3.13.76 \(\int \frac {A+B x}{\sqrt {d+e x} (a-c x^2)} \, dx\)

Optimal. Leaf size=152 \[ \frac {\left (B-\frac {A \sqrt {c}}{\sqrt {a}}\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d-\sqrt {a} e}}\right )}{c^{3/4} \sqrt {\sqrt {c} d-\sqrt {a} e}}+\frac {\left (\frac {A \sqrt {c}}{\sqrt {a}}+B\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {a} e+\sqrt {c} d}}\right )}{c^{3/4} \sqrt {\sqrt {a} e+\sqrt {c} d}} \]

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Rubi [A]  time = 0.16, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {827, 1166, 208} \begin {gather*} \frac {\left (B-\frac {A \sqrt {c}}{\sqrt {a}}\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d-\sqrt {a} e}}\right )}{c^{3/4} \sqrt {\sqrt {c} d-\sqrt {a} e}}+\frac {\left (\frac {A \sqrt {c}}{\sqrt {a}}+B\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {a} e+\sqrt {c} d}}\right )}{c^{3/4} \sqrt {\sqrt {a} e+\sqrt {c} d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((B - (A*Sqrt[c])/Sqrt[a])*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]])/(c^(3/4)*Sqrt[Sqrt[c]
*d - Sqrt[a]*e]) + ((B + (A*Sqrt[c])/Sqrt[a])*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d + Sqrt[a]*e]])/(c
^(3/4)*Sqrt[Sqrt[c]*d + Sqrt[a]*e])

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 827

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2, Subst[Int[(e*f
- d*g + g*x^2)/(c*d^2 + a*e^2 - 2*c*d*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + 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} \left (a-c x^2\right )} \, dx &=2 \operatorname {Subst}\left (\int \frac {-B d+A e+B x^2}{-c d^2+a e^2+2 c d x^2-c x^4} \, dx,x,\sqrt {d+e x}\right )\\ &=\left (B-\frac {A \sqrt {c}}{\sqrt {a}}\right ) \operatorname {Subst}\left (\int \frac {1}{c d-\sqrt {a} \sqrt {c} e-c x^2} \, dx,x,\sqrt {d+e x}\right )+\left (B+\frac {A \sqrt {c}}{\sqrt {a}}\right ) \operatorname {Subst}\left (\int \frac {1}{c d+\sqrt {a} \sqrt {c} e-c x^2} \, dx,x,\sqrt {d+e x}\right )\\ &=\frac {\left (B-\frac {A \sqrt {c}}{\sqrt {a}}\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d-\sqrt {a} e}}\right )}{c^{3/4} \sqrt {\sqrt {c} d-\sqrt {a} e}}+\frac {\left (B+\frac {A \sqrt {c}}{\sqrt {a}}\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d+\sqrt {a} e}}\right )}{c^{3/4} \sqrt {\sqrt {c} d+\sqrt {a} e}}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 155, normalized size = 1.02 \begin {gather*} \frac {\frac {\left (\sqrt {a} B-A \sqrt {c}\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d-\sqrt {a} e}}\right )}{\sqrt {\sqrt {c} d-\sqrt {a} e}}+\frac {\left (\sqrt {a} B+A \sqrt {c}\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {a} e+\sqrt {c} d}}\right )}{\sqrt {\sqrt {a} e+\sqrt {c} d}}}{\sqrt {a} c^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((Sqrt[a]*B - A*Sqrt[c])*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]])/Sqrt[Sqrt[c]*d - Sqrt[
a]*e] + ((Sqrt[a]*B + A*Sqrt[c])*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d + Sqrt[a]*e]])/Sqrt[Sqrt[c]*d
+ Sqrt[a]*e])/(Sqrt[a]*c^(3/4))

