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

Optimal. Leaf size=128 \[ \frac {\sqrt {b x+c x^2} (B d-A e)}{d (d+e x) (c d-b e)}-\frac {(A b e-2 A c d+b B d) \tanh ^{-1}\left (\frac {x (2 c d-b e)+b d}{2 \sqrt {d} \sqrt {b x+c x^2} \sqrt {c d-b e}}\right )}{2 d^{3/2} (c d-b e)^{3/2}} \]

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Rubi [A]  time = 0.08, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {806, 724, 206} \begin {gather*} \frac {\sqrt {b x+c x^2} (B d-A e)}{d (d+e x) (c d-b e)}-\frac {(A b e-2 A c d+b B d) \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 )}{2 d^{3/2} (c d-b e)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rubi steps

\begin {align*} \int \frac {A+B x}{(d+e x)^2 \sqrt {b x+c x^2}} \, dx &=\frac {(B d-A e) \sqrt {b x+c x^2}}{d (c d-b e) (d+e x)}-\frac {(b B d-2 A c d+A b e) \int \frac {1}{(d+e x) \sqrt {b x+c x^2}} \, dx}{2 d (c d-b e)}\\ &=\frac {(B d-A e) \sqrt {b x+c x^2}}{d (c d-b e) (d+e x)}+\frac {(b B d-2 A c d+A b e) \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 )}{d (c d-b e)}\\ &=\frac {(B d-A e) \sqrt {b x+c x^2}}{d (c d-b e) (d+e x)}-\frac {(b B d-2 A c d+A b e) \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 )}{2 d^{3/2} (c d-b e)^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 133, normalized size = 1.04 \begin {gather*} \frac {\sqrt {x} \left (\frac {\sqrt {d} \sqrt {x} (b+c x) (A e-B d)}{d+e x}+\frac {\sqrt {b+c x} (A b e-2 A c d+b B d) \tanh ^{-1}\left (\frac {\sqrt {x} \sqrt {c d-b e}}{\sqrt {d} \sqrt {b+c x}}\right )}{\sqrt {c d-b e}}\right )}{d^{3/2} \sqrt {x (b+c x)} (b e-c d)} \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]*((Sqrt[d]*(-(B*d) + A*e)*Sqrt[x]*(b + c*x))/(d + e*x) + ((b*B*d - 2*A*c*d + A*b*e)*Sqrt[b + c*x]*ArcT
anh[(Sqrt[c*d - b*e]*Sqrt[x])/(Sqrt[d]*Sqrt[b + c*x])])/Sqrt[c*d - b*e]))/(d^(3/2)*(-(c*d) + b*e)*Sqrt[x*(b +
c*x)])

