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

Optimal. Leaf size=189 \[ \frac{5 b^5 (b+2 c x) \sqrt{b x+c x^2} (9 b B-16 A c)}{16384 c^5}-\frac{5 b^3 (b+2 c x) \left (b x+c x^2\right )^{3/2} (9 b B-16 A c)}{6144 c^4}-\frac{5 b^7 (9 b B-16 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{16384 c^{11/2}}+\frac{b (b+2 c x) \left (b x+c x^2\right )^{5/2} (9 b B-16 A c)}{384 c^3}-\frac{\left (b x+c x^2\right )^{7/2} (-16 A c+9 b B-14 B c x)}{112 c^2} \]

[Out]

(5*b^5*(9*b*B - 16*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(16384*c^5) - (5*b^3*(9*b*B - 16*A*c)*(b + 2*c*x)*(b*x
+ c*x^2)^(3/2))/(6144*c^4) + (b*(9*b*B - 16*A*c)*(b + 2*c*x)*(b*x + c*x^2)^(5/2))/(384*c^3) - ((9*b*B - 16*A*c
- 14*B*c*x)*(b*x + c*x^2)^(7/2))/(112*c^2) - (5*b^7*(9*b*B - 16*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/
(16384*c^(11/2))

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Rubi [A]  time = 0.0866696, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {779, 612, 620, 206} \[ \frac{5 b^5 (b+2 c x) \sqrt{b x+c x^2} (9 b B-16 A c)}{16384 c^5}-\frac{5 b^3 (b+2 c x) \left (b x+c x^2\right )^{3/2} (9 b B-16 A c)}{6144 c^4}-\frac{5 b^7 (9 b B-16 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{16384 c^{11/2}}+\frac{b (b+2 c x) \left (b x+c x^2\right )^{5/2} (9 b B-16 A c)}{384 c^3}-\frac{\left (b x+c x^2\right )^{7/2} (-16 A c+9 b B-14 B c x)}{112 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*b^5*(9*b*B - 16*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(16384*c^5) - (5*b^3*(9*b*B - 16*A*c)*(b + 2*c*x)*(b*x
+ c*x^2)^(3/2))/(6144*c^4) + (b*(9*b*B - 16*A*c)*(b + 2*c*x)*(b*x + c*x^2)^(5/2))/(384*c^3) - ((9*b*B - 16*A*c
- 14*B*c*x)*(b*x + c*x^2)^(7/2))/(112*c^2) - (5*b^7*(9*b*B - 16*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/
(16384*c^(11/2))

