3.1457 \(\int \frac{A+B x}{(d+e x)^{3/2} (a-c x^2)^2} \, dx\)

Optimal. Leaf size=303 \[ -\frac{\left (-5 \sqrt{a} A \sqrt{c} e+3 a B e+2 A c d\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{4 a^{3/2} \sqrt [4]{c} \left (\sqrt{c} d-\sqrt{a} e\right )^{5/2}}+\frac{\left (5 \sqrt{a} A \sqrt{c} e+3 a B e+2 A c d\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{4 a^{3/2} \sqrt [4]{c} \left (\sqrt{a} e+\sqrt{c} d\right )^{5/2}}+\frac{x (A c d-a B e)+a (B d-A e)}{2 a \left (a-c x^2\right ) \sqrt{d+e x} \left (c d^2-a e^2\right )}-\frac{e \left (5 a A e^2-6 a B d e+A c d^2\right )}{2 a \sqrt{d+e x} \left (c d^2-a e^2\right )^2} \]

[Out]

-(e*(A*c*d^2 - 6*a*B*d*e + 5*a*A*e^2))/(2*a*(c*d^2 - a*e^2)^2*Sqrt[d + e*x]) + (a*(B*d - A*e) + (A*c*d - a*B*e
)*x)/(2*a*(c*d^2 - a*e^2)*Sqrt[d + e*x]*(a - c*x^2)) - ((2*A*c*d + 3*a*B*e - 5*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^(1/4)*(Sqrt[c]*d - Sqrt[a]*e)^(5/2)) + ((2*A*
c*d + 3*a*B*e + 5*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^(1/4)*(Sqrt[c]*d + Sqrt[a]*e)^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 0.685795, antiderivative size = 303, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {823, 829, 827, 1166, 208} \[ -\frac{\left (-5 \sqrt{a} A \sqrt{c} e+3 a B e+2 A c d\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{4 a^{3/2} \sqrt [4]{c} \left (\sqrt{c} d-\sqrt{a} e\right )^{5/2}}+\frac{\left (5 \sqrt{a} A \sqrt{c} e+3 a B e+2 A c d\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{4 a^{3/2} \sqrt [4]{c} \left (\sqrt{a} e+\sqrt{c} d\right )^{5/2}}+\frac{x (A c d-a B e)+a (B d-A e)}{2 a \left (a-c x^2\right ) \sqrt{d+e x} \left (c d^2-a e^2\right )}-\frac{e \left (5 a A e^2-6 a B d e+A c d^2\right )}{2 a \sqrt{d+e x} \left (c d^2-a e^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/((d + e*x)^(3/2)*(a - c*x^2)^2),x]

[Out]

-(e*(A*c*d^2 - 6*a*B*d*e + 5*a*A*e^2))/(2*a*(c*d^2 - a*e^2)^2*Sqrt[d + e*x]) + (a*(B*d - A*e) + (A*c*d - a*B*e
)*x)/(2*a*(c*d^2 - a*e^2)*Sqrt[d + e*x]*(a - c*x^2)) - ((2*A*c*d + 3*a*B*e - 5*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^(1/4)*(Sqrt[c]*d - Sqrt[a]*e)^(5/2)) + ((2*A*
c*d + 3*a*B*e + 5*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^(1/4)*(Sqrt[c]*d + Sqrt[a]*e)^(5/2))

