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

Optimal. Leaf size=238 \[ \frac {9 b^7 (11 b B-16 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{16384 c^{13/2}}-\frac {9 b^5 (b+2 c x) \sqrt {b x+c x^2} (11 b B-16 A c)}{16384 c^6}+\frac {3 b^3 (b+2 c x) \left (b x+c x^2\right )^{3/2} (11 b B-16 A c)}{2048 c^5}-\frac {3 b^2 \left (b x+c x^2\right )^{5/2} (11 b B-16 A c)}{640 c^4}+\frac {3 b x \left (b x+c x^2\right )^{5/2} (11 b B-16 A c)}{448 c^3}-\frac {x^2 \left (b x+c x^2\right )^{5/2} (11 b B-16 A c)}{112 c^2}+\frac {B x^3 \left (b x+c x^2\right )^{5/2}}{8 c} \]

________________________________________________________________________________________

Rubi [A]  time = 0.25, antiderivative size = 238, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {794, 670, 640, 612, 620, 206} \begin {gather*} -\frac {9 b^5 (b+2 c x) \sqrt {b x+c x^2} (11 b B-16 A c)}{16384 c^6}+\frac {3 b^3 (b+2 c x) \left (b x+c x^2\right )^{3/2} (11 b B-16 A c)}{2048 c^5}-\frac {3 b^2 \left (b x+c x^2\right )^{5/2} (11 b B-16 A c)}{640 c^4}+\frac {9 b^7 (11 b B-16 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{16384 c^{13/2}}+\frac {3 b x \left (b x+c x^2\right )^{5/2} (11 b B-16 A c)}{448 c^3}-\frac {x^2 \left (b x+c x^2\right )^{5/2} (11 b B-16 A c)}{112 c^2}+\frac {B x^3 \left (b x+c x^2\right )^{5/2}}{8 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*b^5*(11*b*B - 16*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(16384*c^6) + (3*b^3*(11*b*B - 16*A*c)*(b + 2*c*x)*(b
*x + c*x^2)^(3/2))/(2048*c^5) - (3*b^2*(11*b*B - 16*A*c)*(b*x + c*x^2)^(5/2))/(640*c^4) + (3*b*(11*b*B - 16*A*
c)*x*(b*x + c*x^2)^(5/2))/(448*c^3) - ((11*b*B - 16*A*c)*x^2*(b*x + c*x^2)^(5/2))/(112*c^2) + (B*x^3*(b*x + c*
x^2)^(5/2))/(8*c) + (9*b^7*(11*b*B - 16*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(16384*c^(13/2))

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])

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 640

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

Rule 670

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[((m + p)*(2*c*d - b*e))/(c*(m + 2*p + 1)), Int[(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] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rule 794

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)
*(2*c*f - b*g))/(c*e*(m + 2*p + 2)), Int[(d + e*x)^m*(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] && NeQ[m + 2*p + 2, 0] && (NeQ[m, 2] || Eq
Q[d, 0])

