3.224 \(\int \frac{(A+B x) (b x+c x^2)^{5/2}}{x^{17/2}} \, dx\)

Optimal. Leaf size=216 \[ -\frac{c^3 \sqrt{b x+c x^2} (10 b B-3 A c)}{128 b^2 x^{3/2}}+\frac{c^4 (10 b B-3 A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{128 b^{5/2}}-\frac{c^2 \sqrt{b x+c x^2} (10 b B-3 A c)}{64 b x^{5/2}}-\frac{c \left (b x+c x^2\right )^{3/2} (10 b B-3 A c)}{48 b x^{9/2}}-\frac{\left (b x+c x^2\right )^{5/2} (10 b B-3 A c)}{40 b x^{13/2}}-\frac{A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}} \]

[Out]

-(c^2*(10*b*B - 3*A*c)*Sqrt[b*x + c*x^2])/(64*b*x^(5/2)) - (c^3*(10*b*B - 3*A*c)*Sqrt[b*x + c*x^2])/(128*b^2*x
^(3/2)) - (c*(10*b*B - 3*A*c)*(b*x + c*x^2)^(3/2))/(48*b*x^(9/2)) - ((10*b*B - 3*A*c)*(b*x + c*x^2)^(5/2))/(40
*b*x^(13/2)) - (A*(b*x + c*x^2)^(7/2))/(5*b*x^(17/2)) + (c^4*(10*b*B - 3*A*c)*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[
b]*Sqrt[x])])/(128*b^(5/2))

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Rubi [A]  time = 0.205995, antiderivative size = 216, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {792, 662, 672, 660, 207} \[ -\frac{c^3 \sqrt{b x+c x^2} (10 b B-3 A c)}{128 b^2 x^{3/2}}+\frac{c^4 (10 b B-3 A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{128 b^{5/2}}-\frac{c^2 \sqrt{b x+c x^2} (10 b B-3 A c)}{64 b x^{5/2}}-\frac{c \left (b x+c x^2\right )^{3/2} (10 b B-3 A c)}{48 b x^{9/2}}-\frac{\left (b x+c x^2\right )^{5/2} (10 b B-3 A c)}{40 b x^{13/2}}-\frac{A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(17/2),x]

[Out]

-(c^2*(10*b*B - 3*A*c)*Sqrt[b*x + c*x^2])/(64*b*x^(5/2)) - (c^3*(10*b*B - 3*A*c)*Sqrt[b*x + c*x^2])/(128*b^2*x
^(3/2)) - (c*(10*b*B - 3*A*c)*(b*x + c*x^2)^(3/2))/(48*b*x^(9/2)) - ((10*b*B - 3*A*c)*(b*x + c*x^2)^(5/2))/(40
*b*x^(13/2)) - (A*(b*x + c*x^2)^(7/2))/(5*b*x^(17/2)) + (c^4*(10*b*B - 3*A*c)*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[
b]*Sqrt[x])])/(128*b^(5/2))

