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

Optimal. Leaf size=142 \[ \frac{a^4 c^2 \sqrt{c x^2}}{b^5}-\frac{a^3 c^2 x \sqrt{c x^2}}{2 b^4}+\frac{a^2 c^2 x^2 \sqrt{c x^2}}{3 b^3}-\frac{a^5 c^2 \sqrt{c x^2} \log (a+b x)}{b^6 x}-\frac{a c^2 x^3 \sqrt{c x^2}}{4 b^2}+\frac{c^2 x^4 \sqrt{c x^2}}{5 b} \]

[Out]

(a^4*c^2*Sqrt[c*x^2])/b^5 - (a^3*c^2*x*Sqrt[c*x^2])/(2*b^4) + (a^2*c^2*x^2*Sqrt[c*x^2])/(3*b^3) - (a*c^2*x^3*S
qrt[c*x^2])/(4*b^2) + (c^2*x^4*Sqrt[c*x^2])/(5*b) - (a^5*c^2*Sqrt[c*x^2]*Log[a + b*x])/(b^6*x)

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Rubi [A]  time = 0.044857, antiderivative size = 142, normalized size of antiderivative = 1., 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} \[ \frac{a^4 c^2 \sqrt{c x^2}}{b^5}-\frac{a^3 c^2 x \sqrt{c x^2}}{2 b^4}+\frac{a^2 c^2 x^2 \sqrt{c x^2}}{3 b^3}-\frac{a^5 c^2 \sqrt{c x^2} \log (a+b x)}{b^6 x}-\frac{a c^2 x^3 \sqrt{c x^2}}{4 b^2}+\frac{c^2 x^4 \sqrt{c x^2}}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[(c*x^2)^(5/2)/(a + b*x),x]

[Out]

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

Mathematica [A]  time = 0.0215217, size = 76, normalized size = 0.54 \[ \frac{c^3 x \left (b x \left (20 a^2 b^2 x^2-30 a^3 b x+60 a^4-15 a b^3 x^3+12 b^4 x^4\right )-60 a^5 \log (a+b x)\right )}{60 b^6 \sqrt{c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*x^2)^(5/2)/(a + b*x),x]

[Out]

(c^3*x*(b*x*(60*a^4 - 30*a^3*b*x + 20*a^2*b^2*x^2 - 15*a*b^3*x^3 + 12*b^4*x^4) - 60*a^5*Log[a + b*x]))/(60*b^6
*Sqrt[c*x^2])

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Maple [A]  time = 0.005, size = 74, normalized size = 0.5 \begin{align*} -{\frac{-12\,{b}^{5}{x}^{5}+15\,a{b}^{4}{x}^{4}-20\,{a}^{2}{b}^{3}{x}^{3}+30\,{a}^{3}{b}^{2}{x}^{2}+60\,{a}^{5}\ln \left ( bx+a \right ) -60\,{a}^{4}bx}{60\,{x}^{5}{b}^{6}} \left ( c{x}^{2} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(c*x^2)^(5/2)*(-12*b^5*x^5+15*a*b^4*x^4-20*a^2*b^3*x^3+30*a^3*b^2*x^2+60*a^5*ln(b*x+a)-60*a^4*b*x)/x^5/b
^6

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.59848, size = 198, normalized size = 1.39 \begin{align*} \frac{{\left (12 \, b^{5} c^{2} x^{5} - 15 \, a b^{4} c^{2} x^{4} + 20 \, a^{2} b^{3} c^{2} x^{3} - 30 \, a^{3} b^{2} c^{2} x^{2} + 60 \, a^{4} b c^{2} x - 60 \, a^{5} c^{2} \log \left (b x + a\right )\right )} \sqrt{c x^{2}}}{60 \, b^{6} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(12*b^5*c^2*x^5 - 15*a*b^4*c^2*x^4 + 20*a^2*b^3*c^2*x^3 - 30*a^3*b^2*c^2*x^2 + 60*a^4*b*c^2*x - 60*a^5*c^
2*log(b*x + a))*sqrt(c*x^2)/(b^6*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c x^{2}\right )^{\frac{5}{2}}}{a + b x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.04936, size = 157, normalized size = 1.11 \begin{align*} -\frac{1}{60} \,{\left (\frac{60 \, a^{5} c^{2} \log \left ({\left | b x + a \right |}\right ) \mathrm{sgn}\left (x\right )}{b^{6}} - \frac{60 \, a^{5} c^{2} \log \left ({\left | a \right |}\right ) \mathrm{sgn}\left (x\right )}{b^{6}} - \frac{12 \, b^{4} c^{2} x^{5} \mathrm{sgn}\left (x\right ) - 15 \, a b^{3} c^{2} x^{4} \mathrm{sgn}\left (x\right ) + 20 \, a^{2} b^{2} c^{2} x^{3} \mathrm{sgn}\left (x\right ) - 30 \, a^{3} b c^{2} x^{2} \mathrm{sgn}\left (x\right ) + 60 \, a^{4} c^{2} x \mathrm{sgn}\left (x\right )}{b^{5}}\right )} \sqrt{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(60*a^5*c^2*log(abs(b*x + a))*sgn(x)/b^6 - 60*a^5*c^2*log(abs(a))*sgn(x)/b^6 - (12*b^4*c^2*x^5*sgn(x) -
15*a*b^3*c^2*x^4*sgn(x) + 20*a^2*b^2*c^2*x^3*sgn(x) - 30*a^3*b*c^2*x^2*sgn(x) + 60*a^4*c^2*x*sgn(x))/b^5)*sqrt
(c)