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

Optimal. Leaf size=267 \[ \frac {\left (b x+c x^2\right )^{3/2} \left (6 c e x (10 A c e-7 b B e+4 B c d)+10 A c e (16 c d-5 b e)+B \left (35 b^2 e^2-100 b c d e+32 c^2 d^2\right )\right )}{240 c^3}-\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right ) \left (10 b^2 c e (A e+2 B d)-16 b c^2 d (2 A e+B d)+32 A c^3 d^2-7 b^3 B e^2\right )}{128 c^{9/2}}+\frac {(b+2 c x) \sqrt {b x+c x^2} \left (10 b^2 c e (A e+2 B d)-16 b c^2 d (2 A e+B d)+32 A c^3 d^2-7 b^3 B e^2\right )}{128 c^4}+\frac {B \left (b x+c x^2\right )^{3/2} (d+e x)^2}{5 c} \]

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Rubi [A]  time = 0.27, antiderivative size = 267, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {832, 779, 612, 620, 206} \begin {gather*} \frac {\left (b x+c x^2\right )^{3/2} \left (6 c e x (10 A c e-7 b B e+4 B c d)+10 A c e (16 c d-5 b e)+B \left (35 b^2 e^2-100 b c d e+32 c^2 d^2\right )\right )}{240 c^3}+\frac {(b+2 c x) \sqrt {b x+c x^2} \left (10 b^2 c e (A e+2 B d)-16 b c^2 d (2 A e+B d)+32 A c^3 d^2-7 b^3 B e^2\right )}{128 c^4}-\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right ) \left (10 b^2 c e (A e+2 B d)-16 b c^2 d (2 A e+B d)+32 A c^3 d^2-7 b^3 B e^2\right )}{128 c^{9/2}}+\frac {B \left (b x+c x^2\right )^{3/2} (d+e x)^2}{5 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((32*A*c^3*d^2 - 7*b^3*B*e^2 + 10*b^2*c*e*(2*B*d + A*e) - 16*b*c^2*d*(B*d + 2*A*e))*(b + 2*c*x)*Sqrt[b*x + c*x
^2])/(128*c^4) + (B*(d + e*x)^2*(b*x + c*x^2)^(3/2))/(5*c) + ((10*A*c*e*(16*c*d - 5*b*e) + B*(32*c^2*d^2 - 100
*b*c*d*e + 35*b^2*e^2) + 6*c*e*(4*B*c*d - 7*b*B*e + 10*A*c*e)*x)*(b*x + c*x^2)^(3/2))/(240*c^3) - (b^2*(32*A*c
^3*d^2 - 7*b^3*B*e^2 + 10*b^2*c*e*(2*B*d + A*e) - 16*b*c^2*d*(B*d + 2*A*e))*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x
^2]])/(128*c^(9/2))

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]

