3.220 \(\int \frac{(A+B x) (b x+c x^2)^{5/2}}{x^{9/2}} \, dx\)

Optimal. Leaf size=160 \[ -b^{3/2} (5 A c+2 b B) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )+\frac{\left (b x+c x^2\right )^{5/2} (5 A c+2 b B)}{5 b x^{5/2}}+\frac{\left (b x+c x^2\right )^{3/2} (5 A c+2 b B)}{3 x^{3/2}}+\frac{b \sqrt{b x+c x^2} (5 A c+2 b B)}{\sqrt{x}}-\frac{A \left (b x+c x^2\right )^{7/2}}{b x^{9/2}} \]

[Out]

(b*(2*b*B + 5*A*c)*Sqrt[b*x + c*x^2])/Sqrt[x] + ((2*b*B + 5*A*c)*(b*x + c*x^2)^(3/2))/(3*x^(3/2)) + ((2*b*B +
5*A*c)*(b*x + c*x^2)^(5/2))/(5*b*x^(5/2)) - (A*(b*x + c*x^2)^(7/2))/(b*x^(9/2)) - b^(3/2)*(2*b*B + 5*A*c)*ArcT
anh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])]

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Rubi [A]  time = 0.160474, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {792, 664, 660, 207} \[ -b^{3/2} (5 A c+2 b B) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )+\frac{\left (b x+c x^2\right )^{5/2} (5 A c+2 b B)}{5 b x^{5/2}}+\frac{\left (b x+c x^2\right )^{3/2} (5 A c+2 b B)}{3 x^{3/2}}+\frac{b \sqrt{b x+c x^2} (5 A c+2 b B)}{\sqrt{x}}-\frac{A \left (b x+c x^2\right )^{7/2}}{b x^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(9/2),x]

[Out]

(b*(2*b*B + 5*A*c)*Sqrt[b*x + c*x^2])/Sqrt[x] + ((2*b*B + 5*A*c)*(b*x + c*x^2)^(3/2))/(3*x^(3/2)) + ((2*b*B +
5*A*c)*(b*x + c*x^2)^(5/2))/(5*b*x^(5/2)) - (A*(b*x + c*x^2)^(7/2))/(b*x^(9/2)) - b^(3/2)*(2*b*B + 5*A*c)*ArcT
anh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])]

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 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 )^{5/2}}{x^{9/2}} \, dx &=-\frac{A \left (b x+c x^2\right )^{7/2}}{b x^{9/2}}+\frac{\left (-\frac{9}{2} (-b B+A c)+\frac{7}{2} (-b B+2 A c)\right ) \int \frac{\left (b x+c x^2\right )^{5/2}}{x^{7/2}} \, dx}{b}\\ &=\frac{(2 b B+5 A c) \left (b x+c x^2\right )^{5/2}}{5 b x^{5/2}}-\frac{A \left (b x+c x^2\right )^{7/2}}{b x^{9/2}}+\frac{1}{2} (2 b B+5 A c) \int \frac{\left (b x+c x^2\right )^{3/2}}{x^{5/2}} \, dx\\ &=\frac{(2 b B+5 A c) \left (b x+c x^2\right )^{3/2}}{3 x^{3/2}}+\frac{(2 b B+5 A c) \left (b x+c x^2\right )^{5/2}}{5 b x^{5/2}}-\frac{A \left (b x+c x^2\right )^{7/2}}{b x^{9/2}}+\frac{1}{2} (b (2 b B+5 A c)) \int \frac{\sqrt{b x+c x^2}}{x^{3/2}} \, dx\\ &=\frac{b (2 b B+5 A c) \sqrt{b x+c x^2}}{\sqrt{x}}+\frac{(2 b B+5 A c) \left (b x+c x^2\right )^{3/2}}{3 x^{3/2}}+\frac{(2 b B+5 A c) \left (b x+c x^2\right )^{5/2}}{5 b x^{5/2}}-\frac{A \left (b x+c x^2\right )^{7/2}}{b x^{9/2}}+\frac{1}{2} \left (b^2 (2 b B+5 A c)\right ) \int \frac{1}{\sqrt{x} \sqrt{b x+c x^2}} \, dx\\ &=\frac{b (2 b B+5 A c) \sqrt{b x+c x^2}}{\sqrt{x}}+\frac{(2 b B+5 A c) \left (b x+c x^2\right )^{3/2}}{3 x^{3/2}}+\frac{(2 b B+5 A c) \left (b x+c x^2\right )^{5/2}}{5 b x^{5/2}}-\frac{A \left (b x+c x^2\right )^{7/2}}{b x^{9/2}}+\left (b^2 (2 b B+5 A c)\right ) \operatorname{Subst}\left (\int \frac{1}{-b+x^2} \, dx,x,\frac{\sqrt{b x+c x^2}}{\sqrt{x}}\right )\\ &=\frac{b (2 b B+5 A c) \sqrt{b x+c x^2}}{\sqrt{x}}+\frac{(2 b B+5 A c) \left (b x+c x^2\right )^{3/2}}{3 x^{3/2}}+\frac{(2 b B+5 A c) \left (b x+c x^2\right )^{5/2}}{5 b x^{5/2}}-\frac{A \left (b x+c x^2\right )^{7/2}}{b x^{9/2}}-b^{3/2} (2 b B+5 A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0805569, size = 118, normalized size = 0.74 \[ \frac{\sqrt{x (b+c x)} \left (\sqrt{b+c x} \left (A \left (-15 b^2+70 b c x+10 c^2 x^2\right )+2 B x \left (23 b^2+11 b c x+3 c^2 x^2\right )\right )-15 b^{3/2} x (5 A c+2 b B) \tanh ^{-1}\left (\frac{\sqrt{b+c x}}{\sqrt{b}}\right )\right )}{15 x^{3/2} \sqrt{b+c x}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(9/2),x]

