3.9.16 \(\int \frac {(c x^2)^{3/2} (a+b x)^2}{x} \, dx\) [816]

Optimal. Leaf size=60 \[ \frac {1}{3} a^2 c x^2 \sqrt {c x^2}+\frac {1}{2} a b c x^3 \sqrt {c x^2}+\frac {1}{5} b^2 c x^4 \sqrt {c x^2} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 60, 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}{3} a^2 c x^2 \sqrt {c x^2}+\frac {1}{2} a b c x^3 \sqrt {c x^2}+\frac {1}{5} b^2 c x^4 \sqrt {c x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((c*x^2)^(3/2)*(a + b*x)^2)/x,x]

[Out]

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

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

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Mathematica [A]
time = 0.00, size = 36, normalized size = 0.60 \begin {gather*} \frac {1}{30} c x^2 \sqrt {c x^2} \left (10 a^2+15 a b x+6 b^2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c*x^2*Sqrt[c*x^2]*(10*a^2 + 15*a*b*x + 6*b^2*x^2))/30

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Maple [A]
time = 0.12, size = 29, normalized size = 0.48

method	result	size
gosper	\(\frac {\left (6 x^{2} b^{2}+15 a b x +10 a^{2}\right ) \left (c \,x^{2}\right )^{\frac {3}{2}}}{30}\)	\(29\)
default	\(\frac {\left (6 x^{2} b^{2}+15 a b x +10 a^{2}\right ) \left (c \,x^{2}\right )^{\frac {3}{2}}}{30}\)	\(29\)
risch	\(\frac {a^{2} c \,x^{2} \sqrt {c \,x^{2}}}{3}+\frac {a b c \,x^{3} \sqrt {c \,x^{2}}}{2}+\frac {b^{2} c \,x^{4} \sqrt {c \,x^{2}}}{5}\)	\(49\)
trager	\(\frac {c \left (6 b^{2} x^{4}+15 a b \,x^{3}+6 b^{2} x^{3}+10 a^{2} x^{2}+15 a b \,x^{2}+6 x^{2} b^{2}+10 a^{2} x +15 a b x +6 b^{2} x +10 a^{2}+15 a b +6 b^{2}\right ) \left (-1+x \right ) \sqrt {c \,x^{2}}}{30 x}\)	\(95\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(6*b^2*x^2+15*a*b*x+10*a^2)*(c*x^2)^(3/2)

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Maxima [A]
time = 0.27, size = 40, normalized size = 0.67 \begin {gather*} \frac {1}{2} \, \left (c x^{2}\right )^{\frac {3}{2}} a b x + \frac {1}{3} \, \left (c x^{2}\right )^{\frac {3}{2}} a^{2} + \frac {\left (c x^{2}\right )^{\frac {5}{2}} b^{2}}{5 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.46, size = 36, normalized size = 0.60 \begin {gather*} \frac {1}{30} \, {\left (6 \, b^{2} c x^{4} + 15 \, a b c x^{3} + 10 \, a^{2} c x^{2}\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,x, algorithm="fricas")

[Out]

1/30*(6*b^2*c*x^4 + 15*a*b*c*x^3 + 10*a^2*c*x^2)*sqrt(c*x^2)

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Sympy [A]
time = 0.17, size = 44, normalized size = 0.73 \begin {gather*} \frac {a^{2} \left (c x^{2}\right )^{\frac {3}{2}}}{3} + \frac {a b x \left (c x^{2}\right )^{\frac {3}{2}}}{2} + \frac {b^{2} x^{2} \left (c x^{2}\right )^{\frac {3}{2}}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.24, size = 35, normalized size = 0.58 \begin {gather*} \frac {1}{30} \, {\left (6 \, b^{2} x^{5} \mathrm {sgn}\left (x\right ) + 15 \, a b x^{4} \mathrm {sgn}\left (x\right ) + 10 \, a^{2} x^{3} \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,x, algorithm="giac")

[Out]

1/30*(6*b^2*x^5*sgn(x) + 15*a*b*x^4*sgn(x) + 10*a^2*x^3*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} \,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,x)

[Out]

int(((c*x^2)^(3/2)*(a + b*x)^2)/x, x)

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