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IntegrateAlgebraic [B]  time = 2.95, size = 930, normalized size = 7.27 \begin {gather*} \frac {B d b^3+B e x b^3+c^{5/2} \sqrt {c x^2+b x} \left (8 B d x^2+8 A d x\right )+c^{3/2} \sqrt {c x^2+b x} \left (-8 b B e x^2-4 A b e x+4 A b d\right )+c \left (8 B e x^2 b^2-A d b^2+4 B d x b^2+3 A e x b^2\right )+c^3 \left (-8 B d x^3-8 A d x^2\right )+c^2 \left (8 b B e x^3-4 b B d x^2+4 A b e x^2-8 A b d x\right )+\sqrt {c} \left (-3 B d b^2-A e b^2-4 B e x b^2\right ) \sqrt {c x^2+b x}}{8 d^2 x^2 (d+e x) c^{7/2}-8 d^2 x (d+e x) \sqrt {c x^2+b x} c^3+d (d+e x) \left (8 b d x-8 b e x^2\right ) c^{5/2}+d (d+e x) (8 b e x-4 b d) \sqrt {c x^2+b x} c^2+d (d+e x) \left (b^2 d-8 b^2 e x\right ) c^{3/2}+4 b^2 d e (d+e x) \sqrt {c x^2+b x} c-b^3 d e (d+e x) \sqrt {c}}+\frac {2 b B \tan ^{-1}\left (\frac {\sqrt {c} x e}{\sqrt {d} \sqrt {b e-c d}}-\frac {\sqrt {c x^2+b x} e}{\sqrt {d} \sqrt {b e-c d}}+\frac {\sqrt {c} \sqrt {d}}{\sqrt {b e-c d}}\right )}{\sqrt {d} (c d-b e) \sqrt {b e-c d}}-\frac {2 B c \sqrt {d} \tan ^{-1}\left (\frac {\sqrt {c} x e}{\sqrt {d} \sqrt {b e-c d}}-\frac {\sqrt {c x^2+b x} e}{\sqrt {d} \sqrt {b e-c d}}+\frac {\sqrt {c} \sqrt {d}}{\sqrt {b e-c d}}\right )}{e (c d-b e) \sqrt {b e-c d}}+\left (\frac {b B}{\sqrt {d} (c d-b e)^{3/2}}+\frac {2 A c}{\sqrt {d} (c d-b e)^{3/2}}\right ) \tanh ^{-1}\left (\frac {\sqrt {c} d+\sqrt {c} e x-e \sqrt {c x^2+b x}}{\sqrt {d} \sqrt {c d-b e}}\right )-\frac {A b e \tanh ^{-1}\left (\frac {\sqrt {c} x e}{\sqrt {d} \sqrt {c d-b e}}-\frac {\sqrt {c x^2+b x} e}{\sqrt {d} \sqrt {c d-b e}}+\frac {\sqrt {c} \sqrt {d}}{\sqrt {c d-b e}}\right )}{d^{3/2} (c d-b e)^{3/2}}-\frac {2 B c \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {c} x e}{\sqrt {d} \sqrt {c d-b e}}-\frac {\sqrt {c x^2+b x} e}{\sqrt {d} \sqrt {c d-b e}}+\frac {\sqrt {c} \sqrt {d}}{\sqrt {c d-b e}}\right )}{e (c d-b e)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b^3*B*d + b^3*B*e*x + Sqrt[c]*(-3*b^2*B*d - A*b^2*e - 4*b^2*B*e*x)*Sqrt[b*x + c*x^2] + c^(5/2)*Sqrt[b*x + c*x
^2]*(8*A*d*x + 8*B*d*x^2) + c^(3/2)*Sqrt[b*x + c*x^2]*(4*A*b*d - 4*A*b*e*x - 8*b*B*e*x^2) + c*(-(A*b^2*d) + 4*
b^2*B*d*x + 3*A*b^2*e*x + 8*b^2*B*e*x^2) + c^3*(-8*A*d*x^2 - 8*B*d*x^3) + c^2*(-8*A*b*d*x - 4*b*B*d*x^2 + 4*A*
b*e*x^2 + 8*b*B*e*x^3))/(-(b^3*Sqrt[c]*d*e*(d + e*x)) + 8*c^(7/2)*d^2*x^2*(d + e*x) + c^(3/2)*d*(d + e*x)*(b^2
*d - 8*b^2*e*x) + 4*b^2*c*d*e*(d + e*x)*Sqrt[b*x + c*x^2] - 8*c^3*d^2*x*(d + e*x)*Sqrt[b*x + c*x^2] + c^2*d*(d
+ e*x)*(-4*b*d + 8*b*e*x)*Sqrt[b*x + c*x^2] + c^(5/2)*d*(d + e*x)*(8*b*d*x - 8*b*e*x^2)) + (2*b*B*ArcTan[(Sqr
t[c]*Sqrt[d])/Sqrt[-(c*d) + b*e] + (Sqrt[c]*e*x)/(Sqrt[d]*Sqrt[-(c*d) + b*e]) - (e*Sqrt[b*x + c*x^2])/(Sqrt[d]
*Sqrt[-(c*d) + b*e])])/(Sqrt[d]*(c*d - b*e)*Sqrt[-(c*d) + b*e]) - (2*B*c*Sqrt[d]*ArcTan[(Sqrt[c]*Sqrt[d])/Sqrt
[-(c*d) + b*e] + (Sqrt[c]*e*x)/(Sqrt[d]*Sqrt[-(c*d) + b*e]) - (e*Sqrt[b*x + c*x^2])/(Sqrt[d]*Sqrt[-(c*d) + b*e
])])/(e*(c*d - b*e)*Sqrt[-(c*d) + b*e]) + ((b*B)/(Sqrt[d]*(c*d - b*e)^(3/2)) + (2*A*c)/(Sqrt[d]*(c*d - b*e)^(3
/2)))*ArcTanh[(Sqrt[c]*d + Sqrt[c]*e*x - e*Sqrt[b*x + c*x^2])/(Sqrt[d]*Sqrt[c*d - b*e])] - (2*B*c*Sqrt[d]*ArcT
anh[(Sqrt[c]*Sqrt[d])/Sqrt[c*d - b*e] + (Sqrt[c]*e*x)/(Sqrt[d]*Sqrt[c*d - b*e]) - (e*Sqrt[b*x + c*x^2])/(Sqrt[
d]*Sqrt[c*d - b*e])])/(e*(c*d - b*e)^(3/2)) - (A*b*e*ArcTanh[(Sqrt[c]*Sqrt[d])/Sqrt[c*d - b*e] + (Sqrt[c]*e*x)
/(Sqrt[d]*Sqrt[c*d - b*e]) - (e*Sqrt[b*x + c*x^2])/(Sqrt[d]*Sqrt[c*d - b*e])])/(d^(3/2)*(c*d - b*e)^(3/2))

