3.11.52 \(\int \frac {A+B x}{(d+e x)^3 \sqrt {b x+c x^2}} \, dx\)

Optimal. Leaf size=216 \[ \frac {\left (b^2 e (3 A e+B d)-4 b c d (2 A e+B d)+8 A c^2 d^2\right ) \tanh ^{-1}\left (\frac {x (2 c d-b e)+b d}{2 \sqrt {d} \sqrt {b x+c x^2} \sqrt {c d-b e}}\right )}{8 d^{5/2} (c d-b e)^{5/2}}-\frac {\sqrt {b x+c x^2} (3 A e (2 c d-b e)-B d (b e+2 c d))}{4 d^2 (d+e x) (c d-b e)^2}+\frac {\sqrt {b x+c x^2} (B d-A e)}{2 d (d+e x)^2 (c d-b e)} \]

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Rubi [A]  time = 0.26, antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {834, 806, 724, 206} \begin {gather*} \frac {\left (b^2 e (3 A e+B d)-4 b c d (2 A e+B d)+8 A c^2 d^2\right ) \tanh ^{-1}\left (\frac {x (2 c d-b e)+b d}{2 \sqrt {d} \sqrt {b x+c x^2} \sqrt {c d-b e}}\right )}{8 d^{5/2} (c d-b e)^{5/2}}-\frac {\sqrt {b x+c x^2} (3 A e (2 c d-b e)-B d (b e+2 c d))}{4 d^2 (d+e x) (c d-b e)^2}+\frac {\sqrt {b x+c x^2} (B d-A e)}{2 d (d+e x)^2 (c d-b e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((B*d - A*e)*Sqrt[b*x + c*x^2])/(2*d*(c*d - b*e)*(d + e*x)^2) - ((3*A*e*(2*c*d - b*e) - B*d*(2*c*d + b*e))*Sqr
t[b*x + c*x^2])/(4*d^2*(c*d - b*e)^2*(d + e*x)) + ((8*A*c^2*d^2 - 4*b*c*d*(B*d + 2*A*e) + b^2*e*(B*d + 3*A*e))
*ArcTanh[(b*d + (2*c*d - b*e)*x)/(2*Sqrt[d]*Sqrt[c*d - b*e]*Sqrt[b*x + c*x^2])])/(8*d^(5/2)*(c*d - b*e)^(5/2))

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 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2)), x] - Dist[(b
*(e*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(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] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[Sim
plify[m + 2*p + 3], 0]

Rule 834

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((m
+ 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*Simp[(c*d*f - f*b*e + a*e*g)*(m + 1)
+ b*(d*g - e*f)*(p + 1) - c*(e*f - d*g)*(m + 2*p + 3)*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] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p])

Rubi steps

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

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Mathematica [A]  time = 0.39, size = 217, normalized size = 1.00 \begin {gather*} \frac {\sqrt {x} \left (-\frac {\sqrt {b+c x} \left (b^2 e (3 A e+B d)-4 b c d (2 A e+B d)+8 A c^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {x} \sqrt {c d-b e}}{\sqrt {d} \sqrt {b+c x}}\right )}{2 d^{3/2} (c d-b e)^{3/2}}-\frac {\sqrt {x} (b+c x) (3 A e (b e-2 c d)+B d (b e+2 c d))}{2 d (d+e x) (c d-b e)}+\frac {\sqrt {x} (b+c x) (A e-B d)}{(d+e x)^2}\right )}{2 d \sqrt {x (b+c x)} (b e-c d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x]*(((-(B*d) + A*e)*Sqrt[x]*(b + c*x))/(d + e*x)^2 - ((3*A*e*(-2*c*d + b*e) + B*d*(2*c*d + b*e))*Sqrt[x]
*(b + c*x))/(2*d*(c*d - b*e)*(d + e*x)) - ((8*A*c^2*d^2 - 4*b*c*d*(B*d + 2*A*e) + b^2*e*(B*d + 3*A*e))*Sqrt[b
+ c*x]*ArcTanh[(Sqrt[c*d - b*e]*Sqrt[x])/(Sqrt[d]*Sqrt[b + c*x])])/(2*d^(3/2)*(c*d - b*e)^(3/2))))/(2*d*(-(c*d
) + b*e)*Sqrt[x*(b + c*x)])