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 829

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

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}{(d+e x)^{3/2} \left (a-c x^2\right )^2} \, dx &=\frac{a (B d-A e)+(A c d-a B e) x}{2 a \left (c d^2-a e^2\right ) \sqrt{d+e x} \left (a-c x^2\right )}-\frac{\int \frac{-\frac{1}{2} c \left (2 A c d^2+3 a B d e-5 a A e^2\right )-\frac{3}{2} c e (A c d-a B e) x}{(d+e x)^{3/2} \left (a-c x^2\right )} \, dx}{2 a c \left (c d^2-a e^2\right )}\\ &=-\frac{e \left (A c d^2-6 a B d e+5 a A e^2\right )}{2 a \left (c d^2-a e^2\right )^2 \sqrt{d+e x}}+\frac{a (B d-A e)+(A c d-a B e) x}{2 a \left (c d^2-a e^2\right ) \sqrt{d+e x} \left (a-c x^2\right )}+\frac{\int \frac{\frac{1}{2} c \left (2 A c d \left (c d^2-4 a e^2\right )+3 a B e \left (c d^2+a e^2\right )\right )+\frac{1}{2} c^2 e \left (A c d^2-6 a B d e+5 a A e^2\right ) x}{\sqrt{d+e x} \left (a-c x^2\right )} \, dx}{2 a c \left (c d^2-a e^2\right )^2}\\ &=-\frac{e \left (A c d^2-6 a B d e+5 a A e^2\right )}{2 a \left (c d^2-a e^2\right )^2 \sqrt{d+e x}}+\frac{a (B d-A e)+(A c d-a B e) x}{2 a \left (c d^2-a e^2\right ) \sqrt{d+e x} \left (a-c x^2\right )}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{1}{2} c^2 d e \left (A c d^2-6 a B d e+5 a A e^2\right )+\frac{1}{2} c e \left (2 A c d \left (c d^2-4 a e^2\right )+3 a B e \left (c d^2+a e^2\right )\right )+\frac{1}{2} c^2 e \left (A c d^2-6 a B d e+5 a A e^2\right ) 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 )^2}\\ &=-\frac{e \left (A c d^2-6 a B d e+5 a A e^2\right )}{2 a \left (c d^2-a e^2\right )^2 \sqrt{d+e x}}+\frac{a (B d-A e)+(A c d-a B e) x}{2 a \left (c d^2-a e^2\right ) \sqrt{d+e x} \left (a-c x^2\right )}-\frac{\left (\sqrt{c} \left (2 A c d+3 a B e-5 \sqrt{a} A \sqrt{c} e\right )\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 )^2}+\frac{\left (\sqrt{c} \left (2 A c d+3 a B e+5 \sqrt{a} A \sqrt{c} e\right )\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 )^2}\\ &=-\frac{e \left (A c d^2-6 a B d e+5 a A e^2\right )}{2 a \left (c d^2-a e^2\right )^2 \sqrt{d+e x}}+\frac{a (B d-A e)+(A c d-a B e) x}{2 a \left (c d^2-a e^2\right ) \sqrt{d+e x} \left (a-c x^2\right )}-\frac{\left (2 A c d+3 a B e-5 \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} \sqrt [4]{c} \left (\sqrt{c} d-\sqrt{a} e\right )^{5/2}}+\frac{\left (2 A c d+3 a B e+5 \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} \sqrt [4]{c} \left (\sqrt{c} d+\sqrt{a} e\right )^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.699346, size = 365, normalized size = 1.2 \[ \frac{-\frac{3 c^{3/4} (A c d-a B e) \left (\frac{\tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{a} e+\sqrt{c} d}}\right )}{\sqrt{\sqrt{a} e+\sqrt{c} d}}-\frac{\tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{d+e x}}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{\sqrt{\sqrt{c} d-\sqrt{a} e}}\right )}{2 \sqrt{a}}+\frac{c \left (5 a A e^2-6 a B d e+A c d^2\right ) \left (\left (\sqrt{a} e+\sqrt{c} d\right ) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-\sqrt{a} e}\right )+\left (\sqrt{a} e-\sqrt{c} d\right ) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{a} e}\right )\right )}{2 \sqrt{a} \sqrt{d+e x} \left (c d^2-a e^2\right )}+\frac{c (-a A e+a B (d-e x)+A c d x)}{\left (c x^2-a\right ) \sqrt{d+e x}}}{2 a c \left (a e^2-c d^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/((d + e*x)^(3/2)*(a - c*x^2)^2),x]

