3.223 \(\int \frac{(A+B x) (b x+c x^2)^{5/2}}{x^{15/2}} \, dx\)

Optimal. Leaf size=179 \[ -\frac{5 c^3 (8 b B-A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{64 b^{3/2}}-\frac{5 c^2 \sqrt{b x+c x^2} (8 b B-A c)}{64 b x^{3/2}}-\frac{\left (b x+c x^2\right )^{5/2} (8 b B-A c)}{24 b x^{11/2}}-\frac{5 c \left (b x+c x^2\right )^{3/2} (8 b B-A c)}{96 b x^{7/2}}-\frac{A \left (b x+c x^2\right )^{7/2}}{4 b x^{15/2}} \]

[Out]

(-5*c^2*(8*b*B - A*c)*Sqrt[b*x + c*x^2])/(64*b*x^(3/2)) - (5*c*(8*b*B - A*c)*(b*x + c*x^2)^(3/2))/(96*b*x^(7/2
)) - ((8*b*B - A*c)*(b*x + c*x^2)^(5/2))/(24*b*x^(11/2)) - (A*(b*x + c*x^2)^(7/2))/(4*b*x^(15/2)) - (5*c^3*(8*
b*B - A*c)*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/(64*b^(3/2))

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Rubi [A]  time = 0.166159, antiderivative size = 179, 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, 662, 660, 207} \[ -\frac{5 c^3 (8 b B-A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{64 b^{3/2}}-\frac{5 c^2 \sqrt{b x+c x^2} (8 b B-A c)}{64 b x^{3/2}}-\frac{\left (b x+c x^2\right )^{5/2} (8 b B-A c)}{24 b x^{11/2}}-\frac{5 c \left (b x+c x^2\right )^{3/2} (8 b B-A c)}{96 b x^{7/2}}-\frac{A \left (b x+c x^2\right )^{7/2}}{4 b x^{15/2}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(15/2),x]

[Out]

(-5*c^2*(8*b*B - A*c)*Sqrt[b*x + c*x^2])/(64*b*x^(3/2)) - (5*c*(8*b*B - A*c)*(b*x + c*x^2)^(3/2))/(96*b*x^(7/2
)) - ((8*b*B - A*c)*(b*x + c*x^2)^(5/2))/(24*b*x^(11/2)) - (A*(b*x + c*x^2)^(7/2))/(4*b*x^(15/2)) - (5*c^3*(8*
b*B - A*c)*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/(64*b^(3/2))

