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

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

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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}{4} a^2 c^2 x^3 \sqrt {c x^2}+\frac {2}{5} a b c^2 x^4 \sqrt {c x^2}+\frac {1}{6} b^2 c^2 x^5 \sqrt {c x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.01, size = 34, normalized size = 0.52 \begin {gather*} \frac {1}{60} c x \left (c x^2\right )^{3/2} \left (15 a^2+24 a b x+10 b^2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.03, size = 35, normalized size = 0.53 \begin {gather*} \frac {\left (c x^2\right )^{5/2} \left (15 a^2+24 a b x+10 b^2 x^2\right )}{60 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.11, size = 42, normalized size = 0.64 \begin {gather*} \frac {1}{60} \, {\left (10 \, b^{2} c^{2} x^{5} + 24 \, a b c^{2} x^{4} + 15 \, a^{2} c^{2} x^{3}\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^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.99, size = 44, normalized size = 0.67 \begin {gather*} \frac {1}{60} \, {\left (10 \, b^{2} c^{2} x^{6} \mathrm {sgn}\relax (x) + 24 \, a b c^{2} x^{5} \mathrm {sgn}\relax (x) + 15 \, a^{2} c^{2} x^{4} \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^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.00, size = 32, normalized size = 0.48 \begin {gather*} \frac {\left (10 b^{2} x^{2}+24 a b x +15 a^{2}\right ) \left (c \,x^{2}\right )^{\frac {5}{2}}}{60 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.37, size = 40, normalized size = 0.61 \begin {gather*} \frac {1}{6} \, \left (c x^{2}\right )^{\frac {5}{2}} b^{2} x + \frac {2}{5} \, \left (c x^{2}\right )^{\frac {5}{2}} a b + \frac {\left (c x^{2}\right )^{\frac {5}{2}} a^{2}}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^2} \,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^2,x)

[Out]

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

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sympy [A]  time = 1.84, size = 54, normalized size = 0.82 \begin {gather*} \frac {a^{2} c^{\frac {5}{2}} \left (x^{2}\right )^{\frac {5}{2}}}{4 x} + \frac {2 a b c^{\frac {5}{2}} \left (x^{2}\right )^{\frac {5}{2}}}{5} + \frac {b^{2} c^{\frac {5}{2}} x \left (x^{2}\right )^{\frac {5}{2}}}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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