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IntegrateAlgebraic [A]  time = 2.03, size = 217, normalized size = 1.00 \begin {gather*} \frac {\left (3 A b^2 e^2-8 A b c d e+8 A c^2 d^2+b^2 B d e-4 b B c d^2\right ) \tanh ^{-1}\left (\frac {-e \sqrt {b x+c x^2}+\sqrt {c} d+\sqrt {c} e x}{\sqrt {d} \sqrt {c d-b e}}\right )}{4 d^{5/2} (c d-b e)^{5/2}}+\frac {\sqrt {b x+c x^2} \left (5 A b d e^2+3 A b e^3 x-8 A c d^2 e-6 A c d e^2 x-b B d^2 e+b B d e^2 x+4 B c d^3+2 B c d^2 e x\right )}{4 d^2 (d+e x)^2 (c d-b e)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((4*B*c*d^3 - b*B*d^2*e - 8*A*c*d^2*e + 5*A*b*d*e^2 + 2*B*c*d^2*e*x + b*B*d*e^2*x - 6*A*c*d*e^2*x + 3*A*b*e^3*
x)*Sqrt[b*x + c*x^2])/(4*d^2*(c*d - b*e)^2*(d + e*x)^2) + ((-4*b*B*c*d^2 + 8*A*c^2*d^2 + b^2*B*d*e - 8*A*b*c*d
*e + 3*A*b^2*e^2)*ArcTanh[(Sqrt[c]*d + Sqrt[c]*e*x - e*Sqrt[b*x + c*x^2])/(Sqrt[d]*Sqrt[c*d - b*e])])/(4*d^(5/
2)*(c*d - b*e)^(5/2))

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fricas [B]  time = 0.46, size = 954, normalized size = 4.42 \begin {gather*} \left [\frac {{\left (3 \, A b^{2} d^{2} e^{2} - 4 \, {\left (B b c - 2 \, A c^{2}\right )} d^{4} + {\left (B b^{2} - 8 \, A b c\right )} d^{3} e + {\left (3 \, A b^{2} e^{4} - 4 \, {\left (B b c - 2 \, A c^{2}\right )} d^{2} e^{2} + {\left (B b^{2} - 8 \, A b c\right )} d e^{3}\right )} x^{2} + 2 \, {\left (3 \, A b^{2} d e^{3} - 4 \, {\left (B b c - 2 \, A c^{2}\right )} d^{3} e + {\left (B b^{2} - 8 \, A b c\right )} d^{2} e^{2}\right )} x\right )} \sqrt {c d^{2} - b d e} \log \left (\frac {b d + {\left (2 \, c d - b e\right )} x + 2 \, \sqrt {c d^{2} - b d e} \sqrt {c x^{2} + b x}}{e x + d}\right ) + 2 \, {\left (4 \, B c^{2} d^{5} - 5 \, A b^{2} d^{2} e^{3} - {\left (5 \, B b c + 8 \, A c^{2}\right )} d^{4} e + {\left (B b^{2} + 13 \, A b c\right )} d^{3} e^{2} + {\left (2 \, B c^{2} d^{4} e - 3 \, A b^{2} d e^{4} - {\left (B b c + 6 \, A c^{2}\right )} d^{3} e^{2} - {\left (B b^{2} - 9 \, A b c\right )} d^{2} e^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{8 \, {\left (c^{3} d^{8} - 3 \, b c^{2} d^{7} e + 3 \, b^{2} c d^{6} e^{2} - b^{3} d^{5} e^{3} + {\left (c^{3} d^{6} e^{2} - 3 \, b c^{2} d^{5} e^{3} + 3 \, b^{2} c d^{4} e^{4} - b^{3} d^{3} e^{5}\right )} x^{2} + 2 \, {\left (c^{3} d^{7} e - 3 \, b c^{2} d^{6} e^{2} + 3 \, b^{2} c d^{5} e^{3} - b^{3} d^{4} e^{4}\right )} x\right )}}, \frac {{\left (3 \, A b^{2} d^{2} e^{2} - 4 \, {\left (B b c - 2 \, A c^{2}\right )} d^{4} + {\left (B b^{2} - 8 \, A b c\right )} d^{3} e + {\left (3 \, A b^{2} e^{4} - 4 \, {\left (B b c - 2 \, A c^{2}\right )} d^{2} e^{2} + {\left (B b^{2} - 8 \, A b c\right )} d e^{3}\right )} x^{2} + 2 \, {\left (3 \, A b^{2} d e^{3} - 4 \, {\left (B b c - 2 \, A c^{2}\right )} d^{3} e + {\left (B b^{2} - 8 \, A b c\right )} d^{2} e^{2}\right )} x\right )} \sqrt {-c d^{2} + b d e} \arctan \left (-\frac {\sqrt {-c d^{2} + b d e} \sqrt {c x^{2} + b x}}{{\left (c d - b e\right )} x}\right ) + {\left (4 \, B c^{2} d^{5} - 5 \, A b^{2} d^{2} e^{3} - {\left (5 \, B b c + 8 \, A c^{2}\right )} d^{4} e + {\left (B b^{2} + 13 \, A b c\right )} d^{3} e^{2} + {\left (2 \, B c^{2} d^{4} e - 3 \, A b^{2} d e^{4} - {\left (B b c + 6 \, A c^{2}\right )} d^{3} e^{2} - {\left (B b^{2} - 9 \, A b c\right )} d^{2} e^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{4 \, {\left (c^{3} d^{8} - 3 \, b c^{2} d^{7} e + 3 \, b^{2} c d^{6} e^{2} - b^{3} d^{5} e^{3} + {\left (c^{3} d^{6} e^{2} - 3 \, b c^{2} d^{5} e^{3} + 3 \, b^{2} c d^{4} e^{4} - b^{3} d^{3} e^{5}\right )} x^{2} + 2 \, {\left (c^{3} d^{7} e - 3 \, b c^{2} d^{6} e^{2} + 3 \, b^{2} c d^{5} e^{3} - b^{3} d^{4} e^{4}\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)^3/(c*x^2+b*x)^(1/2),x, algorithm="fricas")

