3.1164 \(\int (A+B x) (d+e x) \sqrt {b x+c x^2} \, dx\)

Optimal. Leaf size=154 \[ -\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right ) \left (-8 b c (A e+B d)+16 A c^2 d+5 b^2 B e\right )}{64 c^{7/2}}+\frac {(b+2 c x) \sqrt {b x+c x^2} \left (-8 b c (A e+B d)+16 A c^2 d+5 b^2 B e\right )}{64 c^3}-\frac {\left (b x+c x^2\right )^{3/2} (-8 c (A e+B d)+5 b B e-6 B c e x)}{24 c^2} \]

[Out]

-1/24*(5*b*B*e-8*c*(A*e+B*d)-6*B*c*e*x)*(c*x^2+b*x)^(3/2)/c^2-1/64*b^2*(16*A*c^2*d+5*b^2*B*e-8*b*c*(A*e+B*d))*
arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(7/2)+1/64*(16*A*c^2*d+5*b^2*B*e-8*b*c*(A*e+B*d))*(2*c*x+b)*(c*x^2+b*x)
^(1/2)/c^3

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Rubi [A]  time = 0.14, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {779, 612, 620, 206} \[ \frac {(b+2 c x) \sqrt {b x+c x^2} \left (-8 b c (A e+B d)+16 A c^2 d+5 b^2 B e\right )}{64 c^3}-\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right ) \left (-8 b c (A e+B d)+16 A c^2 d+5 b^2 B e\right )}{64 c^{7/2}}-\frac {\left (b x+c x^2\right )^{3/2} (-8 c (A e+B d)+5 b B e-6 B c e x)}{24 c^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)*(d + e*x)*Sqrt[b*x + c*x^2],x]

[Out]

((16*A*c^2*d + 5*b^2*B*e - 8*b*c*(B*d + A*e))*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(64*c^3) - ((5*b*B*e - 8*c*(B*d +
A*e) - 6*B*c*e*x)*(b*x + c*x^2)^(3/2))/(24*c^2) - (b^2*(16*A*c^2*d + 5*b^2*B*e - 8*b*c*(B*d + A*e))*ArcTanh[(
Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(64*c^(7/2))

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 (A+B x) (d+e x) \sqrt {b x+c x^2} \, dx &=-\frac {(5 b B e-8 c (B d+A e)-6 B c e x) \left (b x+c x^2\right )^{3/2}}{24 c^2}+\frac {\left (\frac {5}{2} b^2 B e+4 c (2 A c d-b (B d+A e))\right ) \int \sqrt {b x+c x^2} \, dx}{8 c^2}\\ &=\frac {\left (16 A c^2 d+5 b^2 B e-8 b c (B d+A e)\right ) (b+2 c x) \sqrt {b x+c x^2}}{64 c^3}-\frac {(5 b B e-8 c (B d+A e)-6 B c e x) \left (b x+c x^2\right )^{3/2}}{24 c^2}-\frac {\left (b^2 \left (16 A c^2 d+5 b^2 B e-8 b c (B d+A e)\right )\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{128 c^3}\\ &=\frac {\left (16 A c^2 d+5 b^2 B e-8 b c (B d+A e)\right ) (b+2 c x) \sqrt {b x+c x^2}}{64 c^3}-\frac {(5 b B e-8 c (B d+A e)-6 B c e x) \left (b x+c x^2\right )^{3/2}}{24 c^2}-\frac {\left (b^2 \left (16 A c^2 d+5 b^2 B e-8 b c (B d+A e)\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{64 c^3}\\ &=\frac {\left (16 A c^2 d+5 b^2 B e-8 b c (B d+A e)\right ) (b+2 c x) \sqrt {b x+c x^2}}{64 c^3}-\frac {(5 b B e-8 c (B d+A e)-6 B c e x) \left (b x+c x^2\right )^{3/2}}{24 c^2}-\frac {b^2 \left (16 A c^2 d+5 b^2 B e-8 b c (B d+A e)\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.43, size = 177, normalized size = 1.15 \[ \frac {\sqrt {x (b+c x)} \left (\sqrt {c} \left (-2 b^2 c (12 A e+12 B d+5 B e x)+8 b c^2 (2 A (3 d+e x)+B x (2 d+e x))+16 c^3 x (A (6 d+4 e x)+B x (4 d+3 e x))+15 b^3 B e\right )-\frac {3 b^{3/2} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right ) \left (-8 b c (A e+B d)+16 A c^2 d+5 b^2 B e\right )}{\sqrt {x} \sqrt {\frac {c x}{b}+1}}\right )}{192 c^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)*(d + e*x)*Sqrt[b*x + c*x^2],x]

