3.1456 \(\int \frac{A+B x}{\sqrt{d+e x} (a-c x^2)^2} \, dx\)

Optimal. Leaf size=250 \[ -\frac{\left (-3 \sqrt{a} A \sqrt{c} e+a B e+2 A c d\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{4 a^{3/2} c^{3/4} \left (\sqrt{c} d-\sqrt{a} e\right )^{3/2}}+\frac{\left (3 \sqrt{a} A \sqrt{c} e+a B e+2 A c d\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{4 a^{3/2} c^{3/4} \left (\sqrt{a} e+\sqrt{c} d\right )^{3/2}}+\frac{\sqrt{d+e x} (x (A c d-a B e)+a (B d-A e))}{2 a \left (a-c x^2\right ) \left (c d^2-a e^2\right )} \]

[Out]

(Sqrt[d + e*x]*(a*(B*d - A*e) + (A*c*d - a*B*e)*x))/(2*a*(c*d^2 - a*e^2)*(a - c*x^2)) - ((2*A*c*d + a*B*e - 3*
Sqrt[a]*A*Sqrt[c]*e)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]])/(4*a^(3/2)*c^(3/4)*(Sqrt[c]
*d - Sqrt[a]*e)^(3/2)) + ((2*A*c*d + a*B*e + 3*Sqrt[a]*A*Sqrt[c]*e)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[
c]*d + Sqrt[a]*e]])/(4*a^(3/2)*c^(3/4)*(Sqrt[c]*d + Sqrt[a]*e)^(3/2))

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Rubi [A]  time = 0.498061, antiderivative size = 250, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {823, 827, 1166, 208} \[ -\frac{\left (-3 \sqrt{a} A \sqrt{c} e+a B e+2 A c d\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{4 a^{3/2} c^{3/4} \left (\sqrt{c} d-\sqrt{a} e\right )^{3/2}}+\frac{\left (3 \sqrt{a} A \sqrt{c} e+a B e+2 A c d\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{4 a^{3/2} c^{3/4} \left (\sqrt{a} e+\sqrt{c} d\right )^{3/2}}+\frac{\sqrt{d+e x} (x (A c d-a B e)+a (B d-A e))}{2 a \left (a-c x^2\right ) \left (c d^2-a e^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d + e*x]*(a*(B*d - A*e) + (A*c*d - a*B*e)*x))/(2*a*(c*d^2 - a*e^2)*(a - c*x^2)) - ((2*A*c*d + a*B*e - 3*
Sqrt[a]*A*Sqrt[c]*e)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]])/(4*a^(3/2)*c^(3/4)*(Sqrt[c]
*d - Sqrt[a]*e)^(3/2)) + ((2*A*c*d + a*B*e + 3*Sqrt[a]*A*Sqrt[c]*e)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[
c]*d + Sqrt[a]*e]])/(4*a^(3/2)*c^(3/4)*(Sqrt[c]*d + Sqrt[a]*e)^(3/2))

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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

Mathematica [A]  time = 0.478988, size = 353, normalized size = 1.41 \[ \frac{-\frac{c^{3/4} \left (-3 a A e^2+2 a B d e+A c d^2\right ) \left (\frac{\tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{\sqrt{\sqrt{a} e+\sqrt{c} d}}-\frac{\tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{2 \sqrt{a}}+\frac{c \sqrt{d+e x} (-a A e+a B (d-e x)+A c d x)}{c x^2-a}+\frac{\sqrt [4]{c} (A c d-a B e) \left (\sqrt{\sqrt{c} d-\sqrt{a} e} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )-\sqrt{\sqrt{a} e+\sqrt{c} d} \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )\right )}{2 \sqrt{a}}}{2 a c \left (a e^2-c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c*Sqrt[d + e*x]*(-(a*A*e) + A*c*d*x + a*B*(d - e*x)))/(-a + c*x^2) - (c^(3/4)*(A*c*d^2 + 2*a*B*d*e - 3*a*A*e
^2)*(-(ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]]/Sqrt[Sqrt[c]*d - Sqrt[a]*e]) + ArcTanh[(c^
(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d + Sqrt[a]*e]]/Sqrt[Sqrt[c]*d + Sqrt[a]*e]))/(2*Sqrt[a]) + (c^(1/4)*(A*c*d
- a*B*e)*(Sqrt[Sqrt[c]*d - Sqrt[a]*e]*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]] - Sqrt[Sqrt
[c]*d + Sqrt[a]*e]*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d + Sqrt[a]*e]]))/(2*Sqrt[a]))/(2*a*c*(-(c*d^2
) + a*e^2))

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Maple [B]  time = 0.066, size = 635, normalized size = 2.5 \begin{align*} \text{result too large to display} \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)^2,x)

[Out]

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

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

[Out]

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

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Fricas [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)^2,x, algorithm="fricas")

[Out]

Timed out

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Sympy [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)**2,x)

[Out]

Timed out

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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)^2,x, algorithm="giac")

[Out]

Timed out