Rule 832

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

Rubi steps

\begin {align*} \int (A+B x) (d+e x)^2 \sqrt {b x+c x^2} \, dx &=\frac {B (d+e x)^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac {\int (d+e x) \left (-\frac {1}{2} (3 b B-10 A c) d+\frac {1}{2} (4 B c d-7 b B e+10 A c e) x\right ) \sqrt {b x+c x^2} \, dx}{5 c}\\ &=\frac {B (d+e x)^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac {\left (10 A c e (16 c d-5 b e)+B \left (32 c^2 d^2-100 b c d e+35 b^2 e^2\right )+6 c e (4 B c d-7 b B e+10 A c e) x\right ) \left (b x+c x^2\right )^{3/2}}{240 c^3}+\frac {\left (32 A c^3 d^2-7 b^3 B e^2+10 b^2 c e (2 B d+A e)-16 b c^2 d (B d+2 A e)\right ) \int \sqrt {b x+c x^2} \, dx}{32 c^3}\\ &=\frac {\left (32 A c^3 d^2-7 b^3 B e^2+10 b^2 c e (2 B d+A e)-16 b c^2 d (B d+2 A e)\right ) (b+2 c x) \sqrt {b x+c x^2}}{128 c^4}+\frac {B (d+e x)^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac {\left (10 A c e (16 c d-5 b e)+B \left (32 c^2 d^2-100 b c d e+35 b^2 e^2\right )+6 c e (4 B c d-7 b B e+10 A c e) x\right ) \left (b x+c x^2\right )^{3/2}}{240 c^3}-\frac {\left (b^2 \left (32 A c^3 d^2-7 b^3 B e^2+10 b^2 c e (2 B d+A e)-16 b c^2 d (B d+2 A e)\right )\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{256 c^4}\\ &=\frac {\left (32 A c^3 d^2-7 b^3 B e^2+10 b^2 c e (2 B d+A e)-16 b c^2 d (B d+2 A e)\right ) (b+2 c x) \sqrt {b x+c x^2}}{128 c^4}+\frac {B (d+e x)^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac {\left (10 A c e (16 c d-5 b e)+B \left (32 c^2 d^2-100 b c d e+35 b^2 e^2\right )+6 c e (4 B c d-7 b B e+10 A c e) x\right ) \left (b x+c x^2\right )^{3/2}}{240 c^3}-\frac {\left (b^2 \left (32 A c^3 d^2-7 b^3 B e^2+10 b^2 c e (2 B d+A e)-16 b c^2 d (B d+2 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 )}{128 c^4}\\ &=\frac {\left (32 A c^3 d^2-7 b^3 B e^2+10 b^2 c e (2 B d+A e)-16 b c^2 d (B d+2 A e)\right ) (b+2 c x) \sqrt {b x+c x^2}}{128 c^4}+\frac {B (d+e x)^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac {\left (10 A c e (16 c d-5 b e)+B \left (32 c^2 d^2-100 b c d e+35 b^2 e^2\right )+6 c e (4 B c d-7 b B e+10 A c e) x\right ) \left (b x+c x^2\right )^{3/2}}{240 c^3}-\frac {b^2 \left (32 A c^3 d^2-7 b^3 B e^2+10 b^2 c e (2 B d+A e)-16 b c^2 d (B d+2 A e)\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.58, size = 293, normalized size = 1.10 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (\frac {15 b^{3/2} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right ) \left (-10 b^2 c e (A e+2 B d)+16 b c^2 d (2 A e+B d)-32 A c^3 d^2+7 b^3 B e^2\right )}{\sqrt {x} \sqrt {\frac {c x}{b}+1}}+\sqrt {c} \left (10 b^3 c e (15 A e+30 B d+7 B e x)-4 b^2 c^2 \left (5 A e (24 d+5 e x)+2 B \left (30 d^2+25 d e x+7 e^2 x^2\right )\right )+16 b c^3 \left (5 A \left (6 d^2+4 d e x+e^2 x^2\right )+B x \left (10 d^2+10 d e x+3 e^2 x^2\right )\right )+32 c^4 x \left (5 A \left (6 d^2+8 d e x+3 e^2 x^2\right )+2 B x \left (10 d^2+15 d e x+6 e^2 x^2\right )\right )-105 b^4 B e^2\right )\right )}{1920 c^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(-105*b^4*B*e^2 + 10*b^3*c*e*(30*B*d + 15*A*e + 7*B*e*x) + 16*b*c^3*(5*A*(6*d^2 +
4*d*e*x + e^2*x^2) + B*x*(10*d^2 + 10*d*e*x + 3*e^2*x^2)) + 32*c^4*x*(5*A*(6*d^2 + 8*d*e*x + 3*e^2*x^2) + 2*B*
x*(10*d^2 + 15*d*e*x + 6*e^2*x^2)) - 4*b^2*c^2*(5*A*e*(24*d + 5*e*x) + 2*B*(30*d^2 + 25*d*e*x + 7*e^2*x^2))) +
(15*b^(3/2)*(-32*A*c^3*d^2 + 7*b^3*B*e^2 - 10*b^2*c*e*(2*B*d + A*e) + 16*b*c^2*d*(B*d + 2*A*e))*ArcSinh[(Sqrt
[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(1920*c^(9/2))

