3.9.4 \(\int x^3 \sqrt {c x^2} (a+b x)^2 \, dx\) [804]

Optimal. Leaf size=57 \[ \frac {1}{5} a^2 x^4 \sqrt {c x^2}+\frac {1}{3} a b x^5 \sqrt {c x^2}+\frac {1}{7} b^2 x^6 \sqrt {c x^2} \]

[Out]

1/5*a^2*x^4*(c*x^2)^(1/2)+1/3*a*b*x^5*(c*x^2)^(1/2)+1/7*b^2*x^6*(c*x^2)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 57, 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}{5} a^2 x^4 \sqrt {c x^2}+\frac {1}{3} a b x^5 \sqrt {c x^2}+\frac {1}{7} b^2 x^6 \sqrt {c x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*Sqrt[c*x^2]*(a + b*x)^2,x]

[Out]

(a^2*x^4*Sqrt[c*x^2])/5 + (a*b*x^5*Sqrt[c*x^2])/3 + (b^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*} \int x^3 \sqrt {c x^2} (a+b x)^2 \, dx &=\frac {\sqrt {c x^2} \int x^4 (a+b x)^2 \, dx}{x}\\ &=\frac {\sqrt {c x^2} \int \left (a^2 x^4+2 a b x^5+b^2 x^6\right ) \, dx}{x}\\ &=\frac {1}{5} a^2 x^4 \sqrt {c x^2}+\frac {1}{3} a b x^5 \sqrt {c x^2}+\frac {1}{7} b^2 x^6 \sqrt {c x^2}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 35, normalized size = 0.61 \begin {gather*} \frac {1}{105} x^4 \sqrt {c x^2} \left (21 a^2+35 a b x+15 b^2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^4*Sqrt[c*x^2]*(21*a^2 + 35*a*b*x + 15*b^2*x^2))/105

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Maple [A]
time = 0.12, size = 32, normalized size = 0.56

method	result	size
gosper	\(\frac {x^{4} \left (15 x^{2} b^{2}+35 a b x +21 a^{2}\right ) \sqrt {c \,x^{2}}}{105}\)	\(32\)
default	\(\frac {x^{4} \left (15 x^{2} b^{2}+35 a b x +21 a^{2}\right ) \sqrt {c \,x^{2}}}{105}\)	\(32\)
risch	\(\frac {a^{2} x^{4} \sqrt {c \,x^{2}}}{5}+\frac {a b \,x^{5} \sqrt {c \,x^{2}}}{3}+\frac {b^{2} x^{6} \sqrt {c \,x^{2}}}{7}\)	\(46\)
trager	\(\frac {\left (15 b^{2} x^{6}+35 a b \,x^{5}+15 b^{2} x^{5}+21 a^{2} x^{4}+35 a b \,x^{4}+15 b^{2} x^{4}+21 a^{2} x^{3}+35 a b \,x^{3}+15 b^{2} x^{3}+21 a^{2} x^{2}+35 a b \,x^{2}+15 x^{2} b^{2}+21 a^{2} x +35 a b x +15 b^{2} x +21 a^{2}+35 a b +15 b^{2}\right ) \left (-1+x \right ) \sqrt {c \,x^{2}}}{105 x}\)	\(140\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(b*x+a)^2*(c*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/105*x^4*(15*b^2*x^2+35*a*b*x+21*a^2)*(c*x^2)^(1/2)

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Maxima [A]
time = 0.29, size = 54, normalized size = 0.95 \begin {gather*} \frac {\left (c x^{2}\right )^{\frac {3}{2}} b^{2} x^{4}}{7 \, c} + \frac {\left (c x^{2}\right )^{\frac {3}{2}} a b x^{3}}{3 \, c} + \frac {\left (c x^{2}\right )^{\frac {3}{2}} a^{2} x^{2}}{5 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x+a)^2*(c*x^2)^(1/2),x, algorithm="maxima")

[Out]

1/7*(c*x^2)^(3/2)*b^2*x^4/c + 1/3*(c*x^2)^(3/2)*a*b*x^3/c + 1/5*(c*x^2)^(3/2)*a^2*x^2/c

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Fricas [A]
time = 0.70, size = 33, normalized size = 0.58 \begin {gather*} \frac {1}{105} \, {\left (15 \, b^{2} x^{6} + 35 \, a b x^{5} + 21 \, a^{2} x^{4}\right )} \sqrt {c x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x+a)^2*(c*x^2)^(1/2),x, algorithm="fricas")

[Out]

1/105*(15*b^2*x^6 + 35*a*b*x^5 + 21*a^2*x^4)*sqrt(c*x^2)

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Sympy [A]
time = 0.14, size = 49, normalized size = 0.86 \begin {gather*} \frac {a^{2} x^{4} \sqrt {c x^{2}}}{5} + \frac {a b x^{5} \sqrt {c x^{2}}}{3} + \frac {b^{2} x^{6} \sqrt {c x^{2}}}{7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(b*x+a)**2*(c*x**2)**(1/2),x)

[Out]

a**2*x**4*sqrt(c*x**2)/5 + a*b*x**5*sqrt(c*x**2)/3 + b**2*x**6*sqrt(c*x**2)/7

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Giac [A]
time = 1.20, size = 35, normalized size = 0.61 \begin {gather*} \frac {1}{105} \, {\left (15 \, b^{2} x^{7} \mathrm {sgn}\left (x\right ) + 35 \, a b x^{6} \mathrm {sgn}\left (x\right ) + 21 \, a^{2} x^{5} \mathrm {sgn}\left (x\right )\right )} \sqrt {c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x+a)^2*(c*x^2)^(1/2),x, algorithm="giac")

[Out]

1/105*(15*b^2*x^7*sgn(x) + 35*a*b*x^6*sgn(x) + 21*a^2*x^5*sgn(x))*sqrt(c)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(c*x^2)^(1/2)*(a + b*x)^2,x)

[Out]

int(x^3*(c*x^2)^(1/2)*(a + b*x)^2, x)

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