[Out]

[1/8*((3*A*b^2*d^2*e^2 - 4*(B*b*c - 2*A*c^2)*d^4 + (B*b^2 - 8*A*b*c)*d^3*e + (3*A*b^2*e^4 - 4*(B*b*c - 2*A*c^2
)*d^2*e^2 + (B*b^2 - 8*A*b*c)*d*e^3)*x^2 + 2*(3*A*b^2*d*e^3 - 4*(B*b*c - 2*A*c^2)*d^3*e + (B*b^2 - 8*A*b*c)*d^
2*e^2)*x)*sqrt(c*d^2 - b*d*e)*log((b*d + (2*c*d - b*e)*x + 2*sqrt(c*d^2 - b*d*e)*sqrt(c*x^2 + b*x))/(e*x + d))
+ 2*(4*B*c^2*d^5 - 5*A*b^2*d^2*e^3 - (5*B*b*c + 8*A*c^2)*d^4*e + (B*b^2 + 13*A*b*c)*d^3*e^2 + (2*B*c^2*d^4*e
- 3*A*b^2*d*e^4 - (B*b*c + 6*A*c^2)*d^3*e^2 - (B*b^2 - 9*A*b*c)*d^2*e^3)*x)*sqrt(c*x^2 + b*x))/(c^3*d^8 - 3*b*
c^2*d^7*e + 3*b^2*c*d^6*e^2 - b^3*d^5*e^3 + (c^3*d^6*e^2 - 3*b*c^2*d^5*e^3 + 3*b^2*c*d^4*e^4 - b^3*d^3*e^5)*x^
2 + 2*(c^3*d^7*e - 3*b*c^2*d^6*e^2 + 3*b^2*c*d^5*e^3 - b^3*d^4*e^4)*x), 1/4*((3*A*b^2*d^2*e^2 - 4*(B*b*c - 2*A
*c^2)*d^4 + (B*b^2 - 8*A*b*c)*d^3*e + (3*A*b^2*e^4 - 4*(B*b*c - 2*A*c^2)*d^2*e^2 + (B*b^2 - 8*A*b*c)*d*e^3)*x^
2 + 2*(3*A*b^2*d*e^3 - 4*(B*b*c - 2*A*c^2)*d^3*e + (B*b^2 - 8*A*b*c)*d^2*e^2)*x)*sqrt(-c*d^2 + b*d*e)*arctan(-
sqrt(-c*d^2 + b*d*e)*sqrt(c*x^2 + b*x)/((c*d - b*e)*x)) + (4*B*c^2*d^5 - 5*A*b^2*d^2*e^3 - (5*B*b*c + 8*A*c^2)
*d^4*e + (B*b^2 + 13*A*b*c)*d^3*e^2 + (2*B*c^2*d^4*e - 3*A*b^2*d*e^4 - (B*b*c + 6*A*c^2)*d^3*e^2 - (B*b^2 - 9*
A*b*c)*d^2*e^3)*x)*sqrt(c*x^2 + b*x))/(c^3*d^8 - 3*b*c^2*d^7*e + 3*b^2*c*d^6*e^2 - b^3*d^5*e^3 + (c^3*d^6*e^2
- 3*b*c^2*d^5*e^3 + 3*b^2*c*d^4*e^4 - b^3*d^3*e^5)*x^2 + 2*(c^3*d^7*e - 3*b*c^2*d^6*e^2 + 3*b^2*c*d^5*e^3 - b^
3*d^4*e^4)*x)]

