3.222 \(\int \frac{(A+B x) (b x+c x^2)^{5/2}}{x^{13/2}} \, dx\)

Optimal. Leaf size=175 \[ \frac{5 c^2 \sqrt{b x+c x^2} (A c+6 b B)}{8 b \sqrt{x}}-\frac{5 c^2 (A c+6 b B) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{8 \sqrt{b}}-\frac{\left (b x+c x^2\right )^{5/2} (A c+6 b B)}{12 b x^{9/2}}-\frac{5 c \left (b x+c x^2\right )^{3/2} (A c+6 b B)}{24 b x^{5/2}}-\frac{A \left (b x+c x^2\right )^{7/2}}{3 b x^{13/2}} \]

[Out]

(5*c^2*(6*b*B + A*c)*Sqrt[b*x + c*x^2])/(8*b*Sqrt[x]) - (5*c*(6*b*B + A*c)*(b*x + c*x^2)^(3/2))/(24*b*x^(5/2))
- ((6*b*B + A*c)*(b*x + c*x^2)^(5/2))/(12*b*x^(9/2)) - (A*(b*x + c*x^2)^(7/2))/(3*b*x^(13/2)) - (5*c^2*(6*b*B
+ A*c)*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/(8*Sqrt[b])

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Rubi [A]  time = 0.16988, antiderivative size = 175, 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} \[ \frac{5 c^2 \sqrt{b x+c x^2} (A c+6 b B)}{8 b \sqrt{x}}-\frac{5 c^2 (A c+6 b B) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{8 \sqrt{b}}-\frac{\left (b x+c x^2\right )^{5/2} (A c+6 b B)}{12 b x^{9/2}}-\frac{5 c \left (b x+c x^2\right )^{3/2} (A c+6 b B)}{24 b x^{5/2}}-\frac{A \left (b x+c x^2\right )^{7/2}}{3 b x^{13/2}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(13/2),x]

[Out]

(5*c^2*(6*b*B + A*c)*Sqrt[b*x + c*x^2])/(8*b*Sqrt[x]) - (5*c*(6*b*B + A*c)*(b*x + c*x^2)^(3/2))/(24*b*x^(5/2))
- ((6*b*B + A*c)*(b*x + c*x^2)^(5/2))/(12*b*x^(9/2)) - (A*(b*x + c*x^2)^(7/2))/(3*b*x^(13/2)) - (5*c^2*(6*b*B
+ A*c)*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/(8*Sqrt[b])

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^{13/2}} \, dx &=-\frac{A \left (b x+c x^2\right )^{7/2}}{3 b x^{13/2}}+\frac{\left (-\frac{13}{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^{11/2}} \, dx}{3 b}\\ &=-\frac{(6 b B+A c) \left (b x+c x^2\right )^{5/2}}{12 b x^{9/2}}-\frac{A \left (b x+c x^2\right )^{7/2}}{3 b x^{13/2}}+\frac{(5 c (6 b B+A c)) \int \frac{\left (b x+c x^2\right )^{3/2}}{x^{7/2}} \, dx}{24 b}\\ &=-\frac{5 c (6 b B+A c) \left (b x+c x^2\right )^{3/2}}{24 b x^{5/2}}-\frac{(6 b B+A c) \left (b x+c x^2\right )^{5/2}}{12 b x^{9/2}}-\frac{A \left (b x+c x^2\right )^{7/2}}{3 b x^{13/2}}+\frac{\left (5 c^2 (6 b B+A c)\right ) \int \frac{\sqrt{b x+c x^2}}{x^{3/2}} \, dx}{16 b}\\ &=\frac{5 c^2 (6 b B+A c) \sqrt{b x+c x^2}}{8 b \sqrt{x}}-\frac{5 c (6 b B+A c) \left (b x+c x^2\right )^{3/2}}{24 b x^{5/2}}-\frac{(6 b B+A c) \left (b x+c x^2\right )^{5/2}}{12 b x^{9/2}}-\frac{A \left (b x+c x^2\right )^{7/2}}{3 b x^{13/2}}+\frac{1}{16} \left (5 c^2 (6 b B+A c)\right ) \int \frac{1}{\sqrt{x} \sqrt{b x+c x^2}} \, dx\\ &=\frac{5 c^2 (6 b B+A c) \sqrt{b x+c x^2}}{8 b \sqrt{x}}-\frac{5 c (6 b B+A c) \left (b x+c x^2\right )^{3/2}}{24 b x^{5/2}}-\frac{(6 b B+A c) \left (b x+c x^2\right )^{5/2}}{12 b x^{9/2}}-\frac{A \left (b x+c x^2\right )^{7/2}}{3 b x^{13/2}}+\frac{1}{8} \left (5 c^2 (6 b B+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{5 c^2 (6 b B+A c) \sqrt{b x+c x^2}}{8 b \sqrt{x}}-\frac{5 c (6 b B+A c) \left (b x+c x^2\right )^{3/2}}{24 b x^{5/2}}-\frac{(6 b B+A c) \left (b x+c x^2\right )^{5/2}}{12 b x^{9/2}}-\frac{A \left (b x+c x^2\right )^{7/2}}{3 b x^{13/2}}-\frac{5 c^2 (6 b B+A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{8 \sqrt{b}}\\ \end{align*}

Mathematica [C]  time = 0.0358071, size = 68, normalized size = 0.39 \[ -\frac{(b+c x)^3 \sqrt{x (b+c x)} \left (7 A b^3+c^2 x^3 (A c+6 b B) \, _2F_1\left (3,\frac{7}{2};\frac{9}{2};\frac{c x}{b}+1\right )\right )}{21 b^4 x^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(13/2),x]

[Out]