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(15*b^3*B*e - 2*b^2*c*(12*B*d + 12*A*e + 5*B*e*x) + 8*b*c^2*(B*x*(2*d + e*x) + 2*A
*(3*d + e*x)) + 16*c^3*x*(B*x*(4*d + 3*e*x) + A*(6*d + 4*e*x))) - (3*b^(3/2)*(16*A*c^2*d + 5*b^2*B*e - 8*b*c*(
B*d + A*e))*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(192*c^(7/2))

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fricas [A]  time = 0.72, size = 408, normalized size = 2.65 \[ \left [\frac {3 \, {\left (8 \, {\left (B b^{3} c - 2 \, A b^{2} c^{2}\right )} d - {\left (5 \, B b^{4} - 8 \, A b^{3} c\right )} e\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (48 \, B c^{4} e x^{3} + 8 \, {\left (8 \, B c^{4} d + {\left (B b c^{3} + 8 \, A c^{4}\right )} e\right )} x^{2} - 24 \, {\left (B b^{2} c^{2} - 2 \, A b c^{3}\right )} d + 3 \, {\left (5 \, B b^{3} c - 8 \, A b^{2} c^{2}\right )} e + 2 \, {\left (8 \, {\left (B b c^{3} + 6 \, A c^{4}\right )} d - {\left (5 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} e\right )} x\right )} \sqrt {c x^{2} + b x}}{384 \, c^{4}}, -\frac {3 \, {\left (8 \, {\left (B b^{3} c - 2 \, A b^{2} c^{2}\right )} d - {\left (5 \, B b^{4} - 8 \, A b^{3} c\right )} e\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (48 \, B c^{4} e x^{3} + 8 \, {\left (8 \, B c^{4} d + {\left (B b c^{3} + 8 \, A c^{4}\right )} e\right )} x^{2} - 24 \, {\left (B b^{2} c^{2} - 2 \, A b c^{3}\right )} d + 3 \, {\left (5 \, B b^{3} c - 8 \, A b^{2} c^{2}\right )} e + 2 \, {\left (8 \, {\left (B b c^{3} + 6 \, A c^{4}\right )} d - {\left (5 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} e\right )} x\right )} \sqrt {c x^{2} + b x}}{192 \, c^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)*(c*x^2+b*x)^(1/2),x, algorithm="fricas")

[Out]

[1/384*(3*(8*(B*b^3*c - 2*A*b^2*c^2)*d - (5*B*b^4 - 8*A*b^3*c)*e)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*
sqrt(c)) + 2*(48*B*c^4*e*x^3 + 8*(8*B*c^4*d + (B*b*c^3 + 8*A*c^4)*e)*x^2 - 24*(B*b^2*c^2 - 2*A*b*c^3)*d + 3*(5
*B*b^3*c - 8*A*b^2*c^2)*e + 2*(8*(B*b*c^3 + 6*A*c^4)*d - (5*B*b^2*c^2 - 8*A*b*c^3)*e)*x)*sqrt(c*x^2 + b*x))/c^
4, -1/192*(3*(8*(B*b^3*c - 2*A*b^2*c^2)*d - (5*B*b^4 - 8*A*b^3*c)*e)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c
)/(c*x)) - (48*B*c^4*e*x^3 + 8*(8*B*c^4*d + (B*b*c^3 + 8*A*c^4)*e)*x^2 - 24*(B*b^2*c^2 - 2*A*b*c^3)*d + 3*(5*B
*b^3*c - 8*A*b^2*c^2)*e + 2*(8*(B*b*c^3 + 6*A*c^4)*d - (5*B*b^2*c^2 - 8*A*b*c^3)*e)*x)*sqrt(c*x^2 + b*x))/c^4]