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 672

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((m + p + 1)*(2*c*d - b*e)), x] + Dist[(c*(m + 2*p + 2))/((m + p + 1)*(2*c*d - b*e)), I
nt[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ
[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, 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^{17/2}} \, dx &=-\frac{A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}}+\frac{\left (-\frac{17}{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^{15/2}} \, dx}{5 b}\\ &=-\frac{(10 b B-3 A c) \left (b x+c x^2\right )^{5/2}}{40 b x^{13/2}}-\frac{A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}}+\frac{(c (10 b B-3 A c)) \int \frac{\left (b x+c x^2\right )^{3/2}}{x^{11/2}} \, dx}{16 b}\\ &=-\frac{c (10 b B-3 A c) \left (b x+c x^2\right )^{3/2}}{48 b x^{9/2}}-\frac{(10 b B-3 A c) \left (b x+c x^2\right )^{5/2}}{40 b x^{13/2}}-\frac{A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}}+\frac{\left (c^2 (10 b B-3 A c)\right ) \int \frac{\sqrt{b x+c x^2}}{x^{7/2}} \, dx}{32 b}\\ &=-\frac{c^2 (10 b B-3 A c) \sqrt{b x+c x^2}}{64 b x^{5/2}}-\frac{c (10 b B-3 A c) \left (b x+c x^2\right )^{3/2}}{48 b x^{9/2}}-\frac{(10 b B-3 A c) \left (b x+c x^2\right )^{5/2}}{40 b x^{13/2}}-\frac{A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}}+\frac{\left (c^3 (10 b B-3 A c)\right ) \int \frac{1}{x^{3/2} \sqrt{b x+c x^2}} \, dx}{128 b}\\ &=-\frac{c^2 (10 b B-3 A c) \sqrt{b x+c x^2}}{64 b x^{5/2}}-\frac{c^3 (10 b B-3 A c) \sqrt{b x+c x^2}}{128 b^2 x^{3/2}}-\frac{c (10 b B-3 A c) \left (b x+c x^2\right )^{3/2}}{48 b x^{9/2}}-\frac{(10 b B-3 A c) \left (b x+c x^2\right )^{5/2}}{40 b x^{13/2}}-\frac{A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}}-\frac{\left (c^4 (10 b B-3 A c)\right ) \int \frac{1}{\sqrt{x} \sqrt{b x+c x^2}} \, dx}{256 b^2}\\ &=-\frac{c^2 (10 b B-3 A c) \sqrt{b x+c x^2}}{64 b x^{5/2}}-\frac{c^3 (10 b B-3 A c) \sqrt{b x+c x^2}}{128 b^2 x^{3/2}}-\frac{c (10 b B-3 A c) \left (b x+c x^2\right )^{3/2}}{48 b x^{9/2}}-\frac{(10 b B-3 A c) \left (b x+c x^2\right )^{5/2}}{40 b x^{13/2}}-\frac{A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}}-\frac{\left (c^4 (10 b B-3 A c)\right ) \operatorname{Subst}\left (\int \frac{1}{-b+x^2} \, dx,x,\frac{\sqrt{b x+c x^2}}{\sqrt{x}}\right )}{128 b^2}\\ &=-\frac{c^2 (10 b B-3 A c) \sqrt{b x+c x^2}}{64 b x^{5/2}}-\frac{c^3 (10 b B-3 A c) \sqrt{b x+c x^2}}{128 b^2 x^{3/2}}-\frac{c (10 b B-3 A c) \left (b x+c x^2\right )^{3/2}}{48 b x^{9/2}}-\frac{(10 b B-3 A c) \left (b x+c x^2\right )^{5/2}}{40 b x^{13/2}}-\frac{A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}}+\frac{c^4 (10 b B-3 A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{128 b^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0362579, size = 69, normalized size = 0.32 \[ -\frac{(b+c x)^3 \sqrt{x (b+c x)} \left (7 A b^5+c^4 x^5 (10 b B-3 A c) \, _2F_1\left (\frac{7}{2},5;\frac{9}{2};\frac{c x}{b}+1\right )\right )}{35 b^6 x^{11/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(17/2),x]

[Out]

-((b + c*x)^3*Sqrt[x*(b + c*x)]*(7*A*b^5 + c^4*(10*b*B - 3*A*c)*x^5*Hypergeometric2F1[7/2, 5, 9/2, 1 + (c*x)/b
]))/(35*b^6*x^(11/2))

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Maple [A]  time = 0.022, size = 223, normalized size = 1. \begin{align*} -{\frac{1}{1920}\sqrt{x \left ( cx+b \right ) } \left ( 45\,A{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ){x}^{5}{c}^{5}-150\,B{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ){x}^{5}b{c}^{4}-45\,A{x}^{4}{c}^{4}\sqrt{b}\sqrt{cx+b}+150\,B{x}^{4}{b}^{3/2}{c}^{3}\sqrt{cx+b}+30\,A{x}^{3}{b}^{3/2}{c}^{3}\sqrt{cx+b}+1180\,B{x}^{3}{b}^{5/2}{c}^{2}\sqrt{cx+b}+744\,A{x}^{2}{b}^{5/2}{c}^{2}\sqrt{cx+b}+1360\,B{x}^{2}{b}^{7/2}c\sqrt{cx+b}+1008\,Ax{b}^{7/2}c\sqrt{cx+b}+480\,Bx{b}^{9/2}\sqrt{cx+b}+384\,A{b}^{9/2}\sqrt{cx+b} \right ){b}^{-{\frac{5}{2}}}{x}^{-{\frac{11}{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^(17/2),x)

[Out]

-1/1920*(x*(c*x+b))^(1/2)/b^(5/2)*(45*A*arctanh((c*x+b)^(1/2)/b^(1/2))*x^5*c^5-150*B*arctanh((c*x+b)^(1/2)/b^(
1/2))*x^5*b*c^4-45*A*x^4*c^4*b^(1/2)*(c*x+b)^(1/2)+150*B*x^4*b^(3/2)*c^3*(c*x+b)^(1/2)+30*A*x^3*b^(3/2)*c^3*(c
*x+b)^(1/2)+1180*B*x^3*b^(5/2)*c^2*(c*x+b)^(1/2)+744*A*x^2*b^(5/2)*c^2*(c*x+b)^(1/2)+1360*B*x^2*b^(7/2)*c*(c*x
+b)^(1/2)+1008*A*x*b^(7/2)*c*(c*x+b)^(1/2)+480*B*x*b^(9/2)*(c*x+b)^(1/2)+384*A*b^(9/2)*(c*x+b)^(1/2))/x^(11/2)
/(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{17}{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^(17/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x)^(5/2)*(B*x + A)/x^(17/2), x)

