}+\frac {1}{4} b^2 c x^3 \sqrt {c x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]
time = 0.00, size = 34, normalized size = 0.59 \begin {gather*} \frac {1}{12} c x \sqrt {c x^2} \left (6 a^2+8 a b x+3 b^2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c*x*Sqrt[c*x^2]*(6*a^2 + 8*a*b*x + 3*b^2*x^2))/12

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Maple [A]
time = 0.11, size = 32, normalized size = 0.55

method	result	size
gosper	\(\frac {\left (3 x^{2} b^{2}+8 a b x +6 a^{2}\right ) \left (c \,x^{2}\right )^{\frac {3}{2}}}{12 x}\)	\(32\)
default	\(\frac {\left (3 x^{2} b^{2}+8 a b x +6 a^{2}\right ) \left (c \,x^{2}\right )^{\frac {3}{2}}}{12 x}\)	\(32\)
risch	\(\frac {a^{2} c x \sqrt {c \,x^{2}}}{2}+\frac {2 a b c \,x^{2} \sqrt {c \,x^{2}}}{3}+\frac {b^{2} c \,x^{3} \sqrt {c \,x^{2}}}{4}\)	\(47\)
trager	\(\frac {c \left (3 b^{2} x^{3}+8 a b \,x^{2}+3 x^{2} b^{2}+6 a^{2} x +8 a b x +3 b^{2} x +6 a^{2}+8 a b +3 b^{2}\right ) \left (-1+x \right ) \sqrt {c \,x^{2}}}{12 x}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12/x*(3*b^2*x^2+8*a*b*x+6*a^2)*(c*x^2)^(3/2)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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Fricas [A]
time = 0.43, size = 34, normalized size = 0.59 \begin {gather*} \frac {1}{12} \, {\left (3 \, b^{2} c x^{3} + 8 \, a b c x^{2} + 6 \, a^{2} c x\right )} \sqrt {c x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*b^2*c*x^3 + 8*a*b*c*x^2 + 6*a^2*c*x)*sqrt(c*x^2)

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Sympy [A]
time = 0.17, size = 44, normalized size = 0.76 \begin {gather*} \frac {a^{2} \left (c x^{2}\right )^{\frac {3}{2}}}{2 x} + \frac {2 a b \left (c x^{2}\right )^{\frac {3}{2}}}{3} + \frac {b^{2} x \left (c x^{2}\right )^{\frac {3}{2}}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.07, size = 35, normalized size = 0.60 \begin {gather*} \frac {1}{12} \, {\left (3 \, b^{2} x^{4} \mathrm {sgn}\left (x\right ) + 8 \, a b x^{3} \mathrm {sgn}\left (x\right ) + 6 \, a^{2} x^{2} \mathrm {sgn}\left (x\right )\right )} c^{\frac {3}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*b^2*x^4*sgn(x) + 8*a*b*x^3*sgn(x) + 6*a^2*x^2*sgn(x))*c^(3/2)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (c\,x^2\right )}^{3/2}\,{\left (a+b\,x\right )}^2}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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