3.221 \(\int \frac{(A+B x) (b x+c x^2)^{5/2}}{x^{11/2}} \, dx\)

Optimal. Leaf size=172 \[ -\frac{\left (b x+c x^2\right )^{5/2} (3 A c+4 b B)}{4 b x^{7/2}}+\frac{5 c \left (b x+c x^2\right )^{3/2} (3 A c+4 b B)}{12 b x^{3/2}}+\frac{5 c \sqrt{b x+c x^2} (3 A c+4 b B)}{4 \sqrt{x}}-\frac{5}{4} \sqrt{b} c (3 A c+4 b B) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )-\frac{A \left (b x+c x^2\right )^{7/2}}{2 b x^{11/2}} \]

[Out]

(5*c*(4*b*B + 3*A*c)*Sqrt[b*x + c*x^2])/(4*Sqrt[x]) + (5*c*(4*b*B + 3*A*c)*(b*x + c*x^2)^(3/2))/(12*b*x^(3/2))
- ((4*b*B + 3*A*c)*(b*x + c*x^2)^(5/2))/(4*b*x^(7/2)) - (A*(b*x + c*x^2)^(7/2))/(2*b*x^(11/2)) - (5*Sqrt[b]*c
*(4*b*B + 3*A*c)*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/4

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Rubi [A]  time = 0.15943, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {792, 662, 664, 660, 207} \[ -\frac{\left (b x+c x^2\right )^{5/2} (3 A c+4 b B)}{4 b x^{7/2}}+\frac{5 c \left (b x+c x^2\right )^{3/2} (3 A c+4 b B)}{12 b x^{3/2}}+\frac{5 c \sqrt{b x+c x^2} (3 A c+4 b B)}{4 \sqrt{x}}-\frac{5}{4} \sqrt{b} c (3 A c+4 b B) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )-\frac{A \left (b x+c x^2\right )^{7/2}}{2 b x^{11/2}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(11/2),x]

[Out]

(5*c*(4*b*B + 3*A*c)*Sqrt[b*x + c*x^2])/(4*Sqrt[x]) + (5*c*(4*b*B + 3*A*c)*(b*x + c*x^2)^(3/2))/(12*b*x^(3/2))
- ((4*b*B + 3*A*c)*(b*x + c*x^2)^(5/2))/(4*b*x^(7/2)) - (A*(b*x + c*x^2)^(7/2))/(2*b*x^(11/2)) - (5*Sqrt[b]*c
*(4*b*B + 3*A*c)*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/4

