3.8.80 \(\int \frac {(c x^2)^{5/2} (a+b x)^2}{x} \, dx\)

Optimal. Leaf size=66 \[ \frac {1}{5} a^2 c^2 x^4 \sqrt {c x^2}+\frac {1}{3} a b c^2 x^5 \sqrt {c x^2}+\frac {1}{7} b^2 c^2 x^6 \sqrt {c x^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 66, 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, 43} \begin {gather*} \frac {1}{5} a^2 c^2 x^4 \sqrt {c x^2}+\frac {1}{3} a b c^2 x^5 \sqrt {c x^2}+\frac {1}{7} b^2 c^2 x^6 \sqrt {c x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((c*x^2)^(5/2)*(a + b*x)^2)/x,x]

[Out]

(a^2*c^2*x^4*Sqrt[c*x^2])/5 + (a*b*c^2*x^5*Sqrt[c*x^2])/3 + (b^2*c^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 43

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

________________________________________________________________________________________

Mathematica [A]  time = 0.00, size = 36, normalized size = 0.55 \begin {gather*} \frac {1}{105} c x^2 \left (c x^2\right )^{3/2} \left (21 a^2+35 a b x+15 b^2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c*x^2*(c*x^2)^(3/2)*(21*a^2 + 35*a*b*x + 15*b^2*x^2))/105

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.03, size = 32, normalized size = 0.48 \begin {gather*} \frac {1}{105} \left (c x^2\right )^{5/2} \left (21 a^2+35 a b x+15 b^2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.82, size = 42, normalized size = 0.64 \begin {gather*} \frac {1}{105} \, {\left (15 \, b^{2} c^{2} x^{6} + 35 \, a b c^{2} x^{5} + 21 \, a^{2} c^{2} x^{4}\right )} \sqrt {c x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.96, size = 44, normalized size = 0.67 \begin {gather*} \frac {1}{105} \, {\left (15 \, b^{2} c^{2} x^{7} \mathrm {sgn}\relax (x) + 35 \, a b c^{2} x^{6} \mathrm {sgn}\relax (x) + 21 \, a^{2} c^{2} x^{5} \mathrm {sgn}\relax (x)\right )} \sqrt {c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.00, size = 29, normalized size = 0.44 \begin {gather*} \frac {\left (15 b^{2} x^{2}+35 a b x +21 a^{2}\right ) \left (c \,x^{2}\right )^{\frac {5}{2}}}{105} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 1.28, size = 40, normalized size = 0.61 \begin {gather*} \frac {1}{3} \, \left (c x^{2}\right )^{\frac {5}{2}} a b x + \frac {1}{5} \, \left (c x^{2}\right )^{\frac {5}{2}} a^{2} + \frac {\left (c x^{2}\right )^{\frac {7}{2}} b^{2}}{7 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (c\,x^2\right )}^{5/2}\,{\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)^(5/2)*(a + b*x)^2)/x,x)

[Out]

int(((c*x^2)^(5/2)*(a + b*x)^2)/x, x)

________________________________________________________________________________________

sympy [A]  time = 1.82, size = 54, normalized size = 0.82 \begin {gather*} \frac {a^{2} c^{\frac {5}{2}} \left (x^{2}\right )^{\frac {5}{2}}}{5} + \frac {a b c^{\frac {5}{2}} x \left (x^{2}\right )^{\frac {5}{2}}}{3} + \frac {b^{2} c^{\frac {5}{2}} x^{2} \left (x^{2}\right )^{\frac {5}{2}}}{7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________