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IntegrateAlgebraic [A]  time = 1.42, size = 361, normalized size = 1.35 \begin {gather*} \frac {\sqrt {b x+c x^2} \left (150 A b^3 c e^2-480 A b^2 c^2 d e-100 A b^2 c^2 e^2 x+480 A b c^3 d^2+320 A b c^3 d e x+80 A b c^3 e^2 x^2+960 A c^4 d^2 x+1280 A c^4 d e x^2+480 A c^4 e^2 x^3-105 b^4 B e^2+300 b^3 B c d e+70 b^3 B c e^2 x-240 b^2 B c^2 d^2-200 b^2 B c^2 d e x-56 b^2 B c^2 e^2 x^2+160 b B c^3 d^2 x+160 b B c^3 d e x^2+48 b B c^3 e^2 x^3+640 B c^4 d^2 x^2+960 B c^4 d e x^3+384 B c^4 e^2 x^4\right )}{1920 c^4}+\frac {\log \left (-2 \sqrt {c} \sqrt {b x+c x^2}+b+2 c x\right ) \left (10 A b^4 c e^2-32 A b^3 c^2 d e+32 A b^2 c^3 d^2-7 b^5 B e^2+20 b^4 B c d e-16 b^3 B c^2 d^2\right )}{256 c^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x + c*x^2]*(-240*b^2*B*c^2*d^2 + 480*A*b*c^3*d^2 + 300*b^3*B*c*d*e - 480*A*b^2*c^2*d*e - 105*b^4*B*e^2
+ 150*A*b^3*c*e^2 + 160*b*B*c^3*d^2*x + 960*A*c^4*d^2*x - 200*b^2*B*c^2*d*e*x + 320*A*b*c^3*d*e*x + 70*b^3*B*
c*e^2*x - 100*A*b^2*c^2*e^2*x + 640*B*c^4*d^2*x^2 + 160*b*B*c^3*d*e*x^2 + 1280*A*c^4*d*e*x^2 - 56*b^2*B*c^2*e^
2*x^2 + 80*A*b*c^3*e^2*x^2 + 960*B*c^4*d*e*x^3 + 48*b*B*c^3*e^2*x^3 + 480*A*c^4*e^2*x^3 + 384*B*c^4*e^2*x^4))/
(1920*c^4) + ((-16*b^3*B*c^2*d^2 + 32*A*b^2*c^3*d^2 + 20*b^4*B*c*d*e - 32*A*b^3*c^2*d*e - 7*b^5*B*e^2 + 10*A*b
^4*c*e^2)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[b*x + c*x^2]])/(256*c^(9/2))