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giac [B]  time = 0.31, size = 769, normalized size = 3.56 \begin {gather*} -\frac {{\left (4 \, B b c d^{2} - 8 \, A c^{2} d^{2} - B b^{2} d e + 8 \, A b c d e - 3 \, A b^{2} e^{2}\right )} \arctan \left (-\frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} e + \sqrt {c} d}{\sqrt {-c d^{2} + b d e}}\right )}{4 \, {\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2}\right )} \sqrt {-c d^{2} + b d e}} + \frac {8 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B c^{\frac {5}{2}} d^{4} - 4 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B b c^{\frac {3}{2}} d^{3} e - 24 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} A c^{\frac {5}{2}} d^{3} e + 8 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} B b c^{2} d^{4} + 4 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} B b c d^{2} e^{2} - 8 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} A c^{2} d^{2} e^{2} - 24 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A b c^{2} d^{3} e + 2 \, B b^{2} c^{\frac {3}{2}} d^{4} + 5 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B b^{2} \sqrt {c} d^{2} e^{2} + 24 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} A b c^{\frac {3}{2}} d^{2} e^{2} + B b^{3} \sqrt {c} d^{3} e - 6 \, A b^{2} c^{\frac {3}{2}} d^{3} e - {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} B b^{2} d e^{3} + 8 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} A b c d e^{3} + {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} B b^{3} d^{2} e^{2} + 20 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A b^{2} c d^{2} e^{2} - 9 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} A b^{2} \sqrt {c} d e^{3} + 3 \, A b^{3} \sqrt {c} d^{2} e^{2} - 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} A b^{2} e^{4} - 5 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A b^{3} d e^{3}}{4 \, {\left (c^{2} d^{4} e - 2 \, b c d^{3} e^{2} + b^{2} d^{2} e^{3}\right )} {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} e + 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} d + b d\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)^3/(c*x^2+b*x)^(1/2),x, algorithm="giac")

[Out]

-1/4*(4*B*b*c*d^2 - 8*A*c^2*d^2 - B*b^2*d*e + 8*A*b*c*d*e - 3*A*b^2*e^2)*arctan(-((sqrt(c)*x - sqrt(c*x^2 + b*
x))*e + sqrt(c)*d)/sqrt(-c*d^2 + b*d*e))/((c^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^2)*sqrt(-c*d^2 + b*d*e)) + 1/4*(8
*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*c^(5/2)*d^4 - 4*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b*c^(3/2)*d^3*e - 24*
(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*A*c^(5/2)*d^3*e + 8*(sqrt(c)*x - sqrt(c*x^2 + b*x))*B*b*c^2*d^4 + 4*(sqrt(c)
*x - sqrt(c*x^2 + b*x))^3*B*b*c*d^2*e^2 - 8*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*A*c^2*d^2*e^2 - 24*(sqrt(c)*x -
sqrt(c*x^2 + b*x))*A*b*c^2*d^3*e + 2*B*b^2*c^(3/2)*d^4 + 5*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^2*sqrt(c)*d^2
*e^2 + 24*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*A*b*c^(3/2)*d^2*e^2 + B*b^3*sqrt(c)*d^3*e - 6*A*b^2*c^(3/2)*d^3*e
- (sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b^2*d*e^3 + 8*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*A*b*c*d*e^3 + (sqrt(c)*x
- sqrt(c*x^2 + b*x))*B*b^3*d^2*e^2 + 20*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^2*c*d^2*e^2 - 9*(sqrt(c)*x - sqrt
(c*x^2 + b*x))^2*A*b^2*sqrt(c)*d*e^3 + 3*A*b^3*sqrt(c)*d^2*e^2 - 3*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*A*b^2*e^4
- 5*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^3*d*e^3)/((c^2*d^4*e - 2*b*c*d^3*e^2 + b^2*d^2*e^3)*((sqrt(c)*x - sqr
t(c*x^2 + b*x))^2*e + 2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c)*d + b*d)^2)