Rubi steps

\begin {align*} \int x^3 (A+B x) \left (b x+c x^2\right )^{3/2} \, dx &=\frac {B x^3 \left (b x+c x^2\right )^{5/2}}{8 c}+\frac {\left (3 (-b B+A c)+\frac {5}{2} (-b B+2 A c)\right ) \int x^3 \left (b x+c x^2\right )^{3/2} \, dx}{8 c}\\ &=-\frac {(11 b B-16 A c) x^2 \left (b x+c x^2\right )^{5/2}}{112 c^2}+\frac {B x^3 \left (b x+c x^2\right )^{5/2}}{8 c}+\frac {(9 b (11 b B-16 A c)) \int x^2 \left (b x+c x^2\right )^{3/2} \, dx}{224 c^2}\\ &=\frac {3 b (11 b B-16 A c) x \left (b x+c x^2\right )^{5/2}}{448 c^3}-\frac {(11 b B-16 A c) x^2 \left (b x+c x^2\right )^{5/2}}{112 c^2}+\frac {B x^3 \left (b x+c x^2\right )^{5/2}}{8 c}-\frac {\left (3 b^2 (11 b B-16 A c)\right ) \int x \left (b x+c x^2\right )^{3/2} \, dx}{128 c^3}\\ &=-\frac {3 b^2 (11 b B-16 A c) \left (b x+c x^2\right )^{5/2}}{640 c^4}+\frac {3 b (11 b B-16 A c) x \left (b x+c x^2\right )^{5/2}}{448 c^3}-\frac {(11 b B-16 A c) x^2 \left (b x+c x^2\right )^{5/2}}{112 c^2}+\frac {B x^3 \left (b x+c x^2\right )^{5/2}}{8 c}+\frac {\left (3 b^3 (11 b B-16 A c)\right ) \int \left (b x+c x^2\right )^{3/2} \, dx}{256 c^4}\\ &=\frac {3 b^3 (11 b B-16 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{2048 c^5}-\frac {3 b^2 (11 b B-16 A c) \left (b x+c x^2\right )^{5/2}}{640 c^4}+\frac {3 b (11 b B-16 A c) x \left (b x+c x^2\right )^{5/2}}{448 c^3}-\frac {(11 b B-16 A c) x^2 \left (b x+c x^2\right )^{5/2}}{112 c^2}+\frac {B x^3 \left (b x+c x^2\right )^{5/2}}{8 c}-\frac {\left (9 b^5 (11 b B-16 A c)\right ) \int \sqrt {b x+c x^2} \, dx}{4096 c^5}\\ &=-\frac {9 b^5 (11 b B-16 A c) (b+2 c x) \sqrt {b x+c x^2}}{16384 c^6}+\frac {3 b^3 (11 b B-16 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{2048 c^5}-\frac {3 b^2 (11 b B-16 A c) \left (b x+c x^2\right )^{5/2}}{640 c^4}+\frac {3 b (11 b B-16 A c) x \left (b x+c x^2\right )^{5/2}}{448 c^3}-\frac {(11 b B-16 A c) x^2 \left (b x+c x^2\right )^{5/2}}{112 c^2}+\frac {B x^3 \left (b x+c x^2\right )^{5/2}}{8 c}+\frac {\left (9 b^7 (11 b B-16 A c)\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{32768 c^6}\\ &=-\frac {9 b^5 (11 b B-16 A c) (b+2 c x) \sqrt {b x+c x^2}}{16384 c^6}+\frac {3 b^3 (11 b B-16 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{2048 c^5}-\frac {3 b^2 (11 b B-16 A c) \left (b x+c x^2\right )^{5/2}}{640 c^4}+\frac {3 b (11 b B-16 A c) x \left (b x+c x^2\right )^{5/2}}{448 c^3}-\frac {(11 b B-16 A c) x^2 \left (b x+c x^2\right )^{5/2}}{112 c^2}+\frac {B x^3 \left (b x+c x^2\right )^{5/2}}{8 c}+\frac {\left (9 b^7 (11 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^6}\\ &=-\frac {9 b^5 (11 b B-16 A c) (b+2 c x) \sqrt {b x+c x^2}}{16384 c^6}+\frac {3 b^3 (11 b B-16 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{2048 c^5}-\frac {3 b^2 (11 b B-16 A c) \left (b x+c x^2\right )^{5/2}}{640 c^4}+\frac {3 b (11 b B-16 A c) x \left (b x+c x^2\right )^{5/2}}{448 c^3}-\frac {(11 b B-16 A c) x^2 \left (b x+c x^2\right )^{5/2}}{112 c^2}+\frac {B x^3 \left (b x+c x^2\right )^{5/2}}{8 c}+\frac {9 b^7 (11 b B-16 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{16384 c^{13/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.43, size = 179, normalized size = 0.75 \begin {gather*} \frac {x^5 \sqrt {x (b+c x)} \left (\frac {11 (11 b B-16 A c) \left (315 b^{13/2} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )-\sqrt {c} \sqrt {x} \sqrt {\frac {c x}{b}+1} \left (315 b^6-210 b^5 c x+168 b^4 c^2 x^2-144 b^3 c^3 x^3+128 b^2 c^4 x^4+6400 b c^5 x^5+5120 c^6 x^6\right )\right )}{71680 c^{11/2} x^{11/2} \sqrt {\frac {c x}{b}+1}}+11 B (b+c x)^2\right )}{88 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^5*Sqrt[x*(b + c*x)]*(11*B*(b + c*x)^2 + (11*(11*b*B - 16*A*c)*(-(Sqrt[c]*Sqrt[x]*Sqrt[1 + (c*x)/b]*(315*b^6
- 210*b^5*c*x + 168*b^4*c^2*x^2 - 144*b^3*c^3*x^3 + 128*b^2*c^4*x^4 + 6400*b*c^5*x^5 + 5120*c^6*x^6)) + 315*b
^(13/2)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]]))/(71680*c^(11/2)*x^(11/2)*Sqrt[1 + (c*x)/b])))/(88*c)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.84, size = 225, normalized size = 0.95 \begin {gather*} \frac {\sqrt {b x+c x^2} \left (5040 A b^6 c-3360 A b^5 c^2 x+2688 A b^4 c^3 x^2-2304 A b^3 c^4 x^3+2048 A b^2 c^5 x^4+102400 A b c^6 x^5+81920 A c^7 x^6-3465 b^7 B+2310 b^6 B c x-1848 b^5 B c^2 x^2+1584 b^4 B c^3 x^3-1408 b^3 B c^4 x^4+1280 b^2 B c^5 x^5+87040 b B c^6 x^6+71680 B c^7 x^7\right )}{573440 c^6}-\frac {9 \left (11 b^8 B-16 A b^7 c\right ) \log \left (-2 \sqrt {c} \sqrt {b x+c x^2}+b+2 c x\right )}{32768 c^{13/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x + c*x^2]*(-3465*b^7*B + 5040*A*b^6*c + 2310*b^6*B*c*x - 3360*A*b^5*c^2*x - 1848*b^5*B*c^2*x^2 + 2688
*A*b^4*c^3*x^2 + 1584*b^4*B*c^3*x^3 - 2304*A*b^3*c^4*x^3 - 1408*b^3*B*c^4*x^4 + 2048*A*b^2*c^5*x^4 + 1280*b^2*
B*c^5*x^5 + 102400*A*b*c^6*x^5 + 87040*b*B*c^6*x^6 + 81920*A*c^7*x^6 + 71680*B*c^7*x^7))/(573440*c^6) - (9*(11
*b^8*B - 16*A*b^7*c)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[b*x + c*x^2]])/(32768*c^(13/2))

