3.1450 \(\int \frac{A+B x}{\sqrt{d+e x} (a-c x^2)} \, dx\)
Optimal. Leaf size=152 \[ \frac{\left (B-\frac{A \sqrt{c}}{\sqrt{a}}\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{c^{3/4} \sqrt{\sqrt{c} d-\sqrt{a} e}}+\frac{\left (\frac{A \sqrt{c}}{\sqrt{a}}+B\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{c^{3/4} \sqrt{\sqrt{a} e+\sqrt{c} d}} \]
[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])
________________________________________________________________________________________
Rubi [A] time = 0.163171, antiderivative size = 152, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used =
{827, 1166, 208} \[ \frac{\left (B-\frac{A \sqrt{c}}{\sqrt{a}}\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{c^{3/4} \sqrt{\sqrt{c} d-\sqrt{a} e}}+\frac{\left (\frac{A \sqrt{c}}{\sqrt{a}}+B\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{c^{3/4} \sqrt{\sqrt{a} e+\sqrt{c} d}} \]
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 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]
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]
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*}
Mathematica [A] time = 0.189578, size = 155, normalized size = 1.02 \[ \frac{\frac{\left (\sqrt{a} B-A \sqrt{c}\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{\sqrt{\sqrt{c} d-\sqrt{a} e}}+\frac{\left (\sqrt{a} B+A \sqrt{c}\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{\sqrt{\sqrt{a} e+\sqrt{c} d}}}{\sqrt{a} c^{3/4}} \]
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))
________________________________________________________________________________________
Maple [A] time = 0.021, size = 203, normalized size = 1.3 \begin{align*}{Ace{\it Artanh} \left ({c\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( cd+\sqrt{ac{e}^{2}} \right ) c}}}} \right ){\frac{1}{\sqrt{ac{e}^{2}}}}{\frac{1}{\sqrt{ \left ( cd+\sqrt{ac{e}^{2}} \right ) c}}}}+{B{\it Artanh} \left ({c\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( cd+\sqrt{ac{e}^{2}} \right ) c}}}} \right ){\frac{1}{\sqrt{ \left ( cd+\sqrt{ac{e}^{2}} \right ) c}}}}+{Ace\arctan \left ({c\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( -cd+\sqrt{ac{e}^{2}} \right ) c}}}} \right ){\frac{1}{\sqrt{ac{e}^{2}}}}{\frac{1}{\sqrt{ \left ( -cd+\sqrt{ac{e}^{2}} \right ) c}}}}-{B\arctan \left ({c\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( -cd+\sqrt{ac{e}^{2}} \right ) c}}}} \right ){\frac{1}{\sqrt{ \left ( -cd+\sqrt{ac{e}^{2}} \right ) c}}}} \end{align*}
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/((c*d+(a*c*e^2)^(1/2))*c)^(1/2))*A*e
+1/((c*d+(a*c*e^2)^(1/2))*c)^(1/2)*arctanh((e*x+d)^(1/2)*c/((c*d+(a*c*e^2)^(1/2))*c)^(1/2))*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/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2))*A*e-1/((-c*d+(a*c*
e^2)^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2))*B
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{B x + A}{{\left (c x^{2} - a\right )} \sqrt{e x + d}}\,{d x} \end{align*}
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)
________________________________________________________________________________________
Fricas [B] time = 2.58205, size = 4594, normalized size = 30.22 \begin{align*} \text{result too large to display} \end{align*}
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)))
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} - \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{align*}
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)
________________________________________________________________________________________
Giac [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
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]
Timed out