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 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^{11/2}} \, dx &=-\frac{A \left (b x+c x^2\right )^{7/2}}{2 b x^{11/2}}+\frac{\left (-\frac{11}{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^{9/2}} \, dx}{2 b}\\ &=-\frac{(4 b B+3 A c) \left (b x+c x^2\right )^{5/2}}{4 b x^{7/2}}-\frac{A \left (b x+c x^2\right )^{7/2}}{2 b x^{11/2}}+\frac{(5 c (4 b B+3 A c)) \int \frac{\left (b x+c x^2\right )^{3/2}}{x^{5/2}} \, dx}{8 b}\\ &=\frac{5 c (4 b B+3 A c) \left (b x+c x^2\right )^{3/2}}{12 b x^{3/2}}-\frac{(4 b B+3 A c) \left (b x+c x^2\right )^{5/2}}{4 b x^{7/2}}-\frac{A \left (b x+c x^2\right )^{7/2}}{2 b x^{11/2}}+\frac{1}{8} (5 c (4 b B+3 A c)) \int \frac{\sqrt{b x+c x^2}}{x^{3/2}} \, dx\\ &=\frac{5 c (4 b B+3 A c) \sqrt{b x+c x^2}}{4 \sqrt{x}}+\frac{5 c (4 b B+3 A c) \left (b x+c x^2\right )^{3/2}}{12 b x^{3/2}}-\frac{(4 b B+3 A c) \left (b x+c x^2\right )^{5/2}}{4 b x^{7/2}}-\frac{A \left (b x+c x^2\right )^{7/2}}{2 b x^{11/2}}+\frac{1}{8} (5 b c (4 b B+3 A c)) \int \frac{1}{\sqrt{x} \sqrt{b x+c x^2}} \, dx\\ &=\frac{5 c (4 b B+3 A c) \sqrt{b x+c x^2}}{4 \sqrt{x}}+\frac{5 c (4 b B+3 A c) \left (b x+c x^2\right )^{3/2}}{12 b x^{3/2}}-\frac{(4 b B+3 A c) \left (b x+c x^2\right )^{5/2}}{4 b x^{7/2}}-\frac{A \left (b x+c x^2\right )^{7/2}}{2 b x^{11/2}}+\frac{1}{4} (5 b c (4 b B+3 A c)) \operatorname{Subst}\left (\int \frac{1}{-b+x^2} \, dx,x,\frac{\sqrt{b x+c x^2}}{\sqrt{x}}\right )\\ &=\frac{5 c (4 b B+3 A c) \sqrt{b x+c x^2}}{4 \sqrt{x}}+\frac{5 c (4 b B+3 A c) \left (b x+c x^2\right )^{3/2}}{12 b x^{3/2}}-\frac{(4 b B+3 A c) \left (b x+c x^2\right )^{5/2}}{4 b x^{7/2}}-\frac{A \left (b x+c x^2\right )^{7/2}}{2 b x^{11/2}}-\frac{5}{4} \sqrt{b} c (4 b B+3 A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )\\ \end{align*}

Mathematica [C]  time = 0.033056, size = 67, normalized size = 0.39 \[ \frac{(b+c x)^3 \sqrt{x (b+c x)} \left (c x^2 (3 A c+4 b B) \, _2F_1\left (2,\frac{7}{2};\frac{9}{2};\frac{c x}{b}+1\right )-7 A b^2\right )}{14 b^3 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(11/2),x]

[Out]

((b + c*x)^3*Sqrt[x*(b + c*x)]*(-7*A*b^2 + c*(4*b*B + 3*A*c)*x^2*Hypergeometric2F1[2, 7/2, 9/2, 1 + (c*x)/b]))
/(14*b^3*x^(5/2))

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Maple [A]  time = 0.018, size = 167, normalized size = 1. \begin{align*} -{\frac{1}{12}\sqrt{x \left ( cx+b \right ) } \left ( -8\,B{x}^{3}{c}^{2}\sqrt{b}\sqrt{cx+b}+45\,A{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ){x}^{2}b{c}^{2}-24\,A{x}^{2}{c}^{2}\sqrt{b}\sqrt{cx+b}+60\,B{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ){x}^{2}{b}^{2}c-56\,B{x}^{2}{b}^{3/2}c\sqrt{cx+b}+27\,Ax{b}^{3/2}c\sqrt{cx+b}+12\,Bx{b}^{5/2}\sqrt{cx+b}+6\,A{b}^{5/2}\sqrt{cx+b} \right ){x}^{-{\frac{5}{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^(11/2),x)

[Out]