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 446, normalized size = 1.87 \begin {gather*} \left [-\frac {315 \, {\left (11 \, 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 (71680 \, B c^{8} x^{7} - 3465 \, B b^{7} c + 5040 \, A b^{6} c^{2} + 5120 \, {\left (17 \, B b c^{7} + 16 \, A c^{8}\right )} x^{6} + 1280 \, {\left (B b^{2} c^{6} + 80 \, A b c^{7}\right )} x^{5} - 128 \, {\left (11 \, B b^{3} c^{5} - 16 \, A b^{2} c^{6}\right )} x^{4} + 144 \, {\left (11 \, B b^{4} c^{4} - 16 \, A b^{3} c^{5}\right )} x^{3} - 168 \, {\left (11 \, B b^{5} c^{3} - 16 \, A b^{4} c^{4}\right )} x^{2} + 210 \, {\left (11 \, B b^{6} c^{2} - 16 \, A b^{5} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{1146880 \, c^{7}}, -\frac {315 \, {\left (11 \, 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 (71680 \, B c^{8} x^{7} - 3465 \, B b^{7} c + 5040 \, A b^{6} c^{2} + 5120 \, {\left (17 \, B b c^{7} + 16 \, A c^{8}\right )} x^{6} + 1280 \, {\left (B b^{2} c^{6} + 80 \, A b c^{7}\right )} x^{5} - 128 \, {\left (11 \, B b^{3} c^{5} - 16 \, A b^{2} c^{6}\right )} x^{4} + 144 \, {\left (11 \, B b^{4} c^{4} - 16 \, A b^{3} c^{5}\right )} x^{3} - 168 \, {\left (11 \, B b^{5} c^{3} - 16 \, A b^{4} c^{4}\right )} x^{2} + 210 \, {\left (11 \, B b^{6} c^{2} - 16 \, A b^{5} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{573440 \, c^{7}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x+A)*(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

[-1/1146880*(315*(11*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*(71680*B*c^8
*x^7 - 3465*B*b^7*c + 5040*A*b^6*c^2 + 5120*(17*B*b*c^7 + 16*A*c^8)*x^6 + 1280*(B*b^2*c^6 + 80*A*b*c^7)*x^5 -
128*(11*B*b^3*c^5 - 16*A*b^2*c^6)*x^4 + 144*(11*B*b^4*c^4 - 16*A*b^3*c^5)*x^3 - 168*(11*B*b^5*c^3 - 16*A*b^4*c
^4)*x^2 + 210*(11*B*b^6*c^2 - 16*A*b^5*c^3)*x)*sqrt(c*x^2 + b*x))/c^7, -1/573440*(315*(11*B*b^8 - 16*A*b^7*c)*
sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (71680*B*c^8*x^7 - 3465*B*b^7*c + 5040*A*b^6*c^2 + 5120*(1
7*B*b*c^7 + 16*A*c^8)*x^6 + 1280*(B*b^2*c^6 + 80*A*b*c^7)*x^5 - 128*(11*B*b^3*c^5 - 16*A*b^2*c^6)*x^4 + 144*(1
1*B*b^4*c^4 - 16*A*b^3*c^5)*x^3 - 168*(11*B*b^5*c^3 - 16*A*b^4*c^4)*x^2 + 210*(11*B*b^6*c^2 - 16*A*b^5*c^3)*x)
*sqrt(c*x^2 + b*x))/c^7]

