3.87 \(\int \frac{(A+B x) (b x+c x^2)^{3/2}}{x^5} \, dx\)

Optimal. Leaf size=95 \[ -\frac{2 A \left (b x+c x^2\right )^{5/2}}{5 b x^5}+2 B c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )-\frac{2 B \left (b x+c x^2\right )^{3/2}}{3 x^3}-\frac{2 B c \sqrt{b x+c x^2}}{x} \]

[Out]

(-2*B*c*Sqrt[b*x + c*x^2])/x - (2*B*(b*x + c*x^2)^(3/2))/(3*x^3) - (2*A*(b*x + c*x^2)^(5/2))/(5*b*x^5) + 2*B*c
^(3/2)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]]

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Rubi [A]  time = 0.103551, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {792, 662, 620, 206} \[ -\frac{2 A \left (b x+c x^2\right )^{5/2}}{5 b x^5}+2 B c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )-\frac{2 B \left (b x+c x^2\right )^{3/2}}{3 x^3}-\frac{2 B c \sqrt{b x+c x^2}}{x} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(3/2))/x^5,x]

[Out]

(-2*B*c*Sqrt[b*x + c*x^2])/x - (2*B*(b*x + c*x^2)^(3/2))/(3*x^3) - (2*A*(b*x + c*x^2)^(5/2))/(5*b*x^5) + 2*B*c
^(3/2)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]]

Rule 792

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d*g - e*f)*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/((2*c*d - b*e)*(m + p + 1)), x] + Dist[(m*(g*(c*d - b*e)
+ c*e*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p,
x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((L
tQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0]

Rule 662

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + p + 1)), x] - Dist[(c*p)/(e^2*(m + p + 1)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2
)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[
p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] && IntegerQ[2*p]

Rule 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{(A+B x) \left (b x+c x^2\right )^{3/2}}{x^5} \, dx &=-\frac{2 A \left (b x+c x^2\right )^{5/2}}{5 b x^5}+B \int \frac{\left (b x+c x^2\right )^{3/2}}{x^4} \, dx\\ &=-\frac{2 B \left (b x+c x^2\right )^{3/2}}{3 x^3}-\frac{2 A \left (b x+c x^2\right )^{5/2}}{5 b x^5}+(B c) \int \frac{\sqrt{b x+c x^2}}{x^2} \, dx\\ &=-\frac{2 B c \sqrt{b x+c x^2}}{x}-\frac{2 B \left (b x+c x^2\right )^{3/2}}{3 x^3}-\frac{2 A \left (b x+c x^2\right )^{5/2}}{5 b x^5}+\left (B c^2\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx\\ &=-\frac{2 B c \sqrt{b x+c x^2}}{x}-\frac{2 B \left (b x+c x^2\right )^{3/2}}{3 x^3}-\frac{2 A \left (b x+c x^2\right )^{5/2}}{5 b x^5}+\left (2 B c^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )\\ &=-\frac{2 B c \sqrt{b x+c x^2}}{x}-\frac{2 B \left (b x+c x^2\right )^{3/2}}{3 x^3}-\frac{2 A \left (b x+c x^2\right )^{5/2}}{5 b x^5}+2 B c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0860153, size = 88, normalized size = 0.93 \[ \frac{2 \sqrt{x (b+c x)} \left ((b+c x)^2 \sqrt{\frac{c x}{b}+1} (b B-A c)-b^3 B \, _2F_1\left (-\frac{5}{2},-\frac{5}{2};-\frac{3}{2};-\frac{c x}{b}\right )\right )}{5 b c x^3 \sqrt{\frac{c x}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(3/2))/x^5,x]

[Out]

(2*Sqrt[x*(b + c*x)]*((b*B - A*c)*(b + c*x)^2*Sqrt[1 + (c*x)/b] - b^3*B*Hypergeometric2F1[-5/2, -5/2, -3/2, -(
(c*x)/b)]))/(5*b*c*x^3*Sqrt[1 + (c*x)/b])