-1/12*(x*(c*x+b))^(1/2)*(-8*B*x^3*c^2*b^(1/2)*(c*x+b)^(1/2)+45*A*arctanh((c*x+b)^(1/2)/b^(1/2))*x^2*b*c^2-24*A
*x^2*c^2*b^(1/2)*(c*x+b)^(1/2)+60*B*arctanh((c*x+b)^(1/2)/b^(1/2))*x^2*b^2*c-56*B*x^2*b^(3/2)*c*(c*x+b)^(1/2)+
27*A*x*b^(3/2)*c*(c*x+b)^(1/2)+12*B*x*b^(5/2)*(c*x+b)^(1/2)+6*A*b^(5/2)*(c*x+b)^(1/2))/x^(5/2)/(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}{3} \,{\left (B c^{2} x + B b c\right )} \sqrt{c x + b} + \int \frac{{\left (A b^{2} +{\left (2 \, B b c + A c^{2}\right )} x^{2} +{\left (B b^{2} + 2 \, A b c\right )} x\right )} \sqrt{c x + b}}{x^{3}}\,{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^(11/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.80914, size = 567, normalized size = 3.3 \begin{align*} \left [\frac{15 \,{\left (4 \, B b c + 3 \, A c^{2}\right )} \sqrt{b} x^{3} \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 (8 \, B c^{2} x^{3} - 6 \, A b^{2} + 8 \,{\left (7 \, B b c + 3 \, A c^{2}\right )} x^{2} - 3 \,{\left (4 \, B b^{2} + 9 \, A b c\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{24 \, x^{3}}, \frac{15 \,{\left (4 \, B b c + 3 \, A c^{2}\right )} \sqrt{-b} x^{3} \arctan \left (\frac{\sqrt{-b} \sqrt{x}}{\sqrt{c x^{2} + b x}}\right ) +{\left (8 \, B c^{2} x^{3} - 6 \, A b^{2} + 8 \,{\left (7 \, B b c + 3 \, A c^{2}\right )} x^{2} - 3 \,{\left (4 \, B b^{2} + 9 \, A b c\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{12 \, 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)^(5/2)/x^(11/2),x, algorithm="fricas")

[Out]

[1/24*(15*(4*B*b*c + 3*A*c^2)*sqrt(b)*x^3*log(-(c*x^2 + 2*b*x - 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*
(8*B*c^2*x^3 - 6*A*b^2 + 8*(7*B*b*c + 3*A*c^2)*x^2 - 3*(4*B*b^2 + 9*A*b*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/x^3,
1/12*(15*(4*B*b*c + 3*A*c^2)*sqrt(-b)*x^3*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (8*B*c^2*x^3 - 6*A*b^2
+ 8*(7*B*b*c + 3*A*c^2)*x^2 - 3*(4*B*b^2 + 9*A*b*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/x^3]

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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**(11/2),x)

[Out]

Timed out

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Giac [A]  time = 1.27291, size = 209, normalized size = 1.22 \begin{align*} \frac{8 \,{\left (c x + b\right )}^{\frac{3}{2}} B c^{2} + 48 \, \sqrt{c x + b} B b c^{2} + 24 \, \sqrt{c x + b} A c^{3} + \frac{15 \,{\left (4 \, B b^{2} c^{2} + 3 \, A b c^{3}\right )} \arctan \left (\frac{\sqrt{c x + b}}{\sqrt{-b}}\right )}{\sqrt{-b}} - \frac{3 \,{\left (4 \,{\left (c x + b\right )}^{\frac{3}{2}} B b^{2} c^{2} - 4 \, \sqrt{c x + b} B b^{3} c^{2} + 9 \,{\left (c x + b\right )}^{\frac{3}{2}} A b c^{3} - 7 \, \sqrt{c x + b} A b^{2} c^{3}\right )}}{c^{2} x^{2}}}{12 \, 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^(11/2),x, algorithm="giac")

[Out]

1/12*(8*(c*x + b)^(3/2)*B*c^2 + 48*sqrt(c*x + b)*B*b*c^2 + 24*sqrt(c*x + b)*A*c^3 + 15*(4*B*b^2*c^2 + 3*A*b*c^
3)*arctan(sqrt(c*x + b)/sqrt(-b))/sqrt(-b) - 3*(4*(c*x + b)^(3/2)*B*b^2*c^2 - 4*sqrt(c*x + b)*B*b^3*c^2 + 9*(c
*x + b)^(3/2)*A*b*c^3 - 7*sqrt(c*x + b)*A*b^2*c^3)/(c^2*x^2))/c