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IntegrateAlgebraic [A]  time = 0.40, size = 207, normalized size = 1.36 \begin {gather*} \frac {\left (\sqrt {a} B+A \sqrt {c}\right ) \tan ^{-1}\left (\frac {\sqrt {d+e x} \sqrt {-\sqrt {a} \sqrt {c} e-c d}}{\sqrt {a} e+\sqrt {c} d}\right )}{\sqrt {a} \sqrt {c} \sqrt {-\sqrt {c} \left (\sqrt {a} e+\sqrt {c} d\right )}}+\frac {\left (\sqrt {a} B-A \sqrt {c}\right ) \tan ^{-1}\left (\frac {\sqrt {d+e x} \sqrt {\sqrt {a} \sqrt {c} e-c d}}{\sqrt {c} d-\sqrt {a} e}\right )}{\sqrt {a} \sqrt {c} \sqrt {-\sqrt {c} \left (\sqrt {c} d-\sqrt {a} e\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[a]*B + A*Sqrt[c])*ArcTan[(Sqrt[-(c*d) - Sqrt[a]*Sqrt[c]*e]*Sqrt[d + e*x])/(Sqrt[c]*d + Sqrt[a]*e)])/(Sq
rt[a]*Sqrt[c]*Sqrt[-(Sqrt[c]*(Sqrt[c]*d + Sqrt[a]*e))]) + ((Sqrt[a]*B - A*Sqrt[c])*ArcTan[(Sqrt[-(c*d) + Sqrt[
a]*Sqrt[c]*e]*Sqrt[d + e*x])/(Sqrt[c]*d - Sqrt[a]*e)])/(Sqrt[a]*Sqrt[c]*Sqrt[-(Sqrt[c]*(Sqrt[c]*d - Sqrt[a]*e)
)])

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fricas [B]  time = 0.51, size = 2385, normalized size = 15.69