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fricas [A]  time = 0.47, size = 684, normalized size = 2.56 \begin {gather*} \left [-\frac {15 \, {\left (16 \, {\left (B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} d^{2} - 4 \, {\left (5 \, B b^{4} c - 8 \, A b^{3} c^{2}\right )} d e + {\left (7 \, B b^{5} - 10 \, A b^{4} c\right )} e^{2}\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (384 \, B c^{5} e^{2} x^{4} + 48 \, {\left (20 \, B c^{5} d e + {\left (B b c^{4} + 10 \, A c^{5}\right )} e^{2}\right )} x^{3} - 240 \, {\left (B b^{2} c^{3} - 2 \, A b c^{4}\right )} d^{2} + 60 \, {\left (5 \, B b^{3} c^{2} - 8 \, A b^{2} c^{3}\right )} d e - 15 \, {\left (7 \, B b^{4} c - 10 \, A b^{3} c^{2}\right )} e^{2} + 8 \, {\left (80 \, B c^{5} d^{2} + 20 \, {\left (B b c^{4} + 8 \, A c^{5}\right )} d e - {\left (7 \, B b^{2} c^{3} - 10 \, A b c^{4}\right )} e^{2}\right )} x^{2} + 10 \, {\left (16 \, {\left (B b c^{4} + 6 \, A c^{5}\right )} d^{2} - 4 \, {\left (5 \, B b^{2} c^{3} - 8 \, A b c^{4}\right )} d e + {\left (7 \, B b^{3} c^{2} - 10 \, A b^{2} c^{3}\right )} e^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{3840 \, c^{5}}, -\frac {15 \, {\left (16 \, {\left (B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} d^{2} - 4 \, {\left (5 \, B b^{4} c - 8 \, A b^{3} c^{2}\right )} d e + {\left (7 \, B b^{5} - 10 \, A b^{4} c\right )} e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (384 \, B c^{5} e^{2} x^{4} + 48 \, {\left (20 \, B c^{5} d e + {\left (B b c^{4} + 10 \, A c^{5}\right )} e^{2}\right )} x^{3} - 240 \, {\left (B b^{2} c^{3} - 2 \, A b c^{4}\right )} d^{2} + 60 \, {\left (5 \, B b^{3} c^{2} - 8 \, A b^{2} c^{3}\right )} d e - 15 \, {\left (7 \, B b^{4} c - 10 \, A b^{3} c^{2}\right )} e^{2} + 8 \, {\left (80 \, B c^{5} d^{2} + 20 \, {\left (B b c^{4} + 8 \, A c^{5}\right )} d e - {\left (7 \, B b^{2} c^{3} - 10 \, A b c^{4}\right )} e^{2}\right )} x^{2} + 10 \, {\left (16 \, {\left (B b c^{4} + 6 \, A c^{5}\right )} d^{2} - 4 \, {\left (5 \, B b^{2} c^{3} - 8 \, A b c^{4}\right )} d e + {\left (7 \, B b^{3} c^{2} - 10 \, A b^{2} c^{3}\right )} e^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{1920 \, c^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/3840*(15*(16*(B*b^3*c^2 - 2*A*b^2*c^3)*d^2 - 4*(5*B*b^4*c - 8*A*b^3*c^2)*d*e + (7*B*b^5 - 10*A*b^4*c)*e^2)
*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(384*B*c^5*e^2*x^4 + 48*(20*B*c^5*d*e + (B*b*c^4 + 1
0*A*c^5)*e^2)*x^3 - 240*(B*b^2*c^3 - 2*A*b*c^4)*d^2 + 60*(5*B*b^3*c^2 - 8*A*b^2*c^3)*d*e - 15*(7*B*b^4*c - 10*
A*b^3*c^2)*e^2 + 8*(80*B*c^5*d^2 + 20*(B*b*c^4 + 8*A*c^5)*d*e - (7*B*b^2*c^3 - 10*A*b*c^4)*e^2)*x^2 + 10*(16*(
B*b*c^4 + 6*A*c^5)*d^2 - 4*(5*B*b^2*c^3 - 8*A*b*c^4)*d*e + (7*B*b^3*c^2 - 10*A*b^2*c^3)*e^2)*x)*sqrt(c*x^2 + b
*x))/c^5, -1/1920*(15*(16*(B*b^3*c^2 - 2*A*b^2*c^3)*d^2 - 4*(5*B*b^4*c - 8*A*b^3*c^2)*d*e + (7*B*b^5 - 10*A*b^
4*c)*e^2)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (384*B*c^5*e^2*x^4 + 48*(20*B*c^5*d*e + (B*b*c^4
+ 10*A*c^5)*e^2)*x^3 - 240*(B*b^2*c^3 - 2*A*b*c^4)*d^2 + 60*(5*B*b^3*c^2 - 8*A*b^2*c^3)*d*e - 15*(7*B*b^4*c -
10*A*b^3*c^2)*e^2 + 8*(80*B*c^5*d^2 + 20*(B*b*c^4 + 8*A*c^5)*d*e - (7*B*b^2*c^3 - 10*A*b*c^4)*e^2)*x^2 + 10*(
16*(B*b*c^4 + 6*A*c^5)*d^2 - 4*(5*B*b^2*c^3 - 8*A*b*c^4)*d*e + (7*B*b^3*c^2 - 10*A*b^2*c^3)*e^2)*x)*sqrt(c*x^2
+ b*x))/c^5]