-((b + c*x)^3*Sqrt[x*(b + c*x)]*(7*A*b^3 + c^2*(6*b*B + A*c)*x^3*Hypergeometric2F1[3, 7/2, 9/2, 1 + (c*x)/b]))
/(21*b^4*x^(7/2))

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Maple [A]  time = 0.02, size = 166, normalized size = 1. \begin{align*} -{\frac{1}{24}\sqrt{x \left ( cx+b \right ) } \left ( 15\,A{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ){x}^{3}{c}^{3}+90\,B{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ){x}^{3}b{c}^{2}-48\,B{x}^{3}{c}^{2}\sqrt{b}\sqrt{cx+b}+33\,A{x}^{2}{c}^{2}\sqrt{b}\sqrt{cx+b}+54\,B{x}^{2}{b}^{3/2}c\sqrt{cx+b}+26\,Ax{b}^{3/2}c\sqrt{cx+b}+12\,Bx{b}^{5/2}\sqrt{cx+b}+8\,A{b}^{5/2}\sqrt{cx+b} \right ){x}^{-{\frac{7}{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^(13/2),x)

[Out]

-1/24*(x*(c*x+b))^(1/2)*(15*A*arctanh((c*x+b)^(1/2)/b^(1/2))*x^3*c^3+90*B*arctanh((c*x+b)^(1/2)/b^(1/2))*x^3*b
*c^2-48*B*x^3*c^2*b^(1/2)*(c*x+b)^(1/2)+33*A*x^2*c^2*b^(1/2)*(c*x+b)^(1/2)+54*B*x^2*b^(3/2)*c*(c*x+b)^(1/2)+26
*A*x*b^(3/2)*c*(c*x+b)^(1/2)+12*B*x*b^(5/2)*(c*x+b)^(1/2)+8*A*b^(5/2)*(c*x+b)^(1/2))/x^(7/2)/(c*x+b)^(1/2)/b^(
1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2} + b x\right )}^{\frac{5}{2}}{\left (B x + A\right )}}{x^{\frac{13}{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^(13/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x)^(5/2)*(B*x + A)/x^(13/2), x)

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Fricas [A]  time = 1.61254, size = 610, normalized size = 3.49 \begin{align*} \left [\frac{15 \,{\left (6 \, B b c^{2} + A c^{3}\right )} \sqrt{b} x^{4} \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 (48 \, B b c^{2} x^{3} - 8 \, A b^{3} - 3 \,{\left (18 \, B b^{2} c + 11 \, A b c^{2}\right )} x^{2} - 2 \,{\left (6 \, B b^{3} + 13 \, A b^{2} c\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{48 \, b x^{4}}, \frac{15 \,{\left (6 \, B b c^{2} + A c^{3}\right )} \sqrt{-b} x^{4} \arctan \left (\frac{\sqrt{-b} \sqrt{x}}{\sqrt{c x^{2} + b x}}\right ) +{\left (48 \, B b c^{2} x^{3} - 8 \, A b^{3} - 3 \,{\left (18 \, B b^{2} c + 11 \, A b c^{2}\right )} x^{2} - 2 \,{\left (6 \, B b^{3} + 13 \, A b^{2} c\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{24 \, b x^{4}}\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^(13/2),x, algorithm="fricas")

[Out]

[1/48*(15*(6*B*b*c^2 + A*c^3)*sqrt(b)*x^4*log(-(c*x^2 + 2*b*x - 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*
(48*B*b*c^2*x^3 - 8*A*b^3 - 3*(18*B*b^2*c + 11*A*b*c^2)*x^2 - 2*(6*B*b^3 + 13*A*b^2*c)*x)*sqrt(c*x^2 + b*x)*sq
rt(x))/(b*x^4), 1/24*(15*(6*B*b*c^2 + A*c^3)*sqrt(-b)*x^4*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (48*B*b
*c^2*x^3 - 8*A*b^3 - 3*(18*B*b^2*c + 11*A*b*c^2)*x^2 - 2*(6*B*b^3 + 13*A*b^2*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/
(b*x^4)]

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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**(13/2),x)

[Out]

Timed out

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Giac [A]  time = 1.35028, size = 204, normalized size = 1.17 \begin{align*} \frac{48 \, \sqrt{c x + b} B c^{3} + \frac{15 \,{\left (6 \, B b c^{3} + A c^{4}\right )} \arctan \left (\frac{\sqrt{c x + b}}{\sqrt{-b}}\right )}{\sqrt{-b}} - \frac{54 \,{\left (c x + b\right )}^{\frac{5}{2}} B b c^{3} - 96 \,{\left (c x + b\right )}^{\frac{3}{2}} B b^{2} c^{3} + 42 \, \sqrt{c x + b} B b^{3} c^{3} + 33 \,{\left (c x + b\right )}^{\frac{5}{2}} A c^{4} - 40 \,{\left (c x + b\right )}^{\frac{3}{2}} A b c^{4} + 15 \, \sqrt{c x + b} A b^{2} c^{4}}{c^{3} x^{3}}}{24 \, 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^(13/2),x, algorithm="giac")

[Out]

1/24*(48*sqrt(c*x + b)*B*c^3 + 15*(6*B*b*c^3 + A*c^4)*arctan(sqrt(c*x + b)/sqrt(-b))/sqrt(-b) - (54*(c*x + b)^
(5/2)*B*b*c^3 - 96*(c*x + b)^(3/2)*B*b^2*c^3 + 42*sqrt(c*x + b)*B*b^3*c^3 + 33*(c*x + b)^(5/2)*A*c^4 - 40*(c*x
+ b)^(3/2)*A*b*c^4 + 15*sqrt(c*x + b)*A*b^2*c^4)/(c^3*x^3))/c