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

Optimal. Leaf size=94 \[ \frac{2 A \sqrt{x}}{b^2 \sqrt{b x+c x^2}}-\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{b^{5/2}}-\frac{2 x^{3/2} (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0766622, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {788, 666, 660, 207} \[ \frac{2 A \sqrt{x}}{b^2 \sqrt{b x+c x^2}}-\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{b^{5/2}}-\frac{2 x^{3/2} (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 788

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

Rule 666

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((2*c*d - b*e)*(d +
e*x)^m*(a + b*x + c*x^2)^(p + 1))/(e*(p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*c*d - b*e)*(m + 2*p + 2))/((p + 1)*
(b^2 - 4*a*c)), 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] && LtQ[p, -1] && LtQ[0, m, 1] && 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{x^{3/2} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx &=-\frac{2 (b B-A c) x^{3/2}}{3 b c \left (b x+c x^2\right )^{3/2}}+\frac{A \int \frac{\sqrt{x}}{\left (b x+c x^2\right )^{3/2}} \, dx}{b}\\ &=-\frac{2 (b B-A c) x^{3/2}}{3 b c \left (b x+c x^2\right )^{3/2}}+\frac{2 A \sqrt{x}}{b^2 \sqrt{b x+c x^2}}+\frac{A \int \frac{1}{\sqrt{x} \sqrt{b x+c x^2}} \, dx}{b^2}\\ &=-\frac{2 (b B-A c) x^{3/2}}{3 b c \left (b x+c x^2\right )^{3/2}}+\frac{2 A \sqrt{x}}{b^2 \sqrt{b x+c x^2}}+\frac{(2 A) \operatorname{Subst}\left (\int \frac{1}{-b+x^2} \, dx,x,\frac{\sqrt{b x+c x^2}}{\sqrt{x}}\right )}{b^2}\\ &=-\frac{2 (b B-A c) x^{3/2}}{3 b c \left (b x+c x^2\right )^{3/2}}+\frac{2 A \sqrt{x}}{b^2 \sqrt{b x+c x^2}}-\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{b^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0284159, size = 62, normalized size = 0.66 \[ \frac{2 x^{3/2} \left (b (A c-b B)+3 A c (b+c x) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{c x}{b}+1\right )\right )}{3 b^2 c (x (b+c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.016, size = 101, normalized size = 1.1 \begin{align*} -{\frac{2}{3\, \left ( cx+b \right ) ^{2}c}\sqrt{x \left ( cx+b \right ) } \left ( 3\,A{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ) x{c}^{2}\sqrt{cx+b}+3\,Ac{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ) b\sqrt{cx+b}-3\,A\sqrt{b}x{c}^{2}-4\,A{b}^{3/2}c+B{b}^{{\frac{5}{2}}} \right ){b}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.94924, size = 583, normalized size = 6.2 \begin{align*} \left [\frac{3 \,{\left (A c^{3} x^{3} + 2 \, A b c^{2} x^{2} + A b^{2} c x\right )} \sqrt{b} \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 \, A b c^{2} x - B b^{3} + 4 \, A b^{2} c\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{3 \,{\left (b^{3} c^{3} x^{3} + 2 \, b^{4} c^{2} x^{2} + b^{5} c x\right )}}, \frac{2 \,{\left (3 \,{\left (A c^{3} x^{3} + 2 \, A b c^{2} x^{2} + A b^{2} c x\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} \sqrt{x}}{\sqrt{c x^{2} + b x}}\right ) +{\left (3 \, A b c^{2} x - B b^{3} + 4 \, A b^{2} c\right )} \sqrt{c x^{2} + b x} \sqrt{x}\right )}}{3 \,{\left (b^{3} c^{3} x^{3} + 2 \, b^{4} c^{2} x^{2} + b^{5} c x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/3*(3*(A*c^3*x^3 + 2*A*b*c^2*x^2 + A*b^2*c*x)*sqrt(b)*log(-(c*x^2 + 2*b*x - 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt
(x))/x^2) + 2*(3*A*b*c^2*x - B*b^3 + 4*A*b^2*c)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^3*c^3*x^3 + 2*b^4*c^2*x^2 + b^5*
c*x), 2/3*(3*(A*c^3*x^3 + 2*A*b*c^2*x^2 + A*b^2*c*x)*sqrt(-b)*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (3*
A*b*c^2*x - B*b^3 + 4*A*b^2*c)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^3*c^3*x^3 + 2*b^4*c^2*x^2 + b^5*c*x)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.19064, size = 149, normalized size = 1.59 \begin{align*} \frac{2 \, A \arctan \left (\frac{\sqrt{c x + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b^{2}} - \frac{2 \,{\left (3 \, A \sqrt{b} c \arctan \left (\frac{\sqrt{b}}{\sqrt{-b}}\right ) - B \sqrt{-b} b + 4 \, A \sqrt{-b} c\right )}}{3 \, \sqrt{-b} b^{\frac{5}{2}} c} - \frac{2 \,{\left (B b^{2} - 3 \,{\left (c x + b\right )} A c - A b c\right )}}{3 \,{\left (c x + b\right )}^{\frac{3}{2}} b^{2} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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