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giac [A]  time = 0.22, size = 349, normalized size = 1.31 \begin {gather*} \frac {1}{1920} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, B x e^{2} + \frac {20 \, B c^{4} d e + B b c^{3} e^{2} + 10 \, A c^{4} e^{2}}{c^{4}}\right )} x + \frac {80 \, B c^{4} d^{2} + 20 \, B b c^{3} d e + 160 \, A c^{4} d e - 7 \, B b^{2} c^{2} e^{2} + 10 \, A b c^{3} e^{2}}{c^{4}}\right )} x + \frac {5 \, {\left (16 \, B b c^{3} d^{2} + 96 \, A c^{4} d^{2} - 20 \, B b^{2} c^{2} d e + 32 \, A b c^{3} d e + 7 \, B b^{3} c e^{2} - 10 \, A b^{2} c^{2} e^{2}\right )}}{c^{4}}\right )} x - \frac {15 \, {\left (16 \, B b^{2} c^{2} d^{2} - 32 \, A b c^{3} d^{2} - 20 \, B b^{3} c d e + 32 \, A b^{2} c^{2} d e + 7 \, B b^{4} e^{2} - 10 \, A b^{3} c e^{2}\right )}}{c^{4}}\right )} - \frac {{\left (16 \, B b^{3} c^{2} d^{2} - 32 \, A b^{2} c^{3} d^{2} - 20 \, B b^{4} c d e + 32 \, A b^{3} c^{2} d e + 7 \, B b^{5} e^{2} - 10 \, A b^{4} c e^{2}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{256 \, c^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1920*sqrt(c*x^2 + b*x)*(2*(4*(6*(8*B*x*e^2 + (20*B*c^4*d*e + B*b*c^3*e^2 + 10*A*c^4*e^2)/c^4)*x + (80*B*c^4*
d^2 + 20*B*b*c^3*d*e + 160*A*c^4*d*e - 7*B*b^2*c^2*e^2 + 10*A*b*c^3*e^2)/c^4)*x + 5*(16*B*b*c^3*d^2 + 96*A*c^4
*d^2 - 20*B*b^2*c^2*d*e + 32*A*b*c^3*d*e + 7*B*b^3*c*e^2 - 10*A*b^2*c^2*e^2)/c^4)*x - 15*(16*B*b^2*c^2*d^2 - 3
2*A*b*c^3*d^2 - 20*B*b^3*c*d*e + 32*A*b^2*c^2*d*e + 7*B*b^4*e^2 - 10*A*b^3*c*e^2)/c^4) - 1/256*(16*B*b^3*c^2*d
^2 - 32*A*b^2*c^3*d^2 - 20*B*b^4*c*d*e + 32*A*b^3*c^2*d*e + 7*B*b^5*e^2 - 10*A*b^4*c*e^2)*log(abs(-2*(sqrt(c)*
x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(9/2)

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maple [B]  time = 0.06, size = 671, normalized size = 2.51 \begin {gather*} -\frac {5 A \,b^{4} e^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{128 c^{\frac {7}{2}}}+\frac {A \,b^{3} d e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {5}{2}}}-\frac {A \,b^{2} d^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}+\frac {7 B \,b^{5} e^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{256 c^{\frac {9}{2}}}-\frac {5 B \,b^{4} d e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{64 c^{\frac {7}{2}}}+\frac {B \,b^{3} d^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{16 c^{\frac {5}{2}}}+\frac {5 \sqrt {c \,x^{2}+b x}\, A \,b^{2} e^{2} x}{32 c^{2}}-\frac {\sqrt {c \,x^{2}+b x}\, A b d e x}{2 c}+\frac {\sqrt {c \,x^{2}+b x}\, A \,d^{2} x}{2}-\frac {7 \sqrt {c \,x^{2}+b x}\, B \,b^{3} e^{2} x}{64 c^{3}}+\frac {5 \sqrt {c \,x^{2}+b x}\, B \,b^{2} d e x}{16 c^{2}}-\frac {\sqrt {c \,x^{2}+b x}\, B b \,d^{2} x}{4 c}+\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} B \,e^{2} x^{2}}{5 c}+\frac {5 \sqrt {c \,x^{2}+b x}\, A \,b^{3} e^{2}}{64 c^{3}}-\frac {\sqrt {c \,x^{2}+b x}\, A \,b^{2} d e}{4 c^{2}}+\frac {\sqrt {c \,x^{2}+b x}\, A b \,d^{2}}{4 c}+\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} A \,e^{2} x}{4 c}-\frac {7 \sqrt {c \,x^{2}+b x}\, B \,b^{4} e^{2}}{128 c^{4}}+\frac {5 \sqrt {c \,x^{2}+b x}\, B \,b^{3} d e}{32 c^{3}}-\frac {\sqrt {c \,x^{2}+b x}\, B \,b^{2} d^{2}}{8 c^{2}}-\frac {7 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} B b \,e^{2} x}{40 c^{2}}+\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} B d e x}{2 c}-\frac {5 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} A b \,e^{2}}{24 c^{2}}+\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} A d e}{3 c}+\frac {7 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} B \,b^{2} e^{2}}{48 c^{3}}-\frac {5 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} B b d e}{12 c^{2}}+\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} B \,d^{2}}{3 c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/24*b/c^2*(c*x^2+b*x)^(3/2)*A*e^2+5/64*b^3/c^3*(c*x^2+b*x)^(1/2)*A*e^2-5/128*b^4/c^(7/2)*ln((c*x+1/2*b)/c^(1
/2)+(c*x^2+b*x)^(1/2))*A*e^2+7/256*B*e^2*b^5/c^(9/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))+7/48*B*e^2*b^2/
c^3*(c*x^2+b*x)^(3/2)-7/128*B*e^2*b^4/c^4*(c*x^2+b*x)^(1/2)+1/5*B*e^2*x^2*(c*x^2+b*x)^(3/2)/c-7/40*B*e^2*b/c^2
*x*(c*x^2+b*x)^(3/2)-7/64*B*e^2*b^3/c^3*x*(c*x^2+b*x)^(1/2)+5/16*b^2/c^2*x*(c*x^2+b*x)^(1/2)*B*d*e-1/2*b/c*x*(
c*x^2+b*x)^(1/2)*A*d*e-5/12*b/c^2*(c*x^2+b*x)^(3/2)*B*d*e+5/32*b^2/c^2*x*(c*x^2+b*x)^(1/2)*A*e^2+5/32*b^3/c^3*
(c*x^2+b*x)^(1/2)*B*d*e+1/2*x*(c*x^2+b*x)^(3/2)/c*B*d*e-5/64*b^4/c^(7/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1
/2))*B*d*e+1/8*b^3/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))*A*d*e-1/4*b^2/c^2*(c*x^2+b*x)^(1/2)*A*d*e
-1/4*b/c*x*(c*x^2+b*x)^(1/2)*B*d^2+1/4*x*(c*x^2+b*x)^(3/2)/c*A*e^2+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^2-1/8*b^2/c^2*(c*x^2+b*x)^(1/2)*B*d^2+2/3*(c*x^2+b*x)^(3/2)/c*A*d*e+1/4*A*d^2/c*(c*x^2+b*x
)^(1/2)*b-1/8*A*d^2*b^2/c^(3/2)*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*B*d^2+1/2*A*
d^2*x*(c*x^2+b*x)^(1/2)