Rule 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rule 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int x (A+B x) \left (b x+c x^2\right )^{5/2} \, dx &=-\frac{(9 b B-16 A c-14 B c x) \left (b x+c x^2\right )^{7/2}}{112 c^2}+\frac{(b (9 b B-16 A c)) \int \left (b x+c x^2\right )^{5/2} \, dx}{32 c^2}\\ &=\frac{b (9 b B-16 A c) (b+2 c x) \left (b x+c x^2\right )^{5/2}}{384 c^3}-\frac{(9 b B-16 A c-14 B c x) \left (b x+c x^2\right )^{7/2}}{112 c^2}-\frac{\left (5 b^3 (9 b B-16 A c)\right ) \int \left (b x+c x^2\right )^{3/2} \, dx}{768 c^3}\\ &=-\frac{5 b^3 (9 b B-16 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{6144 c^4}+\frac{b (9 b B-16 A c) (b+2 c x) \left (b x+c x^2\right )^{5/2}}{384 c^3}-\frac{(9 b B-16 A c-14 B c x) \left (b x+c x^2\right )^{7/2}}{112 c^2}+\frac{\left (5 b^5 (9 b B-16 A c)\right ) \int \sqrt{b x+c x^2} \, dx}{4096 c^4}\\ &=\frac{5 b^5 (9 b B-16 A c) (b+2 c x) \sqrt{b x+c x^2}}{16384 c^5}-\frac{5 b^3 (9 b B-16 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{6144 c^4}+\frac{b (9 b B-16 A c) (b+2 c x) \left (b x+c x^2\right )^{5/2}}{384 c^3}-\frac{(9 b B-16 A c-14 B c x) \left (b x+c x^2\right )^{7/2}}{112 c^2}-\frac{\left (5 b^7 (9 b B-16 A c)\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{32768 c^5}\\ &=\frac{5 b^5 (9 b B-16 A c) (b+2 c x) \sqrt{b x+c x^2}}{16384 c^5}-\frac{5 b^3 (9 b B-16 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{6144 c^4}+\frac{b (9 b B-16 A c) (b+2 c x) \left (b x+c x^2\right )^{5/2}}{384 c^3}-\frac{(9 b B-16 A c-14 B c x) \left (b x+c x^2\right )^{7/2}}{112 c^2}-\frac{\left (5 b^7 (9 b B-16 A c)\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )}{16384 c^5}\\ &=\frac{5 b^5 (9 b B-16 A c) (b+2 c x) \sqrt{b x+c x^2}}{16384 c^5}-\frac{5 b^3 (9 b B-16 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{6144 c^4}+\frac{b (9 b B-16 A c) (b+2 c x) \left (b x+c x^2\right )^{5/2}}{384 c^3}-\frac{(9 b B-16 A c-14 B c x) \left (b x+c x^2\right )^{7/2}}{112 c^2}-\frac{5 b^7 (9 b B-16 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{16384 c^{11/2}}\\ \end{align*}

Mathematica [A]  time = 0.417858, size = 182, normalized size = 0.96 \[ \frac{(x (b+c x))^{9/2} \left (9 B (b+c x)^3-\frac{3 (9 b B-16 A c) \left (\sqrt{c} \sqrt{x} \sqrt{\frac{c x}{b}+1} \left (-56 b^4 c^2 x^2+48 b^3 c^3 x^3+4736 b^2 c^4 x^4+70 b^5 c x-105 b^6+7424 b c^5 x^5+3072 c^6 x^6\right )+105 b^{13/2} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )\right )}{14336 c^{9/2} x^{9/2} \sqrt{\frac{c x}{b}+1}}\right )}{72 c (b+c x)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((x*(b + c*x))^(9/2)*(9*B*(b + c*x)^3 - (3*(9*b*B - 16*A*c)*(Sqrt[c]*Sqrt[x]*Sqrt[1 + (c*x)/b]*(-105*b^6 + 70*
b^5*c*x - 56*b^4*c^2*x^2 + 48*b^3*c^3*x^3 + 4736*b^2*c^4*x^4 + 7424*b*c^5*x^5 + 3072*c^6*x^6) + 105*b^(13/2)*A
rcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]]))/(14336*c^(9/2)*x^(9/2)*Sqrt[1 + (c*x)/b])))/(72*c*(b + c*x)^4)