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giac [A]  time = 0.23, size = 205, normalized size = 1.33 \[ \frac {1}{192} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (6 \, B x e + \frac {8 \, B c^{3} d + B b c^{2} e + 8 \, A c^{3} e}{c^{3}}\right )} x + \frac {8 \, B b c^{2} d + 48 \, A c^{3} d - 5 \, B b^{2} c e + 8 \, A b c^{2} e}{c^{3}}\right )} x - \frac {3 \, {\left (8 \, B b^{2} c d - 16 \, A b c^{2} d - 5 \, B b^{3} e + 8 \, A b^{2} c e\right )}}{c^{3}}\right )} - \frac {{\left (8 \, B b^{3} c d - 16 \, A b^{2} c^{2} d - 5 \, B b^{4} e + 8 \, A b^{3} c e\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{128 \, c^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)*(c*x^2+b*x)^(1/2),x, algorithm="giac")

[Out]

1/192*sqrt(c*x^2 + b*x)*(2*(4*(6*B*x*e + (8*B*c^3*d + B*b*c^2*e + 8*A*c^3*e)/c^3)*x + (8*B*b*c^2*d + 48*A*c^3*
d - 5*B*b^2*c*e + 8*A*b*c^2*e)/c^3)*x - 3*(8*B*b^2*c*d - 16*A*b*c^2*d - 5*B*b^3*e + 8*A*b^2*c*e)/c^3) - 1/128*
(8*B*b^3*c*d - 16*A*b^2*c^2*d - 5*B*b^4*e + 8*A*b^3*c*e)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) -
b))/c^(7/2)

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maple [B]  time = 0.05, size = 372, normalized size = 2.42 \[ \frac {A \,b^{3} e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{16 c^{\frac {5}{2}}}-\frac {A \,b^{2} d \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}-\frac {5 B \,b^{4} e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{128 c^{\frac {7}{2}}}+\frac {B \,b^{3} d \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{16 c^{\frac {5}{2}}}-\frac {\sqrt {c \,x^{2}+b x}\, A b e x}{4 c}+\frac {\sqrt {c \,x^{2}+b x}\, A d x}{2}+\frac {5 \sqrt {c \,x^{2}+b x}\, B \,b^{2} e x}{32 c^{2}}-\frac {\sqrt {c \,x^{2}+b x}\, B b d x}{4 c}-\frac {\sqrt {c \,x^{2}+b x}\, A \,b^{2} e}{8 c^{2}}+\frac {\sqrt {c \,x^{2}+b x}\, A b d}{4 c}+\frac {5 \sqrt {c \,x^{2}+b x}\, B \,b^{3} e}{64 c^{3}}-\frac {\sqrt {c \,x^{2}+b x}\, B \,b^{2} d}{8 c^{2}}+\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} B e x}{4 c}+\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} A e}{3 c}-\frac {5 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} B b e}{24 c^{2}}+\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} B d}{3 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)*(c*x^2+b*x)^(1/2),x)

[Out]

1/4*B*e*x*(c*x^2+b*x)^(3/2)/c-5/24*B*e*b/c^2*(c*x^2+b*x)^(3/2)+5/32*B*e*b^2/c^2*x*(c*x^2+b*x)^(1/2)+5/64*B*e*b
^3/c^3*(c*x^2+b*x)^(1/2)-5/128*B*e*b^4/c^(7/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))+1/3*(c*x^2+b*x)^(3/2)
/c*A*e+1/3*(c*x^2+b*x)^(3/2)/c*B*d-1/4*b/c*x*(c*x^2+b*x)^(1/2)*A*e-1/4*b/c*x*(c*x^2+b*x)^(1/2)*B*d-1/8*b^2/c^2
*(c*x^2+b*x)^(1/2)*A*e-1/8*b^2/c^2*(c*x^2+b*x)^(1/2)*B*d+1/16*b^3/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(
1/2))*A*e+1/16*b^3/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))*B*d+1/2*A*d*x*(c*x^2+b*x)^(1/2)+1/4*A*d/c
*(c*x^2+b*x)^(1/2)*b-1/8*A*d*b^2/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))

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maxima [B]  time = 0.54, size = 295, normalized size = 1.92 \[ \frac {1}{2} \, \sqrt {c x^{2} + b x} A d x + \frac {5 \, \sqrt {c x^{2} + b x} B b^{2} e x}{32 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B e x}{4 \, c} - \frac {A b^{2} d \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {3}{2}}} - \frac {5 \, B b^{4} e \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{128 \, c^{\frac {7}{2}}} + \frac {\sqrt {c x^{2} + b x} A b d}{4 \, c} + \frac {5 \, \sqrt {c x^{2} + b x} B b^{3} e}{64 \, c^{3}} - \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b e}{24 \, c^{2}} - \frac {\sqrt {c x^{2} + b x} {\left (B d + A e\right )} b x}{4 \, c} + \frac {{\left (B d + A e\right )} b^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{16 \, c^{\frac {5}{2}}} - \frac {\sqrt {c x^{2} + b x} {\left (B d + A e\right )} b^{2}}{8 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} {\left (B d + A e\right )}}{3 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)*(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