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[b + c*x]*(2*B*x*(23*b^2 + 11*b*c*x + 3*c^2*x^2) + A*(-15*b^2 + 70*b*c*x + 10*c^2*x^2)
) - 15*b^(3/2)*(2*b*B + 5*A*c)*x*ArcTanh[Sqrt[b + c*x]/Sqrt[b]]))/(15*x^(3/2)*Sqrt[b + c*x])

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Maple [A]  time = 0.017, size = 162, normalized size = 1. \begin{align*} -{\frac{1}{15}\sqrt{x \left ( cx+b \right ) } \left ( -6\,B{x}^{3}{c}^{2}\sqrt{b}\sqrt{cx+b}-10\,A{x}^{2}{c}^{2}\sqrt{b}\sqrt{cx+b}-22\,B{x}^{2}{b}^{3/2}c\sqrt{cx+b}+75\,A{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ) x{b}^{2}c-70\,Ax{b}^{3/2}c\sqrt{cx+b}+30\,B{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ) x{b}^{3}-46\,Bx{b}^{5/2}\sqrt{cx+b}+15\,A{b}^{5/2}\sqrt{cx+b} \right ){x}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{cx+b}}}{\frac{1}{\sqrt{b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^(5/2)/x^(9/2),x)

[Out]

-1/15*(x*(c*x+b))^(1/2)*(-6*B*x^3*c^2*b^(1/2)*(c*x+b)^(1/2)-10*A*x^2*c^2*b^(1/2)*(c*x+b)^(1/2)-22*B*x^2*b^(3/2
)*c*(c*x+b)^(1/2)+75*A*arctanh((c*x+b)^(1/2)/b^(1/2))*x*b^2*c-70*A*x*b^(3/2)*c*(c*x+b)^(1/2)+30*B*arctanh((c*x
+b)^(1/2)/b^(1/2))*x*b^3-46*B*x*b^(5/2)*(c*x+b)^(1/2)+15*A*b^(5/2)*(c*x+b)^(1/2))/x^(3/2)/(c*x+b)^(1/2)/b^(1/2
)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^(9/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.65076, size = 578, normalized size = 3.61 \begin{align*} \left [\frac{15 \,{\left (2 \, B b^{2} + 5 \, A b c\right )} \sqrt{b} x^{2} \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 (6 \, B c^{2} x^{3} - 15 \, A b^{2} + 2 \,{\left (11 \, B b c + 5 \, A c^{2}\right )} x^{2} + 2 \,{\left (23 \, B b^{2} + 35 \, A b c\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{30 \, x^{2}}, \frac{15 \,{\left (2 \, B b^{2} + 5 \, A b c\right )} \sqrt{-b} x^{2} \arctan \left (\frac{\sqrt{-b} \sqrt{x}}{\sqrt{c x^{2} + b x}}\right ) +{\left (6 \, B c^{2} x^{3} - 15 \, A b^{2} + 2 \,{\left (11 \, B b c + 5 \, A c^{2}\right )} x^{2} + 2 \,{\left (23 \, B b^{2} + 35 \, A b c\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{15 \, x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^(9/2),x, algorithm="fricas")

[Out]

[1/30*(15*(2*B*b^2 + 5*A*b*c)*sqrt(b)*x^2*log(-(c*x^2 + 2*b*x - 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*
(6*B*c^2*x^3 - 15*A*b^2 + 2*(11*B*b*c + 5*A*c^2)*x^2 + 2*(23*B*b^2 + 35*A*b*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/x
^2, 1/15*(15*(2*B*b^2 + 5*A*b*c)*sqrt(-b)*x^2*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (6*B*c^2*x^3 - 15*A
*b^2 + 2*(11*B*b*c + 5*A*c^2)*x^2 + 2*(23*B*b^2 + 35*A*b*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/x^2]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**(5/2)/x**(9/2),x)

[Out]

Timed out

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Giac [A]  time = 1.28758, size = 169, normalized size = 1.06 \begin{align*} \frac{6 \,{\left (c x + b\right )}^{\frac{5}{2}} B c + 10 \,{\left (c x + b\right )}^{\frac{3}{2}} B b c + 30 \, \sqrt{c x + b} B b^{2} c + 10 \,{\left (c x + b\right )}^{\frac{3}{2}} A c^{2} + 60 \, \sqrt{c x + b} A b c^{2} - \frac{15 \, \sqrt{c x + b} A b^{2} c}{x} + \frac{15 \,{\left (2 \, B b^{3} c + 5 \, A b^{2} c^{2}\right )} \arctan \left (\frac{\sqrt{c x + b}}{\sqrt{-b}}\right )}{\sqrt{-b}}}{15 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^(9/2),x, algorithm="giac")

[Out]

1/15*(6*(c*x + b)^(5/2)*B*c + 10*(c*x + b)^(3/2)*B*b*c + 30*sqrt(c*x + b)*B*b^2*c + 10*(c*x + b)^(3/2)*A*c^2 +
60*sqrt(c*x + b)*A*b*c^2 - 15*sqrt(c*x + b)*A*b^2*c/x + 15*(2*B*b^3*c + 5*A*b^2*c^2)*arctan(sqrt(c*x + b)/sqr
t(-b))/sqrt(-b))/c