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fricas [A]  time = 0.44, size = 399, normalized size = 3.12 \begin {gather*} \left [-\frac {{\left (A b d e + {\left (B b - 2 \, A c\right )} d^{2} + {\left (A b e^{2} + {\left (B b - 2 \, A c\right )} d e\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 (B c d^{3} + A b d e^{2} - {\left (B b + A c\right )} d^{2} e\right )} \sqrt {c x^{2} + b x}}{2 \, {\left (c^{2} d^{5} - 2 \, b c d^{4} e + b^{2} d^{3} e^{2} + {\left (c^{2} d^{4} e - 2 \, b c d^{3} e^{2} + b^{2} d^{2} e^{3}\right )} x\right )}}, -\frac {{\left (A b d e + {\left (B b - 2 \, A c\right )} d^{2} + {\left (A b e^{2} + {\left (B b - 2 \, A c\right )} d e\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 (B c d^{3} + A b d e^{2} - {\left (B b + A c\right )} d^{2} e\right )} \sqrt {c x^{2} + b x}}{c^{2} d^{5} - 2 \, b c d^{4} e + b^{2} d^{3} e^{2} + {\left (c^{2} d^{4} e - 2 \, b c d^{3} e^{2} + b^{2} d^{2} e^{3}\right )} x}\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]

[-1/2*((A*b*d*e + (B*b - 2*A*c)*d^2 + (A*b*e^2 + (B*b - 2*A*c)*d*e)*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*(B*c*d^3 + A*b*d*e^2 - (B*b + A*c)*d^2*e)*sq
rt(c*x^2 + b*x))/(c^2*d^5 - 2*b*c*d^4*e + b^2*d^3*e^2 + (c^2*d^4*e - 2*b*c*d^3*e^2 + b^2*d^2*e^3)*x), -((A*b*d
*e + (B*b - 2*A*c)*d^2 + (A*b*e^2 + (B*b - 2*A*c)*d*e)*x)*sqrt(-c*d^2 + b*d*e)*arctan(-sqrt(-c*d^2 + b*d*e)*sq
rt(c*x^2 + b*x)/((c*d - b*e)*x)) - (B*c*d^3 + A*b*d*e^2 - (B*b + A*c)*d^2*e)*sqrt(c*x^2 + b*x))/(c^2*d^5 - 2*b
*c*d^4*e + b^2*d^3*e^2 + (c^2*d^4*e - 2*b*c*d^3*e^2 + b^2*d^2*e^3)*x)]

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giac [B]  time = 1.10, size = 509, normalized size = 3.98 \begin {gather*} -\frac {1}{2} \, {\left (\frac {{\left (B b d e^{2} \log \left ({\left | 2 \, c d - b e - 2 \, \sqrt {c d^{2} - b d e} \sqrt {c} \right |}\right ) - 2 \, A c d e^{2} \log \left ({\left | 2 \, c d - b e - 2 \, \sqrt {c d^{2} - b d e} \sqrt {c} \right |}\right ) + 2 \, \sqrt {c d^{2} - b d e} B \sqrt {c} d e + A b e^{3} \log \left ({\left | 2 \, c d - b e - 2 \, \sqrt {c d^{2} - b d e} \sqrt {c} \right |}\right ) - 2 \, \sqrt {c d^{2} - b d e} A \sqrt {c} e^{2}\right )} \mathrm {sgn}\left (\frac {1}{x e + d}\right )}{\sqrt {c d^{2} - b d e} c d^{2} - \sqrt {c d^{2} - b d e} b d e} - \frac {2 \, {\left (B d e \mathrm {sgn}\left (\frac {1}{x e + d}\right ) - A e^{2} \mathrm {sgn}\left (\frac {1}{x e + d}\right )\right )} \sqrt {c - \frac {2 \, c d}{x e + d} + \frac {c d^{2}}{{\left (x e + d\right )}^{2}} + \frac {b e}{x e + d} - \frac {b d e}{{\left (x e + d\right )}^{2}}}}{c d^{2} \mathrm {sgn}\left (\frac {1}{x e + d}\right )^{2} - b d e \mathrm {sgn}\left (\frac {1}{x e + d}\right )^{2}} - \frac {{\left (B b d e^{3} - 2 \, A c d e^{3} + A b e^{4}\right )} \log \left ({\left | 2 \, c d - b e - 2 \, \sqrt {c d^{2} - b d e} {\left (\sqrt {c - \frac {2 \, c d}{x e + d} + \frac {c d^{2}}{{\left (x e + d\right )}^{2}} + \frac {b e}{x e + d} - \frac {b d e}{{\left (x e + d\right )}^{2}}} + \frac {\sqrt {c d^{2} e^{2} - b d e^{3}} e^{\left (-1\right )}}{x e + d}\right )} \right |}\right )}{{\left (c d^{2} e - b d e^{2}\right )} \sqrt {c d^{2} - b d e} \mathrm {sgn}\left (\frac {1}{x e + d}\right )}\right )} e^{\left (-2\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="giac")