________________________________________________________________________________________

giac [A]  time = 0.22, size = 249, normalized size = 1.05 \begin {gather*} \frac {1}{573440} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (10 \, {\left (4 \, {\left (14 \, B c x + \frac {17 \, B b c^{7} + 16 \, A c^{8}}{c^{7}}\right )} x + \frac {B b^{2} c^{6} + 80 \, A b c^{7}}{c^{7}}\right )} x - \frac {11 \, B b^{3} c^{5} - 16 \, A b^{2} c^{6}}{c^{7}}\right )} x + \frac {9 \, {\left (11 \, B b^{4} c^{4} - 16 \, A b^{3} c^{5}\right )}}{c^{7}}\right )} x - \frac {21 \, {\left (11 \, B b^{5} c^{3} - 16 \, A b^{4} c^{4}\right )}}{c^{7}}\right )} x + \frac {105 \, {\left (11 \, B b^{6} c^{2} - 16 \, A b^{5} c^{3}\right )}}{c^{7}}\right )} x - \frac {315 \, {\left (11 \, B b^{7} c - 16 \, A b^{6} c^{2}\right )}}{c^{7}}\right )} - \frac {9 \, {\left (11 \, 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 {13}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x+A)*(c*x^2+b*x)^(3/2),x, algorithm="giac")

[Out]

1/573440*sqrt(c*x^2 + b*x)*(2*(4*(2*(8*(10*(4*(14*B*c*x + (17*B*b*c^7 + 16*A*c^8)/c^7)*x + (B*b^2*c^6 + 80*A*b
*c^7)/c^7)*x - (11*B*b^3*c^5 - 16*A*b^2*c^6)/c^7)*x + 9*(11*B*b^4*c^4 - 16*A*b^3*c^5)/c^7)*x - 21*(11*B*b^5*c^
3 - 16*A*b^4*c^4)/c^7)*x + 105*(11*B*b^6*c^2 - 16*A*b^5*c^3)/c^7)*x - 315*(11*B*b^7*c - 16*A*b^6*c^2)/c^7) - 9
/32768*(11*B*b^8 - 16*A*b^7*c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(13/2)

________________________________________________________________________________________

maple [A]  time = 0.05, size = 373, normalized size = 1.57 \begin {gather*} -\frac {9 A \,b^{7} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2048 c^{\frac {11}{2}}}+\frac {99 B \,b^{8} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{32768 c^{\frac {13}{2}}}+\frac {9 \sqrt {c \,x^{2}+b x}\, A \,b^{5} x}{512 c^{4}}-\frac {99 \sqrt {c \,x^{2}+b x}\, B \,b^{6} x}{8192 c^{5}}+\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}} B \,x^{3}}{8 c}+\frac {9 \sqrt {c \,x^{2}+b x}\, A \,b^{6}}{1024 c^{5}}-\frac {3 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} A \,b^{3} x}{64 c^{3}}+\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}} A \,x^{2}}{7 c}-\frac {99 \sqrt {c \,x^{2}+b x}\, B \,b^{7}}{16384 c^{6}}+\frac {33 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} B \,b^{4} x}{1024 c^{4}}-\frac {11 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} B b \,x^{2}}{112 c^{2}}-\frac {3 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} A \,b^{4}}{128 c^{4}}-\frac {3 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} A b x}{28 c^{2}}+\frac {33 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} B \,b^{5}}{2048 c^{5}}+\frac {33 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} B \,b^{2} x}{448 c^{3}}+\frac {3 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} A \,b^{2}}{40 c^{3}}-\frac {33 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} B \,b^{3}}{640 c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*B*x^3*(c*x^2+b*x)^(5/2)/c-11/112*B*b/c^2*x^2*(c*x^2+b*x)^(5/2)+33/448*B*b^2/c^3*x*(c*x^2+b*x)^(5/2)-33/640
*B*b^3/c^4*(c*x^2+b*x)^(5/2)+33/1024*B*b^4/c^4*(c*x^2+b*x)^(3/2)*x+33/2048*B*b^5/c^5*(c*x^2+b*x)^(3/2)-99/8192
*B*b^6/c^5*(c*x^2+b*x)^(1/2)*x-99/16384*B*b^7/c^6*(c*x^2+b*x)^(1/2)+99/32768*B*b^8/c^(13/2)*ln((c*x+1/2*b)/c^(
1/2)+(c*x^2+b*x)^(1/2))+1/7*A*x^2*(c*x^2+b*x)^(5/2)/c-3/28*A*b/c^2*x*(c*x^2+b*x)^(5/2)+3/40*A*b^2/c^3*(c*x^2+b
*x)^(5/2)-3/64*A*b^3/c^3*(c*x^2+b*x)^(3/2)*x-3/128*A*b^4/c^4*(c*x^2+b*x)^(3/2)+9/512*A*b^5/c^4*(c*x^2+b*x)^(1/
2)*x+9/1024*A*b^6/c^5*(c*x^2+b*x)^(1/2)-9/2048*A*b^7/c^(11/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))