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maple [B]  time = 0.06, size = 1821, normalized size = 8.43

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(e*x+d)^3/(c*x^2+b*x)^(1/2),x)

[Out]

B/e/(b*e-c*d)/d/(x+d/e)*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2)-1/2*B/e/(b*e-c*d)/d/(-(b*e-c
*d)*d/e^2)^(1/2)*ln((-2*(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e+2*(-(b*e-c*d)*d/e^2)^(1/2)*((x+d/e)^2*c-(b*e-c*d
)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2))/(x+d/e))*b+3/2*B/e^2/(b*e-c*d)/(-(b*e-c*d)*d/e^2)^(1/2)*ln((-2*(b*e-c*d)
*d/e^2+(b*e-2*c*d)*(x+d/e)/e+2*(-(b*e-c*d)*d/e^2)^(1/2)*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1
/2))/(x+d/e))*c+1/2/e/(b*e-c*d)/d/(x+d/e)^2*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2)*A-1/2/e^
2/(b*e-c*d)/(x+d/e)^2*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2)*B+3/4*e/(b*e-c*d)^2/d^2/(x+d/e
)*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2)*b*A-3/4/(b*e-c*d)^2/d/(x+d/e)*((x+d/e)^2*c-(b*e-c*
d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2)*b*B-3/2/(b*e-c*d)^2/d/(x+d/e)*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(
x+d/e)/e)^(1/2)*c*A+3/2/e/(b*e-c*d)^2/(x+d/e)*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2)*c*B-3/
8*e/(b*e-c*d)^2/d^2/(-(b*e-c*d)*d/e^2)^(1/2)*ln((-2*(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e+2*(-(b*e-c*d)*d/e^2)
^(1/2)*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2))/(x+d/e))*b^2*A+3/8/(b*e-c*d)^2/d/(-(b*e-c*d)
*d/e^2)^(1/2)*ln((-2*(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e+2*(-(b*e-c*d)*d/e^2)^(1/2)*((x+d/e)^2*c-(b*e-c*d)*d
/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2))/(x+d/e))*b^2*B+3/2/(b*e-c*d)^2/d/(-(b*e-c*d)*d/e^2)^(1/2)*ln((-2*(b*e-c*d)*
d/e^2+(b*e-2*c*d)*(x+d/e)/e+2*(-(b*e-c*d)*d/e^2)^(1/2)*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/
2))/(x+d/e))*b*c*A-3/2/e/(b*e-c*d)^2/(-(b*e-c*d)*d/e^2)^(1/2)*ln((-2*(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e+2*(
-(b*e-c*d)*d/e^2)^(1/2)*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2))/(x+d/e))*b*c*B-3/2/e/(b*e-c
*d)^2/(-(b*e-c*d)*d/e^2)^(1/2)*ln((-2*(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e+2*(-(b*e-c*d)*d/e^2)^(1/2)*((x+d/e
)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2))/(x+d/e))*c^2*A+3/2/e^2/(b*e-c*d)^2/(-(b*e-c*d)*d/e^2)^(1/2
)*ln((-2*(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e+2*(-(b*e-c*d)*d/e^2)^(1/2)*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*
c*d)*(x+d/e)/e)^(1/2))/(x+d/e))*c^2*B*d-1/2/e*c/(b*e-c*d)/d/(-(b*e-c*d)*d/e^2)^(1/2)*ln((-2*(b*e-c*d)*d/e^2+(b
*e-2*c*d)*(x+d/e)/e+2*(-(b*e-c*d)*d/e^2)^(1/2)*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2))/(x+d
/e))*A

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)^3/(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b*e-c*d>0)', see `assume?` for
more details)Is b*e-c*d positive, negative or zero?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/((b*x + c*x^2)^(1/2)*(d + e*x)^3),x)

[Out]

int((A + B*x)/((b*x + c*x^2)^(1/2)*(d + e*x)^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)**3/(c*x**2+b*x)**(1/2),x)

[Out]

Integral((A + B*x)/(sqrt(x*(b + c*x))*(d + e*x)**3), x)

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