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

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

[Out]

-5/6144*b^3*(-16*A*c+9*B*b)*(2*c*x+b)*(c*x^2+b*x)^(3/2)/c^4+1/384*b*(-16*A*c+9*B*b)*(2*c*x+b)*(c*x^2+b*x)^(5/2
)/c^3-1/112*(-14*B*c*x-16*A*c+9*B*b)*(c*x^2+b*x)^(7/2)/c^2-5/16384*b^7*(-16*A*c+9*B*b)*arctanh(x*c^(1/2)/(c*x^
2+b*x)^(1/2))/c^(11/2)+5/16384*b^5*(-16*A*c+9*B*b)*(2*c*x+b)*(c*x^2+b*x)^(1/2)/c^5

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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 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]

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 = 0.41, size = 182, normalized size = 0.96 \[ \frac {(x (b+c x))^{9/2} \left (9 B (b+c x)^3-\frac {3 (9 b B-16 A c) \left (105 b^{13/2} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )+\sqrt {c} \sqrt {x} \sqrt {\frac {c x}{b}+1} \left (-105 b^6+70 b^5 c x-56 b^4 c^2 x^2+48 b^3 c^3 x^3+4736 b^2 c^4 x^4+7424 b c^5 x^5+3072 c^6 x^6\right )\right )}{14336 c^{9/2} x^{9/2} \sqrt {\frac {c x}{b}+1}}\right )}{72 c (b+c x)^4} \]

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)

________________________________________________________________________________________

fricas [A]  time = 1.08, size = 447, normalized size = 2.37 \[ \left [-\frac {105 \, {\left (9 \, B b^{8} - 16 \, A b^{7} c\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (43008 \, B c^{8} x^{7} + 945 \, B b^{7} c - 1680 \, A b^{6} c^{2} + 3072 \, {\left (33 \, B b c^{7} + 16 \, A c^{8}\right )} x^{6} + 256 \, {\left (243 \, B b^{2} c^{6} + 464 \, A b c^{7}\right )} x^{5} + 128 \, {\left (3 \, B b^{3} c^{5} + 592 \, A b^{2} c^{6}\right )} x^{4} - 48 \, {\left (9 \, B b^{4} c^{4} - 16 \, A b^{3} c^{5}\right )} x^{3} + 56 \, {\left (9 \, B b^{5} c^{3} - 16 \, A b^{4} c^{4}\right )} x^{2} - 70 \, {\left (9 \, B b^{6} c^{2} - 16 \, A b^{5} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{688128 \, c^{6}}, \frac {105 \, {\left (9 \, B b^{8} - 16 \, A b^{7} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (43008 \, B c^{8} x^{7} + 945 \, B b^{7} c - 1680 \, A b^{6} c^{2} + 3072 \, {\left (33 \, B b c^{7} + 16 \, A c^{8}\right )} x^{6} + 256 \, {\left (243 \, B b^{2} c^{6} + 464 \, A b c^{7}\right )} x^{5} + 128 \, {\left (3 \, B b^{3} c^{5} + 592 \, A b^{2} c^{6}\right )} x^{4} - 48 \, {\left (9 \, B b^{4} c^{4} - 16 \, A b^{3} c^{5}\right )} x^{3} + 56 \, {\left (9 \, B b^{5} c^{3} - 16 \, A b^{4} c^{4}\right )} x^{2} - 70 \, {\left (9 \, B b^{6} c^{2} - 16 \, A b^{5} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{344064 \, c^{6}}\right ] \]

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]