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(-(2*A*B*a*e - (B^2*a + A^2*c)*d + (a*c^2*d^2 - a^2*c*e^2)*sqrt((4*A^2*B^2*c^2*d^2 - 4*(A*B^3*a*c + A^
3*B*c^2)*d*e + (B^4*a^2 + 2*A^2*B^2*a*c + A^4*c^2)*e^2)/(a*c^5*d^4 - 2*a^2*c^4*d^2*e^2 + a^3*c^3*e^4)))/(a*c^2
*d^2 - a^2*c*e^2))*log((2*(A*B^3*a*c - A^3*B*c^2)*d - (B^4*a^2 - A^4*c^2)*e)*sqrt(e*x + d) + (2*A*B^2*a*c^2*d^
2 - (B^3*a^2*c + 3*A^2*B*a*c^2)*d*e + (A*B^2*a^2*c + A^3*a*c^2)*e^2 + (A*a*c^4*d^3 - B*a^2*c^3*d^2*e - A*a^2*c
^3*d*e^2 + B*a^3*c^2*e^3)*sqrt((4*A^2*B^2*c^2*d^2 - 4*(A*B^3*a*c + A^3*B*c^2)*d*e + (B^4*a^2 + 2*A^2*B^2*a*c +
A^4*c^2)*e^2)/(a*c^5*d^4 - 2*a^2*c^4*d^2*e^2 + a^3*c^3*e^4)))*sqrt(-(2*A*B*a*e - (B^2*a + A^2*c)*d + (a*c^2*d
^2 - a^2*c*e^2)*sqrt((4*A^2*B^2*c^2*d^2 - 4*(A*B^3*a*c + A^3*B*c^2)*d*e + (B^4*a^2 + 2*A^2*B^2*a*c + A^4*c^2)*
e^2)/(a*c^5*d^4 - 2*a^2*c^4*d^2*e^2 + a^3*c^3*e^4)))/(a*c^2*d^2 - a^2*c*e^2))) - 1/2*sqrt(-(2*A*B*a*e - (B^2*a
+ A^2*c)*d + (a*c^2*d^2 - a^2*c*e^2)*sqrt((4*A^2*B^2*c^2*d^2 - 4*(A*B^3*a*c + A^3*B*c^2)*d*e + (B^4*a^2 + 2*A
^2*B^2*a*c + A^4*c^2)*e^2)/(a*c^5*d^4 - 2*a^2*c^4*d^2*e^2 + a^3*c^3*e^4)))/(a*c^2*d^2 - a^2*c*e^2))*log((2*(A*
B^3*a*c - A^3*B*c^2)*d - (B^4*a^2 - A^4*c^2)*e)*sqrt(e*x + d) - (2*A*B^2*a*c^2*d^2 - (B^3*a^2*c + 3*A^2*B*a*c^
2)*d*e + (A*B^2*a^2*c + A^3*a*c^2)*e^2 + (A*a*c^4*d^3 - B*a^2*c^3*d^2*e - A*a^2*c^3*d*e^2 + B*a^3*c^2*e^3)*sqr
t((4*A^2*B^2*c^2*d^2 - 4*(A*B^3*a*c + A^3*B*c^2)*d*e + (B^4*a^2 + 2*A^2*B^2*a*c + A^4*c^2)*e^2)/(a*c^5*d^4 - 2
*a^2*c^4*d^2*e^2 + a^3*c^3*e^4)))*sqrt(-(2*A*B*a*e - (B^2*a + A^2*c)*d + (a*c^2*d^2 - a^2*c*e^2)*sqrt((4*A^2*B
^2*c^2*d^2 - 4*(A*B^3*a*c + A^3*B*c^2)*d*e + (B^4*a^2 + 2*A^2*B^2*a*c + A^4*c^2)*e^2)/(a*c^5*d^4 - 2*a^2*c^4*d
^2*e^2 + a^3*c^3*e^4)))/(a*c^2*d^2 - a^2*c*e^2))) + 1/2*sqrt(-(2*A*B*a*e - (B^2*a + A^2*c)*d - (a*c^2*d^2 - a^
2*c*e^2)*sqrt((4*A^2*B^2*c^2*d^2 - 4*(A*B^3*a*c + A^3*B*c^2)*d*e + (B^4*a^2 + 2*A^2*B^2*a*c + A^4*c^2)*e^2)/(a
*c^5*d^4 - 2*a^2*c^4*d^2*e^2 + a^3*c^3*e^4)))/(a*c^2*d^2 - a^2*c*e^2))*log((2*(A*B^3*a*c - A^3*B*c^2)*d - (B^4
*a^2 - A^4*c^2)*e)*sqrt(e*x + d) + (2*A*B^2*a*c^2*d^2 - (B^3*a^2*c + 3*A^2*B*a*c^2)*d*e + (A*B^2*a^2*c + A^3*a
*c^2)*e^2 - (A*a*c^4*d^3 - B*a^2*c^3*d^2*e - A*a^2*c^3*d*e^2 + B*a^3*c^2*e^3)*sqrt((4*A^2*B^2*c^2*d^2 - 4*(A*B
^3*a*c + A^3*B*c^2)*d*e + (B^4*a^2 + 2*A^2*B^2*a*c + A^4*c^2)*e^2)/(a*c^5*d^4 - 2*a^2*c^4*d^2*e^2 + a^3*c^3*e^
4)))*sqrt(-(2*A*B*a*e - (B^2*a + A^2*c)*d - (a*c^2*d^2 - a^2*c*e^2)*sqrt((4*A^2*B^2*c^2*d^2 - 4*(A*B^3*a*c + A
^3*B*c^2)*d*e + (B^4*a^2 + 2*A^2*B^2*a*c + A^4*c^2)*e^2)/(a*c^5*d^4 - 2*a^2*c^4*d^2*e^2 + a^3*c^3*e^4)))/(a*c^
2*d^2 - a^2*c*e^2))) - 1/2*sqrt(-(2*A*B*a*e - (B^2*a + A^2*c)*d - (a*c^2*d^2 - a^2*c*e^2)*sqrt((4*A^2*B^2*c^2*
d^2 - 4*(A*B^3*a*c + A^3*B*c^2)*d*e + (B^4*a^2 + 2*A^2*B^2*a*c + A^4*c^2)*e^2)/(a*c^5*d^4 - 2*a^2*c^4*d^2*e^2
+ a^3*c^3*e^4)))/(a*c^2*d^2 - a^2*c*e^2))*log((2*(A*B^3*a*c - A^3*B*c^2)*d - (B^4*a^2 - A^4*c^2)*e)*sqrt(e*x +
d) - (2*A*B^2*a*c^2*d^2 - (B^3*a^2*c + 3*A^2*B*a*c^2)*d*e + (A*B^2*a^2*c + A^3*a*c^2)*e^2 - (A*a*c^4*d^3 - B*
a^2*c^3*d^2*e - A*a^2*c^3*d*e^2 + B*a^3*c^2*e^3)*sqrt((4*A^2*B^2*c^2*d^2 - 4*(A*B^3*a*c + A^3*B*c^2)*d*e + (B^
4*a^2 + 2*A^2*B^2*a*c + A^4*c^2)*e^2)/(a*c^5*d^4 - 2*a^2*c^4*d^2*e^2 + a^3*c^3*e^4)))*sqrt(-(2*A*B*a*e - (B^2*
a + A^2*c)*d - (a*c^2*d^2 - a^2*c*e^2)*sqrt((4*A^2*B^2*c^2*d^2 - 4*(A*B^3*a*c + A^3*B*c^2)*d*e + (B^4*a^2 + 2*
A^2*B^2*a*c + A^4*c^2)*e^2)/(a*c^5*d^4 - 2*a^2*c^4*d^2*e^2 + a^3*c^3*e^4)))/(a*c^2*d^2 - a^2*c*e^2)))