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Maple [B]  time = 0.007, size = 365, normalized size = 1.9 \begin{align*}{\frac{Bx}{8\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{7}{2}}}}-{\frac{9\,bB}{112\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{7}{2}}}}+{\frac{3\,{b}^{2}Bx}{64\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}+{\frac{3\,{b}^{3}B}{128\,{c}^{3}} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}-{\frac{15\,{b}^{4}Bx}{1024\,{c}^{3}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{15\,B{b}^{5}}{2048\,{c}^{4}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{45\,{b}^{6}Bx}{8192\,{c}^{4}}\sqrt{c{x}^{2}+bx}}+{\frac{45\,B{b}^{7}}{16384\,{c}^{5}}\sqrt{c{x}^{2}+bx}}-{\frac{45\,B{b}^{8}}{32768}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{11}{2}}}}+{\frac{A}{7\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{7}{2}}}}-{\frac{Abx}{12\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}-{\frac{A{b}^{2}}{24\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}+{\frac{5\,A{b}^{3}x}{192\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{5\,A{b}^{4}}{384\,{c}^{3}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{5\,A{b}^{5}x}{512\,{c}^{3}}\sqrt{c{x}^{2}+bx}}-{\frac{5\,A{b}^{6}}{1024\,{c}^{4}}\sqrt{c{x}^{2}+bx}}+{\frac{5\,A{b}^{7}}{2048}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*B*x*(c*x^2+b*x)^(7/2)/c-9/112*B*b/c^2*(c*x^2+b*x)^(7/2)+3/64*B*b^2/c^2*x*(c*x^2+b*x)^(5/2)+3/128*B*b^3/c^3
*(c*x^2+b*x)^(5/2)-15/1024*B*b^4/c^3*(c*x^2+b*x)^(3/2)*x-15/2048*B*b^5/c^4*(c*x^2+b*x)^(3/2)+45/8192*B*b^6/c^4
*(c*x^2+b*x)^(1/2)*x+45/16384*B*b^7/c^5*(c*x^2+b*x)^(1/2)-45/32768*B*b^8/c^(11/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^
2+b*x)^(1/2))+1/7*A*(c*x^2+b*x)^(7/2)/c-1/12*A*b/c*x*(c*x^2+b*x)^(5/2)-1/24*A*b^2/c^2*(c*x^2+b*x)^(5/2)+5/192*
A*b^3/c^2*(c*x^2+b*x)^(3/2)*x+5/384*A*b^4/c^3*(c*x^2+b*x)^(3/2)-5/512*A*b^5/c^3*(c*x^2+b*x)^(1/2)*x-5/1024*A*b
^6/c^4*(c*x^2+b*x)^(1/2)+5/2048*A*b^7/c^(9/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x+A)*(c*x^2+b*x)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.94691, size = 1064, normalized size = 5.63 \begin{align*} \left [-\frac{105 \,{\left (9 \, B b^{8} - 16 \, A b^{7} c\right )} \sqrt{c} \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) - 2 \,{\left (43008 \, B c^{8} x^{7} + 945 \, B b^{7} c - 1680 \, A b^{6} c^{2} + 3072 \,{\left (33 \, B b c^{7} + 16 \, A c^{8}\right )} x^{6} + 256 \,{\left (243 \, B b^{2} c^{6} + 464 \, A b c^{7}\right )} x^{5} + 128 \,{\left (3 \, B b^{3} c^{5} + 592 \, A b^{2} c^{6}\right )} x^{4} - 48 \,{\left (9 \, B b^{4} c^{4} - 16 \, A b^{3} c^{5}\right )} x^{3} + 56 \,{\left (9 \, B b^{5} c^{3} - 16 \, A b^{4} c^{4}\right )} x^{2} - 70 \,{\left (9 \, B b^{6} c^{2} - 16 \, A b^{5} c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{688128 \, c^{6}}, \frac{105 \,{\left (9 \, B b^{8} - 16 \, A b^{7} c\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) +{\left (43008 \, B c^{8} x^{7} + 945 \, B b^{7} c - 1680 \, A b^{6} c^{2} + 3072 \,{\left (33 \, B b c^{7} + 16 \, A c^{8}\right )} x^{6} + 256 \,{\left (243 \, B b^{2} c^{6} + 464 \, A b c^{7}\right )} x^{5} + 128 \,{\left (3 \, B b^{3} c^{5} + 592 \, A b^{2} c^{6}\right )} x^{4} - 48 \,{\left (9 \, B b^{4} c^{4} - 16 \, A b^{3} c^{5}\right )} x^{3} + 56 \,{\left (9 \, B b^{5} c^{3} - 16 \, A b^{4} c^{4}\right )} x^{2} - 70 \,{\left (9 \, B b^{6} c^{2} - 16 \, A b^{5} c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{344064 \, c^{6}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x+A)*(c*x^2+b*x)^(5/2),x, algorithm="fricas")

[Out]

