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

Optimal. Leaf size=105 \[ -2 A b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )+\frac{2 A \left (b x+c x^2\right )^{3/2}}{3 x^{3/2}}+\frac{2 A b \sqrt{b x+c x^2}}{\sqrt{x}}+\frac{2 B \left (b x+c x^2\right )^{5/2}}{5 c x^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.0857289, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {794, 664, 660, 207} \[ -2 A b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )+\frac{2 A \left (b x+c x^2\right )^{3/2}}{3 x^{3/2}}+\frac{2 A b \sqrt{b x+c x^2}}{\sqrt{x}}+\frac{2 B \left (b x+c x^2\right )^{5/2}}{5 c x^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 794

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)
*(2*c*f - b*g))/(c*e*(m + 2*p + 2)), Int[(d + e*x)^m*(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] && NeQ[m + 2*p + 2, 0] && (NeQ[m, 2] || Eq
Q[d, 0])

Rule 664

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 + 2*p + 1)), x] - Dist[(p*(2*c*d - b*e))/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*
(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] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rule 660

Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e, Subst[Int[1/(
2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*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]

Rule 207

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

Rubi steps

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

Mathematica [A]  time = 0.0844592, size = 100, normalized size = 0.95 \[ \frac{2 \sqrt{x} \sqrt{b+c x} \left (\sqrt{b+c x} \left (b (20 A c+6 B c x)+c^2 x (5 A+3 B x)+3 b^2 B\right )-15 A b^{3/2} c \tanh ^{-1}\left (\frac{\sqrt{b+c x}}{\sqrt{b}}\right )\right )}{15 c \sqrt{x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x]*Sqrt[b + c*x]*(Sqrt[b + c*x]*(3*b^2*B + c^2*x*(5*A + 3*B*x) + b*(20*A*c + 6*B*c*x)) - 15*A*b^(3/2)*
c*ArcTanh[Sqrt[b + c*x]/Sqrt[b]]))/(15*c*Sqrt[x*(b + c*x)])

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Maple [A]  time = 0.011, size = 113, normalized size = 1.1 \begin{align*} -{\frac{2}{15\,c}\sqrt{x \left ( cx+b \right ) } \left ( -3\,B{x}^{2}{c}^{2}\sqrt{cx+b}+15\,A{b}^{3/2}c{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ) -5\,Ax{c}^{2}\sqrt{cx+b}-6\,Bxbc\sqrt{cx+b}-20\,Abc\sqrt{cx+b}-3\,B{b}^{2}\sqrt{cx+b} \right ){\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{cx+b}}}} \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^(5/2),x)

[Out]

-2/15*(x*(c*x+b))^(1/2)*(-3*B*x^2*c^2*(c*x+b)^(1/2)+15*A*b^(3/2)*c*arctanh((c*x+b)^(1/2)/b^(1/2))-5*A*x*c^2*(c
*x+b)^(1/2)-6*B*x*b*c*(c*x+b)^(1/2)-20*A*b*c*(c*x+b)^(1/2)-3*B*b^2*(c*x+b)^(1/2))/x^(1/2)/(c*x+b)^(1/2)/c

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} A b \int \frac{\sqrt{c x + b}}{x}\,{d x} + \frac{2 \,{\left (5 \,{\left (B b c + A c^{2}\right )} x^{2} +{\left (3 \, B c^{2} x^{2} + B b c x - 2 \, B b^{2}\right )} x + 5 \,{\left (B b^{2} + A b c\right )} x\right )} \sqrt{c x + b}}{15 \, c x} \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^(5/2),x, algorithm="maxima")

[Out]

A*b*integrate(sqrt(c*x + b)/x, x) + 2/15*(5*(B*b*c + A*c^2)*x^2 + (3*B*c^2*x^2 + B*b*c*x - 2*B*b^2)*x + 5*(B*b
^2 + A*b*c)*x)*sqrt(c*x + b)/(c*x)

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Fricas [A]  time = 1.63185, size = 475, normalized size = 4.52 \begin{align*} \left [\frac{15 \, A b^{\frac{3}{2}} c x \log \left (-\frac{c x^{2} + 2 \, b x - 2 \, \sqrt{c x^{2} + b x} \sqrt{b} \sqrt{x}}{x^{2}}\right ) + 2 \,{\left (3 \, B c^{2} x^{2} + 3 \, B b^{2} + 20 \, A b c +{\left (6 \, B b c + 5 \, A c^{2}\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{15 \, c x}, \frac{2 \,{\left (15 \, A \sqrt{-b} b c x \arctan \left (\frac{\sqrt{-b} \sqrt{x}}{\sqrt{c x^{2} + b x}}\right ) +{\left (3 \, B c^{2} x^{2} + 3 \, B b^{2} + 20 \, A b c +{\left (6 \, B b c + 5 \, A c^{2}\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{x}\right )}}{15 \, c x}\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^(5/2),x, algorithm="fricas")

[Out]

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

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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^{\frac{5}{2}}}\, 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**(5/2),x)

[Out]

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

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Giac [A]  time = 1.173, size = 166, normalized size = 1.58 \begin{align*} \frac{2 \, A b^{2} \arctan \left (\frac{\sqrt{c x + b}}{\sqrt{-b}}\right )}{\sqrt{-b}} - \frac{2 \,{\left (15 \, A b^{2} c \arctan \left (\frac{\sqrt{b}}{\sqrt{-b}}\right ) + 3 \, B \sqrt{-b} b^{\frac{5}{2}} + 20 \, A \sqrt{-b} b^{\frac{3}{2}} c\right )}}{15 \, \sqrt{-b} c} + \frac{2 \,{\left (3 \,{\left (c x + b\right )}^{\frac{5}{2}} B c^{4} + 5 \,{\left (c x + b\right )}^{\frac{3}{2}} A c^{5} + 15 \, \sqrt{c x + b} A b c^{5}\right )}}{15 \, c^{5}} \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^(5/2),x, algorithm="giac")

[Out]

2*A*b^2*arctan(sqrt(c*x + b)/sqrt(-b))/sqrt(-b) - 2/15*(15*A*b^2*c*arctan(sqrt(b)/sqrt(-b)) + 3*B*sqrt(-b)*b^(
5/2) + 20*A*sqrt(-b)*b^(3/2)*c)/(sqrt(-b)*c) + 2/15*(3*(c*x + b)^(5/2)*B*c^4 + 5*(c*x + b)^(3/2)*A*c^5 + 15*sq
rt(c*x + b)*A*b*c^5)/c^5