1/2*sqrt(c*x^2 + b*x)*A*d*x + 5/32*sqrt(c*x^2 + b*x)*B*b^2*e*x/c^2 + 1/4*(c*x^2 + b*x)^(3/2)*B*e*x/c - 1/8*A*b
^2*d*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(3/2) - 5/128*B*b^4*e*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*
sqrt(c))/c^(7/2) + 1/4*sqrt(c*x^2 + b*x)*A*b*d/c + 5/64*sqrt(c*x^2 + b*x)*B*b^3*e/c^3 - 5/24*(c*x^2 + b*x)^(3/
2)*B*b*e/c^2 - 1/4*sqrt(c*x^2 + b*x)*(B*d + A*e)*b*x/c + 1/16*(B*d + A*e)*b^3*log(2*c*x + b + 2*sqrt(c*x^2 + b
*x)*sqrt(c))/c^(5/2) - 1/8*sqrt(c*x^2 + b*x)*(B*d + A*e)*b^2/c^2 + 1/3*(c*x^2 + b*x)^(3/2)*(B*d + A*e)/c

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mupad [B]  time = 2.44, size = 299, normalized size = 1.94 \[ A\,d\,\sqrt {c\,x^2+b\,x}\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )-\frac {5\,B\,b\,e\,\left (\frac {b^3\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x}\right )}{16\,c^{5/2}}+\frac {\sqrt {c\,x^2+b\,x}\,\left (-3\,b^2+2\,b\,c\,x+8\,c^2\,x^2\right )}{24\,c^2}\right )}{8\,c}-\frac {A\,b^2\,d\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{8\,c^{3/2}}+\frac {A\,b^3\,e\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x}\right )}{16\,c^{5/2}}+\frac {B\,b^3\,d\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x}\right )}{16\,c^{5/2}}+\frac {A\,e\,\sqrt {c\,x^2+b\,x}\,\left (-3\,b^2+2\,b\,c\,x+8\,c^2\,x^2\right )}{24\,c^2}+\frac {B\,d\,\sqrt {c\,x^2+b\,x}\,\left (-3\,b^2+2\,b\,c\,x+8\,c^2\,x^2\right )}{24\,c^2}+\frac {B\,e\,x\,{\left (c\,x^2+b\,x\right )}^{3/2}}{4\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(1/2)*(A + B*x)*(d + e*x),x)

[Out]

A*d*(b*x + c*x^2)^(1/2)*(x/2 + b/(4*c)) - (5*B*b*e*((b^3*log((b + 2*c*x)/c^(1/2) + 2*(b*x + c*x^2)^(1/2)))/(16
*c^(5/2)) + ((b*x + c*x^2)^(1/2)*(8*c^2*x^2 - 3*b^2 + 2*b*c*x))/(24*c^2)))/(8*c) - (A*b^2*d*log((b/2 + c*x)/c^
(1/2) + (b*x + c*x^2)^(1/2)))/(8*c^(3/2)) + (A*b^3*e*log((b + 2*c*x)/c^(1/2) + 2*(b*x + c*x^2)^(1/2)))/(16*c^(
5/2)) + (B*b^3*d*log((b + 2*c*x)/c^(1/2) + 2*(b*x + c*x^2)^(1/2)))/(16*c^(5/2)) + (A*e*(b*x + c*x^2)^(1/2)*(8*
c^2*x^2 - 3*b^2 + 2*b*c*x))/(24*c^2) + (B*d*(b*x + c*x^2)^(1/2)*(8*c^2*x^2 - 3*b^2 + 2*b*c*x))/(24*c^2) + (B*e
*x*(b*x + c*x^2)^(3/2))/(4*c)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {x \left (b + c x\right )} \left (A + B x\right ) \left (d + e x\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)*(c*x**2+b*x)**(1/2),x)

[Out]

Integral(sqrt(x*(b + c*x))*(A + B*x)*(d + e*x), x)

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