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Maple [B]  time = 0.01, size = 176, normalized size = 1.9 \begin{align*} -{\frac{2\,B}{3\,b{x}^{4}} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}-{\frac{4\,Bc}{3\,{b}^{2}{x}^{3}} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}+{\frac{16\,B{c}^{2}}{3\,{b}^{3}{x}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}-{\frac{16\,B{c}^{3}}{3\,{b}^{3}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-4\,{\frac{B{c}^{3}\sqrt{c{x}^{2}+bx}x}{{b}^{2}}}-2\,{\frac{B{c}^{2}\sqrt{c{x}^{2}+bx}}{b}}+B{c}^{{\frac{3}{2}}}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ) -{\frac{2\,A}{5\,b{x}^{5}} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^(3/2)/x^5,x)

[Out]

-2/3*B/b/x^4*(c*x^2+b*x)^(5/2)-4/3*B/b^2*c/x^3*(c*x^2+b*x)^(5/2)+16/3*B/b^3*c^2/x^2*(c*x^2+b*x)^(5/2)-16/3*B/b
^3*c^3*(c*x^2+b*x)^(3/2)-4*B/b^2*c^3*(c*x^2+b*x)^(1/2)*x-2*B/b*c^2*(c*x^2+b*x)^(1/2)+B*c^(3/2)*ln((1/2*b+c*x)/
c^(1/2)+(c*x^2+b*x)^(1/2))-2/5*A*(c*x^2+b*x)^(5/2)/b/x^5

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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((B*x+A)*(c*x^2+b*x)^(3/2)/x^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.94417, size = 443, normalized size = 4.66 \begin{align*} \left [\frac{15 \, B b c^{\frac{3}{2}} x^{3} \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) - 2 \,{\left (3 \, A b^{2} +{\left (20 \, B b c + 3 \, A c^{2}\right )} x^{2} +{\left (5 \, B b^{2} + 6 \, A b c\right )} x\right )} \sqrt{c x^{2} + b x}}{15 \, b x^{3}}, -\frac{2 \,{\left (15 \, B b \sqrt{-c} c x^{3} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) +{\left (3 \, A b^{2} +{\left (20 \, B b c + 3 \, A c^{2}\right )} x^{2} +{\left (5 \, B b^{2} + 6 \, A b c\right )} x\right )} \sqrt{c x^{2} + b x}\right )}}{15 \, b x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(3/2)/x^5,x, algorithm="fricas")

[Out]

[1/15*(15*B*b*c^(3/2)*x^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(3*A*b^2 + (20*B*b*c + 3*A*c^2)*x^2
+ (5*B*b^2 + 6*A*b*c)*x)*sqrt(c*x^2 + b*x))/(b*x^3), -2/15*(15*B*b*sqrt(-c)*c*x^3*arctan(sqrt(c*x^2 + b*x)*sq
rt(-c)/(c*x)) + (3*A*b^2 + (20*B*b*c + 3*A*c^2)*x^2 + (5*B*b^2 + 6*A*b*c)*x)*sqrt(c*x^2 + b*x))/(b*x^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (x \left (b + c x\right )\right )^{\frac{3}{2}} \left (A + B x\right )}{x^{5}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**(3/2)/x**5,x)

[Out]

Integral((x*(b + c*x))**(3/2)*(A + B*x)/x**5, x)

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Giac [B]  time = 1.20063, size = 365, normalized size = 3.84 \begin{align*} -B c^{\frac{3}{2}} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right ) + \frac{2 \,{\left (30 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{4} B b c^{\frac{3}{2}} + 15 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{4} A c^{\frac{5}{2}} + 15 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} B b^{2} c + 30 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} A b c^{2} + 5 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} B b^{3} \sqrt{c} + 30 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} A b^{2} c^{\frac{3}{2}} + 15 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} A b^{3} c + 3 \, A b^{4} \sqrt{c}\right )}}{15 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{5} \sqrt{c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(3/2)/x^5,x, algorithm="giac")

[Out]

-B*c^(3/2)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b)) + 2/15*(30*(sqrt(c)*x - sqrt(c*x^2 + b*x))
^4*B*b*c^(3/2) + 15*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*A*c^(5/2) + 15*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b^2*c
+ 30*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*A*b*c^2 + 5*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^3*sqrt(c) + 30*(sqrt
(c)*x - sqrt(c*x^2 + b*x))^2*A*b^2*c^(3/2) + 15*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^3*c + 3*A*b^4*sqrt(c))/((s
qrt(c)*x - sqrt(c*x^2 + b*x))^5*sqrt(c))