________________________________________________________________________________________

giac [A]  time = 0.26, size = 253, normalized size = 1.34 \[ \frac {1}{344064} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (2 \, {\left (12 \, {\left (14 \, B c^{2} x + \frac {33 \, B b c^{8} + 16 \, A c^{9}}{c^{7}}\right )} x + \frac {243 \, B b^{2} c^{7} + 464 \, A b c^{8}}{c^{7}}\right )} x + \frac {3 \, B b^{3} c^{6} + 592 \, A b^{2} c^{7}}{c^{7}}\right )} x - \frac {3 \, {\left (9 \, B b^{4} c^{5} - 16 \, A b^{3} c^{6}\right )}}{c^{7}}\right )} x + \frac {7 \, {\left (9 \, B b^{5} c^{4} - 16 \, A b^{4} c^{5}\right )}}{c^{7}}\right )} x - \frac {35 \, {\left (9 \, B b^{6} c^{3} - 16 \, A b^{5} c^{4}\right )}}{c^{7}}\right )} x + \frac {105 \, {\left (9 \, B b^{7} c^{2} - 16 \, A b^{6} c^{3}\right )}}{c^{7}}\right )} + \frac {5 \, {\left (9 \, B b^{8} - 16 \, A b^{7} c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{32768 \, c^{\frac {11}{2}}} \]

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)

________________________________________________________________________________________

maple [B]  time = 0.06, size = 365, normalized size = 1.93 \[ \frac {5 A \,b^{7} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2048 c^{\frac {9}{2}}}-\frac {45 B \,b^{8} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{32768 c^{\frac {11}{2}}}-\frac {5 \sqrt {c \,x^{2}+b x}\, A \,b^{5} x}{512 c^{3}}+\frac {45 \sqrt {c \,x^{2}+b x}\, B \,b^{6} x}{8192 c^{4}}-\frac {5 \sqrt {c \,x^{2}+b x}\, A \,b^{6}}{1024 c^{4}}+\frac {5 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} A \,b^{3} x}{192 c^{2}}+\frac {45 \sqrt {c \,x^{2}+b x}\, B \,b^{7}}{16384 c^{5}}-\frac {15 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} B \,b^{4} x}{1024 c^{3}}+\frac {5 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} A \,b^{4}}{384 c^{3}}-\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}} A b x}{12 c}-\frac {15 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} B \,b^{5}}{2048 c^{4}}+\frac {3 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} B \,b^{2} x}{64 c^{2}}-\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}} A \,b^{2}}{24 c^{2}}+\frac {3 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} B \,b^{3}}{128 c^{3}}+\frac {\left (c \,x^{2}+b x \right )^{\frac {7}{2}} B x}{8 c}+\frac {\left (c \,x^{2}+b x \right )^{\frac {7}{2}} A}{7 c}-\frac {9 \left (c \,x^{2}+b x \right )^{\frac {7}{2}} B b}{112 c^{2}} \]

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((c*x+1/2*b)/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((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))

________________________________________________________________________________________

maxima [B]  time = 1.00, size = 362, normalized size = 1.92 \[ \frac {45 \, \sqrt {c x^{2} + b x} B b^{6} x}{8192 \, c^{4}} - \frac {15 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b^{4} x}{1024 \, c^{3}} - \frac {5 \, \sqrt {c x^{2} + b x} A b^{5} x}{512 \, c^{3}} + \frac {3 \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} B b^{2} x}{64 \, c^{2}} + \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b^{3} x}{192 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {7}{2}} B x}{8 \, c} - \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} A b x}{12 \, c} - \frac {45 \, B b^{8} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{32768 \, c^{\frac {11}{2}}} + \frac {5 \, A b^{7} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2048 \, c^{\frac {9}{2}}} + \frac {45 \, \sqrt {c x^{2} + b x} B b^{7}}{16384 \, c^{5}} - \frac {15 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b^{5}}{2048 \, c^{4}} - \frac {5 \, \sqrt {c x^{2} + b x} A b^{6}}{1024 \, c^{4}} + \frac {3 \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} B b^{3}}{128 \, c^{3}} + \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b^{4}}{384 \, c^{3}} - \frac {9 \, {\left (c x^{2} + b x\right )}^{\frac {7}{2}} B b}{112 \, c^{2}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} A b^{2}}{24 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {7}{2}} A}{7 \, c} \]

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]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,{\left (c\,x^2+b\,x\right )}^{5/2}\,\left (A+B\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \left (x \left (b + c x\right )\right )^{\frac {5}{2}} \left (A + B x\right )\, dx \]

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)

________________________________________________________________________________________