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maxima [B]  time = 0.61, size = 512, normalized size = 1.92 \begin {gather*} \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B e^{2} x^{2}}{5 \, c} + \frac {1}{2} \, \sqrt {c x^{2} + b x} A d^{2} x - \frac {7 \, \sqrt {c x^{2} + b x} B b^{3} e^{2} x}{64 \, c^{3}} - \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b e^{2} x}{40 \, c^{2}} - \frac {A b^{2} d^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {3}{2}}} + \frac {7 \, B b^{5} e^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{256 \, c^{\frac {9}{2}}} + \frac {\sqrt {c x^{2} + b x} A b d^{2}}{4 \, c} - \frac {7 \, \sqrt {c x^{2} + b x} B b^{4} e^{2}}{128 \, c^{4}} + \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b^{2} e^{2}}{48 \, c^{3}} + \frac {5 \, {\left (2 \, B d e + A e^{2}\right )} \sqrt {c x^{2} + b x} b^{2} x}{32 \, c^{2}} + \frac {{\left (2 \, B d e + A e^{2}\right )} {\left (c x^{2} + b x\right )}^{\frac {3}{2}} x}{4 \, c} - \frac {{\left (B d^{2} + 2 \, A d e\right )} \sqrt {c x^{2} + b x} b x}{4 \, c} - \frac {5 \, {\left (2 \, B d e + A e^{2}\right )} b^{4} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{128 \, c^{\frac {7}{2}}} + \frac {{\left (B d^{2} + 2 \, A d 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 {5 \, {\left (2 \, B d e + A e^{2}\right )} \sqrt {c x^{2} + b x} b^{3}}{64 \, c^{3}} - \frac {5 \, {\left (2 \, B d e + A e^{2}\right )} {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b}{24 \, c^{2}} - \frac {{\left (B d^{2} + 2 \, A d e\right )} \sqrt {c x^{2} + b x} b^{2}}{8 \, c^{2}} + \frac {{\left (B d^{2} + 2 \, A d e\right )} {\left (c x^{2} + b x\right )}^{\frac {3}{2}}}{3 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(c*x^2 + b*x)^(3/2)*B*e^2*x^2/c + 1/2*sqrt(c*x^2 + b*x)*A*d^2*x - 7/64*sqrt(c*x^2 + b*x)*B*b^3*e^2*x/c^3 -
7/40*(c*x^2 + b*x)^(3/2)*B*b*e^2*x/c^2 - 1/8*A*b^2*d^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(3/2) +
7/256*B*b^5*e^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(9/2) + 1/4*sqrt(c*x^2 + b*x)*A*b*d^2/c - 7/12
8*sqrt(c*x^2 + b*x)*B*b^4*e^2/c^4 + 7/48*(c*x^2 + b*x)^(3/2)*B*b^2*e^2/c^3 + 5/32*(2*B*d*e + A*e^2)*sqrt(c*x^2
+ b*x)*b^2*x/c^2 + 1/4*(2*B*d*e + A*e^2)*(c*x^2 + b*x)^(3/2)*x/c - 1/4*(B*d^2 + 2*A*d*e)*sqrt(c*x^2 + b*x)*b*
x/c - 5/128*(2*B*d*e + A*e^2)*b^4*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(7/2) + 1/16*(B*d^2 + 2*A*d*e
)*b^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(5/2) + 5/64*(2*B*d*e + A*e^2)*sqrt(c*x^2 + b*x)*b^3/c^3
- 5/24*(2*B*d*e + A*e^2)*(c*x^2 + b*x)^(3/2)*b/c^2 - 1/8*(B*d^2 + 2*A*d*e)*sqrt(c*x^2 + b*x)*b^2/c^2 + 1/3*(B*
d^2 + 2*A*d*e)*(c*x^2 + b*x)^(3/2)/c

