3.1244 \(\int \frac{A+B x}{(d+e x)^{3/2} (b x+c x^2)^2} \, dx\)

Optimal. Leaf size=254 \[ -\frac{e \left (b^2 (-e) (2 B d-3 A e)-b c d (2 A e+B d)+2 A c^2 d^2\right )}{b^2 d^2 \sqrt{d+e x} (c d-b e)^2}+\frac{c^{3/2} \left (7 A b c e-4 A c^2 d-5 b^2 B e+2 b B c d\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b^3 (c d-b e)^{5/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right ) (-3 A b e-4 A c d+2 b B d)}{b^3 d^{5/2}}-\frac{c x (2 A c d-b (A e+B d))+A b (c d-b e)}{b^2 d \left (b x+c x^2\right ) \sqrt{d+e x} (c d-b e)} \]

[Out]

-((e*(2*A*c^2*d^2 - b^2*e*(2*B*d - 3*A*e) - b*c*d*(B*d + 2*A*e)))/(b^2*d^2*(c*d - b*e)^2*Sqrt[d + e*x])) - (A*
b*(c*d - b*e) + c*(2*A*c*d - b*(B*d + A*e))*x)/(b^2*d*(c*d - b*e)*Sqrt[d + e*x]*(b*x + c*x^2)) - ((2*b*B*d - 4
*A*c*d - 3*A*b*e)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(b^3*d^(5/2)) + (c^(3/2)*(2*b*B*c*d - 4*A*c^2*d - 5*b^2*B*e
+ 7*A*b*c*e)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(b^3*(c*d - b*e)^(5/2))

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Rubi [A]  time = 0.555835, antiderivative size = 254, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {822, 828, 826, 1166, 208} \[ -\frac{e \left (b^2 (-e) (2 B d-3 A e)-b c d (2 A e+B d)+2 A c^2 d^2\right )}{b^2 d^2 \sqrt{d+e x} (c d-b e)^2}+\frac{c^{3/2} \left (7 A b c e-4 A c^2 d-5 b^2 B e+2 b B c d\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b^3 (c d-b e)^{5/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right ) (-3 A b e-4 A c d+2 b B d)}{b^3 d^{5/2}}-\frac{c x (2 A c d-b (A e+B d))+A b (c d-b e)}{b^2 d \left (b x+c x^2\right ) \sqrt{d+e x} (c d-b e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((e*(2*A*c^2*d^2 - b^2*e*(2*B*d - 3*A*e) - b*c*d*(B*d + 2*A*e)))/(b^2*d^2*(c*d - b*e)^2*Sqrt[d + e*x])) - (A*
b*(c*d - b*e) + c*(2*A*c*d - b*(B*d + A*e))*x)/(b^2*d*(c*d - b*e)*Sqrt[d + e*x]*(b*x + c*x^2)) - ((2*b*B*d - 4
*A*c*d - 3*A*b*e)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(b^3*d^(5/2)) + (c^(3/2)*(2*b*B*c*d - 4*A*c^2*d - 5*b^2*B*e
+ 7*A*b*c*e)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(b^3*(c*d - b*e)^(5/2))

Rule 822

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

Rule 828

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

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]

