3.212 \(\int \frac{(A+B x) (b x+c x^2)^{3/2}}{x^{13/2}} \, dx\)

Optimal. Leaf size=179 \[ -\frac{c^2 \sqrt{b x+c x^2} (8 b B-3 A c)}{64 b^2 x^{3/2}}+\frac{c^3 (8 b B-3 A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{64 b^{5/2}}-\frac{c \sqrt{b x+c x^2} (8 b B-3 A c)}{32 b x^{5/2}}-\frac{\left (b x+c x^2\right )^{3/2} (8 b B-3 A c)}{24 b x^{9/2}}-\frac{A \left (b x+c x^2\right )^{5/2}}{4 b x^{13/2}} \]

[Out]

-(c*(8*b*B - 3*A*c)*Sqrt[b*x + c*x^2])/(32*b*x^(5/2)) - (c^2*(8*b*B - 3*A*c)*Sqrt[b*x + c*x^2])/(64*b^2*x^(3/2
)) - ((8*b*B - 3*A*c)*(b*x + c*x^2)^(3/2))/(24*b*x^(9/2)) - (A*(b*x + c*x^2)^(5/2))/(4*b*x^(13/2)) + (c^3*(8*b
*B - 3*A*c)*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/(64*b^(5/2))

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Rubi [A]  time = 0.169471, antiderivative size = 179, 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, 672, 660, 207} \[ -\frac{c^2 \sqrt{b x+c x^2} (8 b B-3 A c)}{64 b^2 x^{3/2}}+\frac{c^3 (8 b B-3 A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{64 b^{5/2}}-\frac{c \sqrt{b x+c x^2} (8 b B-3 A c)}{32 b x^{5/2}}-\frac{\left (b x+c x^2\right )^{3/2} (8 b B-3 A c)}{24 b x^{9/2}}-\frac{A \left (b x+c x^2\right )^{5/2}}{4 b x^{13/2}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(3/2))/x^(13/2),x]

[Out]

-(c*(8*b*B - 3*A*c)*Sqrt[b*x + c*x^2])/(32*b*x^(5/2)) - (c^2*(8*b*B - 3*A*c)*Sqrt[b*x + c*x^2])/(64*b^2*x^(3/2
)) - ((8*b*B - 3*A*c)*(b*x + c*x^2)^(3/2))/(24*b*x^(9/2)) - (A*(b*x + c*x^2)^(5/2))/(4*b*x^(13/2)) + (c^3*(8*b
*B - 3*A*c)*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/(64*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 )^{3/2}}{x^{13/2}} \, dx &=-\frac{A \left (b x+c x^2\right )^{5/2}}{4 b x^{13/2}}+\frac{\left (-\frac{13}{2} (-b B+A c)+\frac{5}{2} (-b B+2 A c)\right ) \int \frac{\left (b x+c x^2\right )^{3/2}}{x^{11/2}} \, dx}{4 b}\\ &=-\frac{(8 b B-3 A c) \left (b x+c x^2\right )^{3/2}}{24 b x^{9/2}}-\frac{A \left (b x+c x^2\right )^{5/2}}{4 b x^{13/2}}+\frac{(c (8 b B-3 A c)) \int \frac{\sqrt{b x+c x^2}}{x^{7/2}} \, dx}{16 b}\\ &=-\frac{c (8 b B-3 A c) \sqrt{b x+c x^2}}{32 b x^{5/2}}-\frac{(8 b B-3 A c) \left (b x+c x^2\right )^{3/2}}{24 b x^{9/2}}-\frac{A \left (b x+c x^2\right )^{5/2}}{4 b x^{13/2}}+\frac{\left (c^2 (8 b B-3 A c)\right ) \int \frac{1}{x^{3/2} \sqrt{b x+c x^2}} \, dx}{64 b}\\ &=-\frac{c (8 b B-3 A c) \sqrt{b x+c x^2}}{32 b x^{5/2}}-\frac{c^2 (8 b B-3 A c) \sqrt{b x+c x^2}}{64 b^2 x^{3/2}}-\frac{(8 b B-3 A c) \left (b x+c x^2\right )^{3/2}}{24 b x^{9/2}}-\frac{A \left (b x+c x^2\right )^{5/2}}{4 b x^{13/2}}-\frac{\left (c^3 (8 b B-3 A c)\right ) \int \frac{1}{\sqrt{x} \sqrt{b x+c x^2}} \, dx}{128 b^2}\\ &=-\frac{c (8 b B-3 A c) \sqrt{b x+c x^2}}{32 b x^{5/2}}-\frac{c^2 (8 b B-3 A c) \sqrt{b x+c x^2}}{64 b^2 x^{3/2}}-\frac{(8 b B-3 A c) \left (b x+c x^2\right )^{3/2}}{24 b x^{9/2}}-\frac{A \left (b x+c x^2\right )^{5/2}}{4 b x^{13/2}}-\frac{\left (c^3 (8 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 )}{64 b^2}\\ &=-\frac{c (8 b B-3 A c) \sqrt{b x+c x^2}}{32 b x^{5/2}}-\frac{c^2 (8 b B-3 A c) \sqrt{b x+c x^2}}{64 b^2 x^{3/2}}-\frac{(8 b B-3 A c) \left (b x+c x^2\right )^{3/2}}{24 b x^{9/2}}-\frac{A \left (b x+c x^2\right )^{5/2}}{4 b x^{13/2}}+\frac{c^3 (8 b B-3 A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{64 b^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0360508, size = 62, normalized size = 0.35 \[ \frac{(x (b+c x))^{5/2} \left (c^3 x^4 (8 b B-3 A c) \, _2F_1\left (\frac{5}{2},4;\frac{7}{2};\frac{c x}{b}+1\right )-5 A b^4\right )}{20 b^5 x^{13/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(3/2))/x^(13/2),x]