________________________________________________________________________________________

maxima [A]  time = 0.95, size = 370, normalized size = 1.55 \begin {gather*} \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} B x^{3}}{8 \, c} - \frac {11 \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} B b x^{2}}{112 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} A x^{2}}{7 \, c} - \frac {99 \, \sqrt {c x^{2} + b x} B b^{6} x}{8192 \, c^{5}} + \frac {33 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b^{4} x}{1024 \, c^{4}} + \frac {9 \, \sqrt {c x^{2} + b x} A b^{5} x}{512 \, c^{4}} + \frac {33 \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} B b^{2} x}{448 \, c^{3}} - \frac {3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b^{3} x}{64 \, c^{3}} - \frac {3 \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} A b x}{28 \, c^{2}} + \frac {99 \, B b^{8} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{32768 \, c^{\frac {13}{2}}} - \frac {9 \, A b^{7} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2048 \, c^{\frac {11}{2}}} - \frac {99 \, \sqrt {c x^{2} + b x} B b^{7}}{16384 \, c^{6}} + \frac {33 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b^{5}}{2048 \, c^{5}} + \frac {9 \, \sqrt {c x^{2} + b x} A b^{6}}{1024 \, c^{5}} - \frac {33 \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} B b^{3}}{640 \, c^{4}} - \frac {3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b^{4}}{128 \, c^{4}} + \frac {3 \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} A b^{2}}{40 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x+A)*(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

1/8*(c*x^2 + b*x)^(5/2)*B*x^3/c - 11/112*(c*x^2 + b*x)^(5/2)*B*b*x^2/c^2 + 1/7*(c*x^2 + b*x)^(5/2)*A*x^2/c - 9
9/8192*sqrt(c*x^2 + b*x)*B*b^6*x/c^5 + 33/1024*(c*x^2 + b*x)^(3/2)*B*b^4*x/c^4 + 9/512*sqrt(c*x^2 + b*x)*A*b^5
*x/c^4 + 33/448*(c*x^2 + b*x)^(5/2)*B*b^2*x/c^3 - 3/64*(c*x^2 + b*x)^(3/2)*A*b^3*x/c^3 - 3/28*(c*x^2 + b*x)^(5
/2)*A*b*x/c^2 + 99/32768*B*b^8*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(13/2) - 9/2048*A*b^7*log(2*c*x
+ b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(11/2) - 99/16384*sqrt(c*x^2 + b*x)*B*b^7/c^6 + 33/2048*(c*x^2 + b*x)^(3/
2)*B*b^5/c^5 + 9/1024*sqrt(c*x^2 + b*x)*A*b^6/c^5 - 33/640*(c*x^2 + b*x)^(5/2)*B*b^3/c^4 - 3/128*(c*x^2 + b*x)
^(3/2)*A*b^4/c^4 + 3/40*(c*x^2 + b*x)^(5/2)*A*b^2/c^3

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^3\,{\left (c\,x^2+b\,x\right )}^{3/2}\,\left (A+B\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{3} \left (x \left (b + c x\right )\right )^{\frac {3}{2}} \left (A + B x\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*(x*(b + c*x))**(3/2)*(A + B*x), x)

________________________________________________________________________________________