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giac [A]  time = 0.26, size = 177, normalized size = 1.16 \begin {gather*} -\frac {{\left (B a {\left | c \right |} + \sqrt {a c} A {\left | c \right |}\right )} \arctan \left (\frac {\sqrt {x e + d}}{\sqrt {-\frac {c d + \sqrt {c^{2} d^{2} - {\left (c d^{2} - a e^{2}\right )} c}}{c}}}\right )}{\sqrt {-c^{2} d - \sqrt {a c} c e} a c} - \frac {{\left (B a {\left | c \right |} - \sqrt {a c} A {\left | c \right |}\right )} \arctan \left (\frac {\sqrt {x e + d}}{\sqrt {-\frac {c d - \sqrt {c^{2} d^{2} - {\left (c d^{2} - a e^{2}\right )} c}}{c}}}\right )}{\sqrt {-c^{2} d + \sqrt {a c} c e} a 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+a),x, algorithm="giac")

[Out]

-(B*a*abs(c) + sqrt(a*c)*A*abs(c))*arctan(sqrt(x*e + d)/sqrt(-(c*d + sqrt(c^2*d^2 - (c*d^2 - a*e^2)*c))/c))/(s
qrt(-c^2*d - sqrt(a*c)*c*e)*a*c) - (B*a*abs(c) - sqrt(a*c)*A*abs(c))*arctan(sqrt(x*e + d)/sqrt(-(c*d - sqrt(c^
2*d^2 - (c*d^2 - a*e^2)*c))/c))/(sqrt(-c^2*d + sqrt(a*c)*c*e)*a*c)

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maple [A]  time = 0.08, size = 203, normalized size = 1.34 \begin {gather*} \frac {A c e \arctanh \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{\sqrt {a c \,e^{2}}\, \sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}}+\frac {A c e \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{\sqrt {a c \,e^{2}}\, \sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}}+\frac {B \arctanh \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{\sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}}-\frac {B \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{\sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) 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+a),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {B x + A}{{\left (c x^{2} - a\right )} \sqrt {e x + d}}\,{d x} \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+a),x, algorithm="maxima")

[Out]

-integrate((B*x + A)/((c*x^2 - a)*sqrt(e*x + d)), x)

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mupad [B]  time = 3.31, size = 2065, normalized size = 13.59