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mupad [B]  time = 3.03, size = 537, normalized size = 2.01 \begin {gather*} A\,d^2\,\sqrt {c\,x^2+b\,x}\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )+\frac {A\,e^2\,x\,{\left (c\,x^2+b\,x\right )}^{3/2}}{4\,c}-\frac {5\,A\,b\,e^2\,\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 {7\,B\,b\,e^2\,\left (\frac {x\,{\left (c\,x^2+b\,x\right )}^{3/2}}{4\,c}-\frac {5\,b\,\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}\right )}{10\,c}+\frac {B\,e^2\,x^2\,{\left (c\,x^2+b\,x\right )}^{3/2}}{5\,c}-\frac {A\,b^2\,d^2\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{8\,c^{3/2}}+\frac {B\,b^3\,d^2\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x}\right )}{16\,c^{5/2}}+\frac {B\,d^2\,\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\,e\,x\,{\left (c\,x^2+b\,x\right )}^{3/2}}{2\,c}-\frac {5\,B\,b\,d\,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 )}{4\,c}+\frac {A\,b^3\,d\,e\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x}\right )}{8\,c^{5/2}}+\frac {A\,d\,e\,\sqrt {c\,x^2+b\,x}\,\left (-3\,b^2+2\,b\,c\,x+8\,c^2\,x^2\right )}{12\,c^2} \end {gather*}

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)^2,x)

[Out]

A*d^2*(b*x + c*x^2)^(1/2)*(x/2 + b/(4*c)) + (A*e^2*x*(b*x + c*x^2)^(3/2))/(4*c) - (5*A*b*e^2*((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) - (7*B*b*e^2*((x*(b*x + c*x^2)^(3/2))/(4*c) - (5*b*((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)))/(10*c) + (B*e^
2*x^2*(b*x + c*x^2)^(3/2))/(5*c) - (A*b^2*d^2*log((b/2 + c*x)/c^(1/2) + (b*x + c*x^2)^(1/2)))/(8*c^(3/2)) + (B
*b^3*d^2*log((b + 2*c*x)/c^(1/2) + 2*(b*x + c*x^2)^(1/2)))/(16*c^(5/2)) + (B*d^2*(b*x + c*x^2)^(1/2)*(8*c^2*x^
2 - 3*b^2 + 2*b*c*x))/(24*c^2) + (B*d*e*x*(b*x + c*x^2)^(3/2))/(2*c) - (5*B*b*d*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)))/(4*c
) + (A*b^3*d*e*log((b + 2*c*x)/c^(1/2) + 2*(b*x + c*x^2)^(1/2)))/(8*c^(5/2)) + (A*d*e*(b*x + c*x^2)^(1/2)*(8*c
^2*x^2 - 3*b^2 + 2*b*c*x))/(12*c^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(x*(b + c*x))*(A + B*x)*(d + e*x)**2, x)

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