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Fricas [A]  time = 1.59197, size = 807, normalized size = 3.74 \begin{align*} \left [-\frac{15 \,{\left (10 \, B b c^{4} - 3 \, A c^{5}\right )} \sqrt{b} x^{6} \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 (384 \, A b^{5} + 15 \,{\left (10 \, B b^{2} c^{3} - 3 \, A b c^{4}\right )} x^{4} + 10 \,{\left (118 \, B b^{3} c^{2} + 3 \, A b^{2} c^{3}\right )} x^{3} + 8 \,{\left (170 \, B b^{4} c + 93 \, A b^{3} c^{2}\right )} x^{2} + 48 \,{\left (10 \, B b^{5} + 21 \, A b^{4} c\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{3840 \, b^{3} x^{6}}, -\frac{15 \,{\left (10 \, B b c^{4} - 3 \, A c^{5}\right )} \sqrt{-b} x^{6} \arctan \left (\frac{\sqrt{-b} \sqrt{x}}{\sqrt{c x^{2} + b x}}\right ) +{\left (384 \, A b^{5} + 15 \,{\left (10 \, B b^{2} c^{3} - 3 \, A b c^{4}\right )} x^{4} + 10 \,{\left (118 \, B b^{3} c^{2} + 3 \, A b^{2} c^{3}\right )} x^{3} + 8 \,{\left (170 \, B b^{4} c + 93 \, A b^{3} c^{2}\right )} x^{2} + 48 \,{\left (10 \, B b^{5} + 21 \, A b^{4} c\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{1920 \, b^{3} x^{6}}\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^(17/2),x, algorithm="fricas")

[Out]

[-1/3840*(15*(10*B*b*c^4 - 3*A*c^5)*sqrt(b)*x^6*log(-(c*x^2 + 2*b*x - 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2
) + 2*(384*A*b^5 + 15*(10*B*b^2*c^3 - 3*A*b*c^4)*x^4 + 10*(118*B*b^3*c^2 + 3*A*b^2*c^3)*x^3 + 8*(170*B*b^4*c +
93*A*b^3*c^2)*x^2 + 48*(10*B*b^5 + 21*A*b^4*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^3*x^6), -1/1920*(15*(10*B*b*c
^4 - 3*A*c^5)*sqrt(-b)*x^6*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (384*A*b^5 + 15*(10*B*b^2*c^3 - 3*A*b*
c^4)*x^4 + 10*(118*B*b^3*c^2 + 3*A*b^2*c^3)*x^3 + 8*(170*B*b^4*c + 93*A*b^3*c^2)*x^2 + 48*(10*B*b^5 + 21*A*b^4
*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^3*x^6)]

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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**(17/2),x)

[Out]

Timed out

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Giac [A]  time = 1.34881, size = 281, normalized size = 1.3 \begin{align*} -\frac{\frac{15 \,{\left (10 \, B b c^{5} - 3 \, A c^{6}\right )} \arctan \left (\frac{\sqrt{c x + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b^{2}} + \frac{150 \,{\left (c x + b\right )}^{\frac{9}{2}} B b c^{5} + 580 \,{\left (c x + b\right )}^{\frac{7}{2}} B b^{2} c^{5} - 1280 \,{\left (c x + b\right )}^{\frac{5}{2}} B b^{3} c^{5} + 700 \,{\left (c x + b\right )}^{\frac{3}{2}} B b^{4} c^{5} - 150 \, \sqrt{c x + b} B b^{5} c^{5} - 45 \,{\left (c x + b\right )}^{\frac{9}{2}} A c^{6} + 210 \,{\left (c x + b\right )}^{\frac{7}{2}} A b c^{6} + 384 \,{\left (c x + b\right )}^{\frac{5}{2}} A b^{2} c^{6} - 210 \,{\left (c x + b\right )}^{\frac{3}{2}} A b^{3} c^{6} + 45 \, \sqrt{c x + b} A b^{4} c^{6}}{b^{2} c^{5} x^{5}}}{1920 \, 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^(17/2),x, algorithm="giac")

[Out]

-1/1920*(15*(10*B*b*c^5 - 3*A*c^6)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^2) + (150*(c*x + b)^(9/2)*B*b*c^
5 + 580*(c*x + b)^(7/2)*B*b^2*c^5 - 1280*(c*x + b)^(5/2)*B*b^3*c^5 + 700*(c*x + b)^(3/2)*B*b^4*c^5 - 150*sqrt(
c*x + b)*B*b^5*c^5 - 45*(c*x + b)^(9/2)*A*c^6 + 210*(c*x + b)^(7/2)*A*b*c^6 + 384*(c*x + b)^(5/2)*A*b^2*c^6 -
210*(c*x + b)^(3/2)*A*b^3*c^6 + 45*sqrt(c*x + b)*A*b^4*c^6)/(b^2*c^5*x^5))/c