[-1/688128*(105*(9*B*b^8 - 16*A*b^7*c)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(43008*B*c^8*x
^7 + 945*B*b^7*c - 1680*A*b^6*c^2 + 3072*(33*B*b*c^7 + 16*A*c^8)*x^6 + 256*(243*B*b^2*c^6 + 464*A*b*c^7)*x^5 +
128*(3*B*b^3*c^5 + 592*A*b^2*c^6)*x^4 - 48*(9*B*b^4*c^4 - 16*A*b^3*c^5)*x^3 + 56*(9*B*b^5*c^3 - 16*A*b^4*c^4)
*x^2 - 70*(9*B*b^6*c^2 - 16*A*b^5*c^3)*x)*sqrt(c*x^2 + b*x))/c^6, 1/344064*(105*(9*B*b^8 - 16*A*b^7*c)*sqrt(-c
)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (43008*B*c^8*x^7 + 945*B*b^7*c - 1680*A*b^6*c^2 + 3072*(33*B*b*c^
7 + 16*A*c^8)*x^6 + 256*(243*B*b^2*c^6 + 464*A*b*c^7)*x^5 + 128*(3*B*b^3*c^5 + 592*A*b^2*c^6)*x^4 - 48*(9*B*b^
4*c^4 - 16*A*b^3*c^5)*x^3 + 56*(9*B*b^5*c^3 - 16*A*b^4*c^4)*x^2 - 70*(9*B*b^6*c^2 - 16*A*b^5*c^3)*x)*sqrt(c*x^
2 + b*x))/c^6]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \left (x \left (b + c x\right )\right )^{\frac{5}{2}} \left (A + B x\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*(x*(b + c*x))**(5/2)*(A + B*x), x)

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Giac [A]  time = 1.18657, size = 342, normalized size = 1.81 \begin{align*} \frac{1}{344064} \, \sqrt{c x^{2} + b x}{\left (2 \,{\left (4 \,{\left (2 \,{\left (8 \,{\left (2 \,{\left (12 \,{\left (14 \, B c^{2} x + \frac{33 \, B b c^{8} + 16 \, A c^{9}}{c^{7}}\right )} x + \frac{243 \, B b^{2} c^{7} + 464 \, A b c^{8}}{c^{7}}\right )} x + \frac{3 \, B b^{3} c^{6} + 592 \, A b^{2} c^{7}}{c^{7}}\right )} x - \frac{3 \,{\left (9 \, B b^{4} c^{5} - 16 \, A b^{3} c^{6}\right )}}{c^{7}}\right )} x + \frac{7 \,{\left (9 \, B b^{5} c^{4} - 16 \, A b^{4} c^{5}\right )}}{c^{7}}\right )} x - \frac{35 \,{\left (9 \, B b^{6} c^{3} - 16 \, A b^{5} c^{4}\right )}}{c^{7}}\right )} x + \frac{105 \,{\left (9 \, B b^{7} c^{2} - 16 \, A b^{6} c^{3}\right )}}{c^{7}}\right )} + \frac{5 \,{\left (9 \, B b^{8} - 16 \, A b^{7} c\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{32768 \, c^{\frac{11}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x+A)*(c*x^2+b*x)^(5/2),x, algorithm="giac")

[Out]

1/344064*sqrt(c*x^2 + b*x)*(2*(4*(2*(8*(2*(12*(14*B*c^2*x + (33*B*b*c^8 + 16*A*c^9)/c^7)*x + (243*B*b^2*c^7 +
464*A*b*c^8)/c^7)*x + (3*B*b^3*c^6 + 592*A*b^2*c^7)/c^7)*x - 3*(9*B*b^4*c^5 - 16*A*b^3*c^6)/c^7)*x + 7*(9*B*b^
5*c^4 - 16*A*b^4*c^5)/c^7)*x - 35*(9*B*b^6*c^3 - 16*A*b^5*c^4)/c^7)*x + 105*(9*B*b^7*c^2 - 16*A*b^6*c^3)/c^7)
+ 5/32768*(9*B*b^8 - 16*A*b^7*c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(11/2)