3.9.19 \(\int \frac {(c x^2)^{3/2}}{a+b x} \, dx\)

Optimal. Leaf size=84 \[ -\frac {a^3 c \sqrt {c x^2} \log (a+b x)}{b^4 x}+\frac {a^2 c \sqrt {c x^2}}{b^3}-\frac {a c x \sqrt {c x^2}}{2 b^2}+\frac {c x^2 \sqrt {c x^2}}{3 b} \]

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Rubi [A]  time = 0.02, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {15, 43} \begin {gather*} \frac {a^2 c \sqrt {c x^2}}{b^3}-\frac {a^3 c \sqrt {c x^2} \log (a+b x)}{b^4 x}-\frac {a c x \sqrt {c x^2}}{2 b^2}+\frac {c x^2 \sqrt {c x^2}}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*x^2)^(3/2)/(a + b*x),x]

[Out]

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

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

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Mathematica [A]  time = 0.01, size = 53, normalized size = 0.63 \begin {gather*} \frac {\left (c x^2\right )^{3/2} \left (b x \left (6 a^2-3 a b x+2 b^2 x^2\right )-6 a^3 \log (a+b x)\right )}{6 b^4 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c*x^2)^(3/2)/(a + b*x),x]

[Out]

((c*x^2)^(3/2)*(b*x*(6*a^2 - 3*a*b*x + 2*b^2*x^2) - 6*a^3*Log[a + b*x]))/(6*b^4*x^3)

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IntegrateAlgebraic [A]  time = 0.04, size = 57, normalized size = 0.68 \begin {gather*} \left (c x^2\right )^{3/2} \left (\frac {6 a^2-3 a b x+2 b^2 x^2}{6 b^3 x^2}-\frac {a^3 \log (a+b x)}{b^4 x^3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c*x^2)^(3/2)*((6*a^2 - 3*a*b*x + 2*b^2*x^2)/(6*b^3*x^2) - (a^3*Log[a + b*x])/(b^4*x^3))

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fricas [A]  time = 1.31, size = 55, normalized size = 0.65 \begin {gather*} \frac {{\left (2 \, b^{3} c x^{3} - 3 \, a b^{2} c x^{2} + 6 \, a^{2} b c x - 6 \, a^{3} c \log \left (b x + a\right )\right )} \sqrt {c x^{2}}}{6 \, b^{4} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*b^3*c*x^3 - 3*a*b^2*c*x^2 + 6*a^2*b*c*x - 6*a^3*c*log(b*x + a))*sqrt(c*x^2)/(b^4*x)

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giac [A]  time = 1.15, size = 69, normalized size = 0.82 \begin {gather*} -\frac {1}{6} \, c^{\frac {3}{2}} {\left (\frac {6 \, a^{3} \log \left ({\left | b x + a \right |}\right ) \mathrm {sgn}\relax (x)}{b^{4}} - \frac {6 \, a^{3} \log \left ({\left | a \right |}\right ) \mathrm {sgn}\relax (x)}{b^{4}} - \frac {2 \, b^{2} x^{3} \mathrm {sgn}\relax (x) - 3 \, a b x^{2} \mathrm {sgn}\relax (x) + 6 \, a^{2} x \mathrm {sgn}\relax (x)}{b^{3}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*c^(3/2)*(6*a^3*log(abs(b*x + a))*sgn(x)/b^4 - 6*a^3*log(abs(a))*sgn(x)/b^4 - (2*b^2*x^3*sgn(x) - 3*a*b*x^
2*sgn(x) + 6*a^2*x*sgn(x))/b^3)

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maple [A]  time = 0.00, size = 52, normalized size = 0.62 \begin {gather*} -\frac {\left (c \,x^{2}\right )^{\frac {3}{2}} \left (-2 b^{3} x^{3}+3 a \,b^{2} x^{2}+6 a^{3} \ln \left (b x +a \right )-6 a^{2} b x \right )}{6 b^{4} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(c*x^2)^(3/2)*(-2*b^3*x^3+3*a*b^2*x^2+6*a^3*ln(b*x+a)-6*a^2*b*x)/x^3/b^4

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maxima [A]  time = 1.54, size = 109, normalized size = 1.30 \begin {gather*} -\frac {\left (-1\right )^{\frac {2 \, c x}{b}} a^{3} c^{\frac {3}{2}} \log \left (\frac {2 \, c x}{b}\right )}{b^{4}} - \frac {\left (-1\right )^{\frac {2 \, a c x}{b}} a^{3} c^{\frac {3}{2}} \log \left (-\frac {2 \, a c x}{b {\left | b x + a \right |}}\right )}{b^{4}} - \frac {\sqrt {c x^{2}} a c x}{2 \, b^{2}} + \frac {\left (c x^{2}\right )^{\frac {3}{2}}}{3 \, b} + \frac {\sqrt {c x^{2}} a^{2} c}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-1)^(2*c*x/b)*a^3*c^(3/2)*log(2*c*x/b)/b^4 - (-1)^(2*a*c*x/b)*a^3*c^(3/2)*log(-2*a*c*x/(b*abs(b*x + a)))/b^4
- 1/2*sqrt(c*x^2)*a*c*x/b^2 + 1/3*(c*x^2)^(3/2)/b + sqrt(c*x^2)*a^2*c/b^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,x^2\right )}^{3/2}}{a+b\,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),x)

[Out]

int((c*x^2)^(3/2)/(a + b*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c x^{2}\right )^{\frac {3}{2}}}{a + b x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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