[Out]

((c*(-(a*A*e) + A*c*d*x + a*B*(d - e*x)))/(Sqrt[d + e*x]*(-a + c*x^2)) - (3*c^(3/4)*(A*c*d - a*B*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]]/Sqrt[Sqrt[c]*d + Sqrt[a]*e]))/(2*Sqrt[a]) + (c*(A*c*d^2 - 6*a*B*d*e + 5*a*
A*e^2)*((Sqrt[c]*d + Sqrt[a]*e)*Hypergeometric2F1[-1/2, 1, 1/2, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[a]*e)] +
(-(Sqrt[c]*d) + Sqrt[a]*e)*Hypergeometric2F1[-1/2, 1, 1/2, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[a]*e)]))/(2*
Sqrt[a]*(c*d^2 - a*e^2)*Sqrt[d + e*x]))/(2*a*c*(-(c*d^2) + a*e^2))

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Maple [B]  time = 0.047, size = 1392, normalized size = 4.6 \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)^(3/2)/(-c*x^2+a)^2,x)

[Out]

-1/2*e^3/(a*e^2-c*d^2)^2/(c*e^2*x^2-a*e^2)*c*(e*x+d)^(3/2)*A-1/2*e/(a*e^2-c*d^2)^2/(c*e^2*x^2-a*e^2)*c^2/a*(e*
x+d)^(3/2)*A*d^2+e^2/(a*e^2-c*d^2)^2/(c*e^2*x^2-a*e^2)*c*(e*x+d)^(3/2)*B*d+3/2*e^3/(a*e^2-c*d^2)^2/(c*e^2*x^2-
a*e^2)*(e*x+d)^(1/2)*A*c*d+1/2*e/(a*e^2-c*d^2)^2/(c*e^2*x^2-a*e^2)/a*(e*x+d)^(1/2)*A*d^3*c^2-1/2*e^4/(a*e^2-c*
d^2)^2/(c*e^2*x^2-a*e^2)*a*(e*x+d)^(1/2)*B-3/2*e^2/(a*e^2-c*d^2)^2/(c*e^2*x^2-a*e^2)*(e*x+d)^(1/2)*B*c*d^2-2*e
^3/(a*e^2-c*d^2)^2*c^2/(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+1/2*e/(a*e^2-c*d^2)^2/a*c^3/(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^3+3/4*e^4/(a*e^2-c*d^2)^2*a*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))*B+3/4*e^2/(a*e^2-c*d^2)^2*c^2/(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*d^2+5/4*e
^3/(a*e^2-c*d^2)^2*c/((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/(a*e^2-c*d^2)^2/a*c^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^2-3/2*e^2/(a*e^2-c*d^2)^2*c/((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*d-2*e^3/(a*e^2-c*d^2)^2*c^2/(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+1/2*e/(a*e^2-c*d^2)^2/a*c^3/(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^3+3/4*e^4/(a*e^2-c*d^2)^2*a*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))*B+3/4*
e^2/(a*e^2-c*d^2)^2*c^2/(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*d^2-5/4*e^3/(a*e^2-c*d^2)^2*c/((-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-1/4*e/(a*e^2-c*d^2)^2/a*c^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^2+3/2*e^2/(a*e^2-c*d^2)^2*c/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2)*arc
tan((e*x+d)^(1/2)*c/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2))*B*d-2*e^3/(a*e^2-c*d^2)^2/(e*x+d)^(1/2)*A+2*e^2/(a*e^2-c
*d^2)^2/(e*x+d)^(1/2)*B*d

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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}{\left (e x + d\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)^(3/2)/(-c*x^2+a)^2,x, algorithm="maxima")

[Out]

integrate((B*x + A)/((c*x^2 - a)^2*(e*x + d)^(3/2)), 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)^(3/2)/(-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)**(3/2)/(-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)^(3/2)/(-c*x^2+a)^2,x, algorithm="giac")

[Out]

Timed out