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 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^{15/2}} \, dx &=-\frac{A \left (b x+c x^2\right )^{7/2}}{4 b x^{15/2}}+\frac{\left (-\frac{15}{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^{13/2}} \, dx}{4 b}\\ &=-\frac{(8 b B-A c) \left (b x+c x^2\right )^{5/2}}{24 b x^{11/2}}-\frac{A \left (b x+c x^2\right )^{7/2}}{4 b x^{15/2}}+\frac{(5 c (8 b B-A c)) \int \frac{\left (b x+c x^2\right )^{3/2}}{x^{9/2}} \, dx}{48 b}\\ &=-\frac{5 c (8 b B-A c) \left (b x+c x^2\right )^{3/2}}{96 b x^{7/2}}-\frac{(8 b B-A c) \left (b x+c x^2\right )^{5/2}}{24 b x^{11/2}}-\frac{A \left (b x+c x^2\right )^{7/2}}{4 b x^{15/2}}+\frac{\left (5 c^2 (8 b B-A c)\right ) \int \frac{\sqrt{b x+c x^2}}{x^{5/2}} \, dx}{64 b}\\ &=-\frac{5 c^2 (8 b B-A c) \sqrt{b x+c x^2}}{64 b x^{3/2}}-\frac{5 c (8 b B-A c) \left (b x+c x^2\right )^{3/2}}{96 b x^{7/2}}-\frac{(8 b B-A c) \left (b x+c x^2\right )^{5/2}}{24 b x^{11/2}}-\frac{A \left (b x+c x^2\right )^{7/2}}{4 b x^{15/2}}+\frac{\left (5 c^3 (8 b B-A c)\right ) \int \frac{1}{\sqrt{x} \sqrt{b x+c x^2}} \, dx}{128 b}\\ &=-\frac{5 c^2 (8 b B-A c) \sqrt{b x+c x^2}}{64 b x^{3/2}}-\frac{5 c (8 b B-A c) \left (b x+c x^2\right )^{3/2}}{96 b x^{7/2}}-\frac{(8 b B-A c) \left (b x+c x^2\right )^{5/2}}{24 b x^{11/2}}-\frac{A \left (b x+c x^2\right )^{7/2}}{4 b x^{15/2}}+\frac{\left (5 c^3 (8 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 )}{64 b}\\ &=-\frac{5 c^2 (8 b B-A c) \sqrt{b x+c x^2}}{64 b x^{3/2}}-\frac{5 c (8 b B-A c) \left (b x+c x^2\right )^{3/2}}{96 b x^{7/2}}-\frac{(8 b B-A c) \left (b x+c x^2\right )^{5/2}}{24 b x^{11/2}}-\frac{A \left (b x+c x^2\right )^{7/2}}{4 b x^{15/2}}-\frac{5 c^3 (8 b B-A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{64 b^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.141085, size = 129, normalized size = 0.72 \[ -\frac{(b+c x) \left (A \left (136 b^2 c x+48 b^3+118 b c^2 x^2+15 c^3 x^3\right )+8 b B x \left (8 b^2+26 b c x+33 c^2 x^2\right )\right )+15 c^3 x^4 \sqrt{\frac{c x}{b}+1} (8 b B-A c) \tanh ^{-1}\left (\sqrt{\frac{c x}{b}+1}\right )}{192 b x^{7/2} \sqrt{x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(15/2),x]

[Out]

-((b + c*x)*(8*b*B*x*(8*b^2 + 26*b*c*x + 33*c^2*x^2) + A*(48*b^3 + 136*b^2*c*x + 118*b*c^2*x^2 + 15*c^3*x^3))
+ 15*c^3*(8*b*B - A*c)*x^4*Sqrt[1 + (c*x)/b]*ArcTanh[Sqrt[1 + (c*x)/b]])/(192*b*x^(7/2)*Sqrt[x*(b + c*x)])

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Maple [A]  time = 0.019, size = 185, normalized size = 1. \begin{align*}{\frac{1}{192}\sqrt{x \left ( cx+b \right ) } \left ( 15\,A{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ){x}^{4}{c}^{4}-120\,B{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ){x}^{4}b{c}^{3}-15\,A{x}^{3}{c}^{3}\sqrt{b}\sqrt{cx+b}-264\,B{x}^{3}{b}^{3/2}{c}^{2}\sqrt{cx+b}-118\,A{x}^{2}{b}^{3/2}{c}^{2}\sqrt{cx+b}-208\,B{x}^{2}{b}^{5/2}c\sqrt{cx+b}-136\,Ax{b}^{5/2}c\sqrt{cx+b}-64\,Bx{b}^{7/2}\sqrt{cx+b}-48\,A{b}^{7/2}\sqrt{cx+b} \right ){b}^{-{\frac{3}{2}}}{x}^{-{\frac{9}{2}}}{\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)^(5/2)/x^(15/2),x)

[Out]