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atan((a^2*c^5*d^3*((B^2*a*e*(a^3*c^3)^(1/2) + A^2*c*e*(a^3*c^3)^(1/2) + A^2*a*c^3*d + B^2*a^2*c^2*d - 2*A*B*a^
2*c^2*e - 2*A*B*c*d*(a^3*c^3)^(1/2))/(4*a^2*c^4*d^2 - 4*a^3*c^3*e^2))^(3/2)*(d + e*x)^(1/2)*8i + A^2*a^2*c^3*e
^2*((B^2*a*e*(a^3*c^3)^(1/2) + A^2*c*e*(a^3*c^3)^(1/2) + A^2*a*c^3*d + B^2*a^2*c^2*d - 2*A*B*a^2*c^2*e - 2*A*B
*c*d*(a^3*c^3)^(1/2))/(4*a^2*c^4*d^2 - 4*a^3*c^3*e^2))^(1/2)*(d + e*x)^(1/2)*2i - B^2*a^2*c^3*d^2*((B^2*a*e*(a
^3*c^3)^(1/2) + A^2*c*e*(a^3*c^3)^(1/2) + A^2*a*c^3*d + B^2*a^2*c^2*d - 2*A*B*a^2*c^2*e - 2*A*B*c*d*(a^3*c^3)^
(1/2))/(4*a^2*c^4*d^2 - 4*a^3*c^3*e^2))^(1/2)*(d + e*x)^(1/2)*2i + B^2*a^3*c^2*e^2*((B^2*a*e*(a^3*c^3)^(1/2) +
A^2*c*e*(a^3*c^3)^(1/2) + A^2*a*c^3*d + B^2*a^2*c^2*d - 2*A*B*a^2*c^2*e - 2*A*B*c*d*(a^3*c^3)^(1/2))/(4*a^2*c
^4*d^2 - 4*a^3*c^3*e^2))^(1/2)*(d + e*x)^(1/2)*2i - A^2*a*c^4*d^2*((B^2*a*e*(a^3*c^3)^(1/2) + A^2*c*e*(a^3*c^3
)^(1/2) + A^2*a*c^3*d + B^2*a^2*c^2*d - 2*A*B*a^2*c^2*e - 2*A*B*c*d*(a^3*c^3)^(1/2))/(4*a^2*c^4*d^2 - 4*a^3*c^
3*e^2))^(1/2)*(d + e*x)^(1/2)*2i - a^3*c^4*d*e^2*((B^2*a*e*(a^3*c^3)^(1/2) + A^2*c*e*(a^3*c^3)^(1/2) + A^2*a*c
^3*d + B^2*a^2*c^2*d - 2*A*B*a^2*c^2*e - 2*A*B*c*d*(a^3*c^3)^(1/2))/(4*a^2*c^4*d^2 - 4*a^3*c^3*e^2))^(3/2)*(d
+ e*x)^(1/2)*8i)/(A^3*c*e^2*(a^3*c^3)^(1/2) - B^3*a^3*c*e^2 - 2*A^2*B*a*c^3*d^2 - B^3*a*d*e*(a^3*c^3)^(1/2) -
A^2*B*a^2*c^2*e^2 + A*B^2*a*e^2*(a^3*c^3)^(1/2) + 2*A*B^2*c*d^2*(a^3*c^3)^(1/2) + A^3*a*c^3*d*e + 3*A*B^2*a^2*
c^2*d*e - 3*A^2*B*c*d*e*(a^3*c^3)^(1/2)))*((B^2*a*e*(a^3*c^3)^(1/2) + A^2*c*e*(a^3*c^3)^(1/2) + A^2*a*c^3*d +
B^2*a^2*c^2*d - 2*A*B*a^2*c^2*e - 2*A*B*c*d*(a^3*c^3)^(1/2))/(4*a^2*c^4*d^2 - 4*a^3*c^3*e^2))^(1/2)*2i - atan(
(a^2*c^5*d^3*(-(B^2*a*e*(a^3*c^3)^(1/2) + A^2*c*e*(a^3*c^3)^(1/2) - A^2*a*c^3*d - B^2*a^2*c^2*d + 2*A*B*a^2*c^
2*e - 2*A*B*c*d*(a^3*c^3)^(1/2))/(4*a^2*c^4*d^2 - 4*a^3*c^3*e^2))^(3/2)*(d + e*x)^(1/2)*8i + A^2*a^2*c^3*e^2*(
