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

Optimal. Leaf size=101 \[ -\frac {2 (b B-A c) \sqrt {c d-b e} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b c^{3/2}}-\frac {2 A \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b}+\frac {2 B \sqrt {d+e x}}{c} \]

________________________________________________________________________________________

Rubi [A]  time = 0.18, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {824, 826, 1166, 208} \begin {gather*} -\frac {2 (b B-A c) \sqrt {c d-b e} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b c^{3/2}}-\frac {2 A \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b}+\frac {2 B \sqrt {d+e x}}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 824

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

Rule 826

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

Rule 1166

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.11, size = 101, normalized size = 1.00 \begin {gather*} \frac {2 (A c-b B) \sqrt {c d-b e} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b c^{3/2}}-\frac {2 A \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b}+\frac {2 B \sqrt {d+e x}}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.18, size = 111, normalized size = 1.10 \begin {gather*} -\frac {2 (A c-b B) \sqrt {b e-c d} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x} \sqrt {b e-c d}}{c d-b e}\right )}{b c^{3/2}}-\frac {2 A \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b}+\frac {2 B \sqrt {d+e x}}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.52, size = 450, normalized size = 4.46 \begin {gather*} \left [\frac {A c \sqrt {d} \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) + 2 \, \sqrt {e x + d} B b - {\left (B b - A c\right )} \sqrt {\frac {c d - b e}{c}} \log \left (\frac {c e x + 2 \, c d - b e + 2 \, \sqrt {e x + d} c \sqrt {\frac {c d - b e}{c}}}{c x + b}\right )}{b c}, \frac {A c \sqrt {d} \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) + 2 \, \sqrt {e x + d} B b - 2 \, {\left (B b - A c\right )} \sqrt {-\frac {c d - b e}{c}} \arctan \left (-\frac {\sqrt {e x + d} c \sqrt {-\frac {c d - b e}{c}}}{c d - b e}\right )}{b c}, \frac {2 \, A c \sqrt {-d} \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + 2 \, \sqrt {e x + d} B b - {\left (B b - A c\right )} \sqrt {\frac {c d - b e}{c}} \log \left (\frac {c e x + 2 \, c d - b e + 2 \, \sqrt {e x + d} c \sqrt {\frac {c d - b e}{c}}}{c x + b}\right )}{b c}, \frac {2 \, {\left (A c \sqrt {-d} \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + \sqrt {e x + d} B b - {\left (B b - A c\right )} \sqrt {-\frac {c d - b e}{c}} \arctan \left (-\frac {\sqrt {e x + d} c \sqrt {-\frac {c d - b e}{c}}}{c d - b e}\right )\right )}}{b c}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(A*c*sqrt(d)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + 2*sqrt(e*x + d)*B*b - (B*b - A*c)*sqrt((c*d - b*e
)/c)*log((c*e*x + 2*c*d - b*e + 2*sqrt(e*x + d)*c*sqrt((c*d - b*e)/c))/(c*x + b)))/(b*c), (A*c*sqrt(d)*log((e*
x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + 2*sqrt(e*x + d)*B*b - 2*(B*b - A*c)*sqrt(-(c*d - b*e)/c)*arctan(-sqrt(
e*x + d)*c*sqrt(-(c*d - b*e)/c)/(c*d - b*e)))/(b*c), (2*A*c*sqrt(-d)*arctan(sqrt(e*x + d)*sqrt(-d)/d) + 2*sqrt
(e*x + d)*B*b - (B*b - A*c)*sqrt((c*d - b*e)/c)*log((c*e*x + 2*c*d - b*e + 2*sqrt(e*x + d)*c*sqrt((c*d - b*e)/
c))/(c*x + b)))/(b*c), 2*(A*c*sqrt(-d)*arctan(sqrt(e*x + d)*sqrt(-d)/d) + sqrt(e*x + d)*B*b - (B*b - A*c)*sqrt
(-(c*d - b*e)/c)*arctan(-sqrt(e*x + d)*c*sqrt(-(c*d - b*e)/c)/(c*d - b*e)))/(b*c)]

________________________________________________________________________________________

giac [A]  time = 0.20, size = 116, normalized size = 1.15 \begin {gather*} \frac {2 \, A d \arctan \left (\frac {\sqrt {x e + d}}{\sqrt {-d}}\right )}{b \sqrt {-d}} + \frac {2 \, \sqrt {x e + d} B}{c} + \frac {2 \, {\left (B b c d - A c^{2} d - B b^{2} e + A b c e\right )} \arctan \left (\frac {\sqrt {x e + d} c}{\sqrt {-c^{2} d + b c e}}\right )}{\sqrt {-c^{2} d + b c e} b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*d*arctan(sqrt(x*e + d)/sqrt(-d))/(b*sqrt(-d)) + 2*sqrt(x*e + d)*B/c + 2*(B*b*c*d - A*c^2*d - B*b^2*e + A*b
*c*e)*arctan(sqrt(x*e + d)*c/sqrt(-c^2*d + b*c*e))/(sqrt(-c^2*d + b*c*e)*b*c)

________________________________________________________________________________________

maple [B]  time = 0.06, size = 196, normalized size = 1.94 \begin {gather*} -\frac {2 A c d \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\sqrt {\left (b e -c d \right ) c}\, b}+\frac {2 A e \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\sqrt {\left (b e -c d \right ) c}}-\frac {2 B b e \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\sqrt {\left (b e -c d \right ) c}\, c}+\frac {2 B d \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\sqrt {\left (b e -c d \right ) c}}-\frac {2 A \sqrt {d}\, \arctanh \left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b}+\frac {2 \sqrt {e x +d}\, B}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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)^(1/2)/(c*x^2+b*x),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 or negative?

________________________________________________________________________________________

mupad [B]  time = 1.91, size = 2368, normalized size = 23.45

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 17.77, size = 104, normalized size = 1.03 \begin {gather*} \frac {2 \left (\frac {A d e \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d}} \right )}}{b \sqrt {- d}} + \frac {B e \sqrt {d + e x}}{c} - \frac {e \left (- A c + B b\right ) \left (b e - c d\right ) \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {b e - c d}{c}}} \right )}}{b c^{2} \sqrt {\frac {b e - c d}{c}}}\right )}{e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(A*d*e*atan(sqrt(d + e*x)/sqrt(-d))/(b*sqrt(-d)) + B*e*sqrt(d + e*x)/c - e*(-A*c + B*b)*(b*e - c*d)*atan(sqr
t(d + e*x)/sqrt((b*e - c*d)/c))/(b*c**2*sqrt((b*e - c*d)/c)))/e

________________________________________________________________________________________