1/192*(x*(c*x+b))^(1/2)/b^(3/2)*(15*A*arctanh((c*x+b)^(1/2)/b^(1/2))*x^4*c^4-120*B*arctanh((c*x+b)^(1/2)/b^(1/
2))*x^4*b*c^3-15*A*x^3*c^3*b^(1/2)*(c*x+b)^(1/2)-264*B*x^3*b^(3/2)*c^2*(c*x+b)^(1/2)-118*A*x^2*b^(3/2)*c^2*(c*
x+b)^(1/2)-208*B*x^2*b^(5/2)*c*(c*x+b)^(1/2)-136*A*x*b^(5/2)*c*(c*x+b)^(1/2)-64*B*x*b^(7/2)*(c*x+b)^(1/2)-48*A
*b^(7/2)*(c*x+b)^(1/2))/x^(9/2)/(c*x+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{15}{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^(15/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.64263, size = 679, normalized size = 3.79 \begin{align*} \left [-\frac{15 \,{\left (8 \, B b c^{3} - A c^{4}\right )} \sqrt{b} x^{5} \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 \, A b^{4} + 3 \,{\left (88 \, B b^{2} c^{2} + 5 \, A b c^{3}\right )} x^{3} + 2 \,{\left (104 \, B b^{3} c + 59 \, A b^{2} c^{2}\right )} x^{2} + 8 \,{\left (8 \, B b^{4} + 17 \, A b^{3} c\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{384 \, b^{2} x^{5}}, \frac{15 \,{\left (8 \, B b c^{3} - A c^{4}\right )} \sqrt{-b} x^{5} \arctan \left (\frac{\sqrt{-b} \sqrt{x}}{\sqrt{c x^{2} + b x}}\right ) -{\left (48 \, A b^{4} + 3 \,{\left (88 \, B b^{2} c^{2} + 5 \, A b c^{3}\right )} x^{3} + 2 \,{\left (104 \, B b^{3} c + 59 \, A b^{2} c^{2}\right )} x^{2} + 8 \,{\left (8 \, B b^{4} + 17 \, A b^{3} c\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{192 \, b^{2} x^{5}}\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^(15/2),x, algorithm="fricas")

[Out]

[-1/384*(15*(8*B*b*c^3 - A*c^4)*sqrt(b)*x^5*log(-(c*x^2 + 2*b*x + 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) +
2*(48*A*b^4 + 3*(88*B*b^2*c^2 + 5*A*b*c^3)*x^3 + 2*(104*B*b^3*c + 59*A*b^2*c^2)*x^2 + 8*(8*B*b^4 + 17*A*b^3*c)
*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^2*x^5), 1/192*(15*(8*B*b*c^3 - A*c^4)*sqrt(-b)*x^5*arctan(sqrt(-b)*sqrt(x)/s
qrt(c*x^2 + b*x)) - (48*A*b^4 + 3*(88*B*b^2*c^2 + 5*A*b*c^3)*x^3 + 2*(104*B*b^3*c + 59*A*b^2*c^2)*x^2 + 8*(8*B
*b^4 + 17*A*b^3*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^2*x^5)]

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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**(15/2),x)

[Out]

Timed out

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Giac [A]  time = 1.31029, size = 239, normalized size = 1.34 \begin{align*} \frac{\frac{15 \,{\left (8 \, B b c^{4} - A c^{5}\right )} \arctan \left (\frac{\sqrt{c x + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b} - \frac{264 \,{\left (c x + b\right )}^{\frac{7}{2}} B b c^{4} - 584 \,{\left (c x + b\right )}^{\frac{5}{2}} B b^{2} c^{4} + 440 \,{\left (c x + b\right )}^{\frac{3}{2}} B b^{3} c^{4} - 120 \, \sqrt{c x + b} B b^{4} c^{4} + 15 \,{\left (c x + b\right )}^{\frac{7}{2}} A c^{5} + 73 \,{\left (c x + b\right )}^{\frac{5}{2}} A b c^{5} - 55 \,{\left (c x + b\right )}^{\frac{3}{2}} A b^{2} c^{5} + 15 \, \sqrt{c x + b} A b^{3} c^{5}}{b c^{4} x^{4}}}{192 \, 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^(15/2),x, algorithm="giac")

[Out]

1/192*(15*(8*B*b*c^4 - A*c^5)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b) - (264*(c*x + b)^(7/2)*B*b*c^4 - 584
*(c*x + b)^(5/2)*B*b^2*c^4 + 440*(c*x + b)^(3/2)*B*b^3*c^4 - 120*sqrt(c*x + b)*B*b^4*c^4 + 15*(c*x + b)^(7/2)*
A*c^5 + 73*(c*x + b)^(5/2)*A*b*c^5 - 55*(c*x + b)^(3/2)*A*b^2*c^5 + 15*sqrt(c*x + b)*A*b^3*c^5)/(b*c^4*x^4))/c