-(B^2*a*e*(a^3*c^3)^(1/2) + A^2*c*e*(a^3*c^3)^(1/2) - A^2*a*c^3*d - B^2*a^2*c^2*d + 2*A*B*a^2*c^2*e - 2*A*B*c*
d*(a^3*c^3)^(1/2))/(4*a^2*c^4*d^2 - 4*a^3*c^3*e^2))^(1/2)*(d + e*x)^(1/2)*2i - B^2*a^2*c^3*d^2*(-(B^2*a*e*(a^3
*c^3)^(1/2) + A^2*c*e*(a^3*c^3)^(1/2) - A^2*a*c^3*d - B^2*a^2*c^2*d + 2*A*B*a^2*c^2*e - 2*A*B*c*d*(a^3*c^3)^(1
/2))/(4*a^2*c^4*d^2 - 4*a^3*c^3*e^2))^(1/2)*(d + e*x)^(1/2)*2i + B^2*a^3*c^2*e^2*(-(B^2*a*e*(a^3*c^3)^(1/2) +
A^2*c*e*(a^3*c^3)^(1/2) - A^2*a*c^3*d - B^2*a^2*c^2*d + 2*A*B*a^2*c^2*e - 2*A*B*c*d*(a^3*c^3)^(1/2))/(4*a^2*c^
4*d^2 - 4*a^3*c^3*e^2))^(1/2)*(d + e*x)^(1/2)*2i - A^2*a*c^4*d^2*(-(B^2*a*e*(a^3*c^3)^(1/2) + A^2*c*e*(a^3*c^3
)^(1/2) - A^2*a*c^3*d - B^2*a^2*c^2*d + 2*A*B*a^2*c^2*e - 2*A*B*c*d*(a^3*c^3)^(1/2))/(4*a^2*c^4*d^2 - 4*a^3*c^
3*e^2))^(1/2)*(d + e*x)^(1/2)*2i - a^3*c^4*d*e^2*(-(B^2*a*e*(a^3*c^3)^(1/2) + A^2*c*e*(a^3*c^3)^(1/2) - A^2*a*
c^3*d - B^2*a^2*c^2*d + 2*A*B*a^2*c^2*e - 2*A*B*c*d*(a^3*c^3)^(1/2))/(4*a^2*c^4*d^2 - 4*a^3*c^3*e^2))^(3/2)*(d
+ e*x)^(1/2)*8i)/(A^3*c*e^2*(a^3*c^3)^(1/2) + B^3*a^3*c*e^2 + 2*A^2*B*a*c^3*d^2 - B^3*a*d*e*(a^3*c^3)^(1/2) +
A^2*B*a^2*c^2*e^2 + A*B^2*a*e^2*(a^3*c^3)^(1/2) + 2*A*B^2*c*d^2*(a^3*c^3)^(1/2) - A^3*a*c^3*d*e - 3*A*B^2*a^2
*c^2*d*e - 3*A^2*B*c*d*e*(a^3*c^3)^(1/2)))*(-(B^2*a*e*(a^3*c^3)^(1/2) + A^2*c*e*(a^3*c^3)^(1/2) - A^2*a*c^3*d
- B^2*a^2*c^2*d + 2*A*B*a^2*c^2*e - 2*A*B*c*d*(a^3*c^3)^(1/2))/(4*a^2*c^4*d^2 - 4*a^3*c^3*e^2))^(1/2)*2i

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {A}{- a \sqrt {d + e x} + c x^{2} \sqrt {d + e x}}\, dx - \int \frac {B x}{- a \sqrt {d + e x} + c x^{2} \sqrt {d + e x}}\, dx \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+a),x)

[Out]

-Integral(A/(-a*sqrt(d + e*x) + c*x**2*sqrt(d + e*x)), x) - Integral(B*x/(-a*sqrt(d + e*x) + c*x**2*sqrt(d + e
*x)), x)

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