[Out]

((x*(b + c*x))^(5/2)*(-5*A*b^4 + c^3*(8*b*B - 3*A*c)*x^4*Hypergeometric2F1[5/2, 4, 7/2, 1 + (c*x)/b]))/(20*b^5
*x^(13/2))

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Maple [A]  time = 0.02, size = 185, normalized size = 1. \begin{align*} -{\frac{1}{192}\sqrt{x \left ( cx+b \right ) } \left ( 9\,A{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ){x}^{4}{c}^{4}-24\,B{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ){x}^{4}b{c}^{3}-9\,A{x}^{3}{c}^{3}\sqrt{b}\sqrt{cx+b}+24\,B{x}^{3}{b}^{3/2}{c}^{2}\sqrt{cx+b}+6\,A{x}^{2}{b}^{3/2}{c}^{2}\sqrt{cx+b}+112\,B{x}^{2}{b}^{5/2}c\sqrt{cx+b}+72\,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{5}{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)^(3/2)/x^(13/2),x)

[Out]

-1/192*(x*(c*x+b))^(1/2)/b^(5/2)*(9*A*arctanh((c*x+b)^(1/2)/b^(1/2))*x^4*c^4-24*B*arctanh((c*x+b)^(1/2)/b^(1/2
))*x^4*b*c^3-9*A*x^3*c^3*b^(1/2)*(c*x+b)^(1/2)+24*B*x^3*b^(3/2)*c^2*(c*x+b)^(1/2)+6*A*x^2*b^(3/2)*c^2*(c*x+b)^
(1/2)+112*B*x^2*b^(5/2)*c*(c*x+b)^(1/2)+72*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{3}{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)^(3/2)/x^(13/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x)^(3/2)*(B*x + A)/x^(13/2), x)

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Fricas [A]  time = 1.68852, size = 672, normalized size = 3.75 \begin{align*} \left [-\frac{3 \,{\left (8 \, B b c^{3} - 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 (8 \, B b^{2} c^{2} - 3 \, A b c^{3}\right )} x^{3} + 2 \,{\left (56 \, B b^{3} c + 3 \, A b^{2} c^{2}\right )} x^{2} + 8 \,{\left (8 \, B b^{4} + 9 \, A b^{3} c\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{384 \, b^{3} x^{5}}, -\frac{3 \,{\left (8 \, B b c^{3} - 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 (8 \, B b^{2} c^{2} - 3 \, A b c^{3}\right )} x^{3} + 2 \,{\left (56 \, B b^{3} c + 3 \, A b^{2} c^{2}\right )} x^{2} + 8 \,{\left (8 \, B b^{4} + 9 \, A b^{3} c\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{192 \, b^{3} 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)^(3/2)/x^(13/2),x, algorithm="fricas")

[Out]

[-1/384*(3*(8*B*b*c^3 - 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*(8*B*b^2*c^2 - 3*A*b*c^3)*x^3 + 2*(56*B*b^3*c + 3*A*b^2*c^2)*x^2 + 8*(8*B*b^4 + 9*A*b^3*c)*x)
*sqrt(c*x^2 + b*x)*sqrt(x))/(b^3*x^5), -1/192*(3*(8*B*b*c^3 - 3*A*c^4)*sqrt(-b)*x^5*arctan(sqrt(-b)*sqrt(x)/sq
rt(c*x^2 + b*x)) + (48*A*b^4 + 3*(8*B*b^2*c^2 - 3*A*b*c^3)*x^3 + 2*(56*B*b^3*c + 3*A*b^2*c^2)*x^2 + 8*(8*B*b^4
+ 9*A*b^3*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^3*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)**(3/2)/x**(13/2),x)

[Out]

Timed out

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Giac [A]  time = 1.29452, size = 238, normalized size = 1.33 \begin{align*} -\frac{\frac{3 \,{\left (8 \, B b c^{4} - 3 \, A c^{5}\right )} \arctan \left (\frac{\sqrt{c x + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b^{2}} + \frac{24 \,{\left (c x + b\right )}^{\frac{7}{2}} B b c^{4} + 40 \,{\left (c x + b\right )}^{\frac{5}{2}} B b^{2} c^{4} - 88 \,{\left (c x + b\right )}^{\frac{3}{2}} B b^{3} c^{4} + 24 \, \sqrt{c x + b} B b^{4} c^{4} - 9 \,{\left (c x + b\right )}^{\frac{7}{2}} A c^{5} + 33 \,{\left (c x + b\right )}^{\frac{5}{2}} A b c^{5} + 33 \,{\left (c x + b\right )}^{\frac{3}{2}} A b^{2} c^{5} - 9 \, \sqrt{c x + b} A b^{3} c^{5}}{b^{2} 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)^(3/2)/x^(13/2),x, algorithm="giac")

[Out]

-1/192*(3*(8*B*b*c^4 - 3*A*c^5)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^2) + (24*(c*x + b)^(7/2)*B*b*c^4 +
40*(c*x + b)^(5/2)*B*b^2*c^4 - 88*(c*x + b)^(3/2)*B*b^3*c^4 + 24*sqrt(c*x + b)*B*b^4*c^4 - 9*(c*x + b)^(7/2)*A
*c^5 + 33*(c*x + b)^(5/2)*A*b*c^5 + 33*(c*x + b)^(3/2)*A*b^2*c^5 - 9*sqrt(c*x + b)*A*b^3*c^5)/(b^2*c^4*x^4))/c