[Out]

-1/2*((B*b*d*e^2*log(abs(2*c*d - b*e - 2*sqrt(c*d^2 - b*d*e)*sqrt(c))) - 2*A*c*d*e^2*log(abs(2*c*d - b*e - 2*s
qrt(c*d^2 - b*d*e)*sqrt(c))) + 2*sqrt(c*d^2 - b*d*e)*B*sqrt(c)*d*e + A*b*e^3*log(abs(2*c*d - b*e - 2*sqrt(c*d^
2 - b*d*e)*sqrt(c))) - 2*sqrt(c*d^2 - b*d*e)*A*sqrt(c)*e^2)*sgn(1/(x*e + d))/(sqrt(c*d^2 - b*d*e)*c*d^2 - sqrt
(c*d^2 - b*d*e)*b*d*e) - 2*(B*d*e*sgn(1/(x*e + d)) - A*e^2*sgn(1/(x*e + d)))*sqrt(c - 2*c*d/(x*e + d) + c*d^2/
(x*e + d)^2 + b*e/(x*e + d) - b*d*e/(x*e + d)^2)/(c*d^2*sgn(1/(x*e + d))^2 - b*d*e*sgn(1/(x*e + d))^2) - (B*b*
d*e^3 - 2*A*c*d*e^3 + A*b*e^4)*log(abs(2*c*d - b*e - 2*sqrt(c*d^2 - b*d*e)*(sqrt(c - 2*c*d/(x*e + d) + c*d^2/(
x*e + d)^2 + b*e/(x*e + d) - b*d*e/(x*e + d)^2) + sqrt(c*d^2*e^2 - b*d*e^3)*e^(-1)/(x*e + d))))/((c*d^2*e - b*
d*e^2)*sqrt(c*d^2 - b*d*e)*sgn(1/(x*e + d))))*e^(-2)

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maple [B]  time = 0.06, size = 849, normalized size = 6.63 \begin {gather*} -\frac {A b \ln \left (\frac {-\frac {2 \left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {-\frac {\left (b e -c d \right ) d}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {\left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}}}{x +\frac {d}{e}}\right )}{2 \left (b e -c d \right ) \sqrt {-\frac {\left (b e -c d \right ) d}{e^{2}}}\, d}+\frac {A c \ln \left (\frac {-\frac {2 \left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {-\frac {\left (b e -c d \right ) d}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {\left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}}}{x +\frac {d}{e}}\right )}{\left (b e -c d \right ) \sqrt {-\frac {\left (b e -c d \right ) d}{e^{2}}}\, e}+\frac {B b \ln \left (\frac {-\frac {2 \left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {-\frac {\left (b e -c d \right ) d}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {\left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}}}{x +\frac {d}{e}}\right )}{2 \left (b e -c d \right ) \sqrt {-\frac {\left (b e -c d \right ) d}{e^{2}}}\, e}-\frac {B c d \ln \left (\frac {-\frac {2 \left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {-\frac {\left (b e -c d \right ) d}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {\left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}}}{x +\frac {d}{e}}\right )}{\left (b e -c d \right ) \sqrt {-\frac {\left (b e -c d \right ) d}{e^{2}}}\, e^{2}}+\frac {\sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {\left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}}\, A}{\left (b e -c d \right ) \left (x +\frac {d}{e}\right ) d}-\frac {\sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {\left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}}\, B}{\left (b e -c d \right ) \left (x +\frac {d}{e}\right ) e}-\frac {B \ln \left (\frac {-\frac {2 \left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {-\frac {\left (b e -c d \right ) d}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {\left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}}}{x +\frac {d}{e}}\right )}{\sqrt {-\frac {\left (b e -c d \right ) d}{e^{2}}}\, e^{2}} \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]

1/(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)*A-1/e/(b*e-c*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-1/2/(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*A+1/2/e/(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))*b*B+1/e/(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*A-1/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*B*d-B/e^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))

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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)^2/(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 = -0.01 \begin {gather*} \int \frac {A+B\,x}{\sqrt {c\,x^2+b\,x}\,{\left (d+e\,x\right )}^2} \,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)^2),x)

[Out]

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

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