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 (b x+c x^2\right )^2} \, dx &=-\frac{A b (c d-b e)+c (2 A c d-b (B d+A e)) x}{b^2 d (c d-b e) \sqrt{d+e x} \left (b x+c x^2\right )}-\frac{\int \frac{-\frac{1}{2} (c d-b e) (2 b B d-4 A c d-3 A b e)-\frac{3}{2} c e (b B d-2 A c d+A b e) x}{(d+e x)^{3/2} \left (b x+c x^2\right )} \, dx}{b^2 d (c d-b e)}\\ &=-\frac{e \left (2 A c^2 d^2-b^2 e (2 B d-3 A e)-b c d (B d+2 A e)\right )}{b^2 d^2 (c d-b e)^2 \sqrt{d+e x}}-\frac{A b (c d-b e)+c (2 A c d-b (B d+A e)) x}{b^2 d (c d-b e) \sqrt{d+e x} \left (b x+c x^2\right )}-\frac{\int \frac{-\frac{1}{2} (c d-b e)^2 (2 b B d-4 A c d-3 A b e)+\frac{1}{2} c e \left (2 A c^2 d^2-b^2 e (2 B d-3 A e)-b c d (B d+2 A e)\right ) x}{\sqrt{d+e x} \left (b x+c x^2\right )} \, dx}{b^2 d^2 (c d-b e)^2}\\ &=-\frac{e \left (2 A c^2 d^2-b^2 e (2 B d-3 A e)-b c d (B d+2 A e)\right )}{b^2 d^2 (c d-b e)^2 \sqrt{d+e x}}-\frac{A b (c d-b e)+c (2 A c d-b (B d+A e)) x}{b^2 d (c d-b e) \sqrt{d+e x} \left (b x+c x^2\right )}-\frac{2 \operatorname{Subst}\left (\int \frac{-\frac{1}{2} e (c d-b e)^2 (2 b B d-4 A c d-3 A b e)-\frac{1}{2} c d e \left (2 A c^2 d^2-b^2 e (2 B d-3 A e)-b c d (B d+2 A e)\right )+\frac{1}{2} c e \left (2 A c^2 d^2-b^2 e (2 B d-3 A e)-b c d (B d+2 A e)\right ) 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 )}{b^2 d^2 (c d-b e)^2}\\ &=-\frac{e \left (2 A c^2 d^2-b^2 e (2 B d-3 A e)-b c d (B d+2 A e)\right )}{b^2 d^2 (c d-b e)^2 \sqrt{d+e x}}-\frac{A b (c d-b e)+c (2 A c d-b (B d+A e)) x}{b^2 d (c d-b e) \sqrt{d+e x} \left (b x+c x^2\right )}+\frac{(c (2 b B d-4 A c d-3 A 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^3 d^2}-\frac{\left (2 \left (\frac{1}{4} c e \left (2 A c^2 d^2-b^2 e (2 B d-3 A e)-b c d (B d+2 A e)\right )-\frac{-\frac{1}{2} c e (-2 c d+b e) \left (2 A c^2 d^2-b^2 e (2 B d-3 A e)-b c d (B d+2 A e)\right )+2 c \left (-\frac{1}{2} e (c d-b e)^2 (2 b B d-4 A c d-3 A b e)-\frac{1}{2} c d e \left (2 A c^2 d^2-b^2 e (2 B d-3 A e)-b c d (B d+2 A e)\right )\right )}{2 b e}\right )\right ) \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^2 d^2 (c d-b e)^2}\\ &=-\frac{e \left (2 A c^2 d^2-b^2 e (2 B d-3 A e)-b c d (B d+2 A e)\right )}{b^2 d^2 (c d-b e)^2 \sqrt{d+e x}}-\frac{A b (c d-b e)+c (2 A c d-b (B d+A e)) x}{b^2 d (c d-b e) \sqrt{d+e x} \left (b x+c x^2\right )}-\frac{(2 b B d-4 A c d-3 A b e) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b^3 d^{5/2}}+\frac{c^{3/2} \left (2 b B c d-4 A c^2 d-5 b^2 B e+7 A b c e\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b^3 (c d-b e)^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.172701, size = 191, normalized size = 0.75 \[ \frac{-x (b+c x) \left (c d^2 \left (b c (7 A e+2 B d)-4 A c^2 d-5 b^2 B e\right ) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{c (d+e x)}{c d-b e}\right )+(c d-b e)^2 \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{e x}{d}+1\right ) (3 A b e+4 A c d-2 b B d)\right )-A b^2 d (c d-b e)^2-b c d x (b e-c d) (A b e-2 A c d+b B d)}{b^3 d^2 x (b+c x) \sqrt{d+e x} (c d-b e)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(A*b^2*d*(c*d - b*e)^2) - b*c*d*(-(c*d) + b*e)*(b*B*d - 2*A*c*d + A*b*e)*x - x*(b + c*x)*(c*d^2*(-4*A*c^2*d
- 5*b^2*B*e + b*c*(2*B*d + 7*A*e))*Hypergeometric2F1[-1/2, 1, 1/2, (c*(d + e*x))/(c*d - b*e)] + (c*d - b*e)^2*
(-2*b*B*d + 4*A*c*d + 3*A*b*e)*Hypergeometric2F1[-1/2, 1, 1/2, 1 + (e*x)/d]))/(b^3*d^2*(c*d - b*e)^2*x*(b + c*
x)*Sqrt[d + e*x])

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Maple [A]  time = 0.029, size = 427, normalized size = 1.7 \begin{align*} -2\,{\frac{{e}^{3}A}{{d}^{2} \left ( be-cd \right ) ^{2}\sqrt{ex+d}}}+2\,{\frac{{e}^{2}B}{d \left ( be-cd \right ) ^{2}\sqrt{ex+d}}}-{\frac{A}{{d}^{2}{b}^{2}x}\sqrt{ex+d}}+3\,{\frac{Ae}{{d}^{5/2}{b}^{2}}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \right ) }+4\,{\frac{Ac}{{d}^{3/2}{b}^{3}}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \right ) }-2\,{\frac{B}{{d}^{3/2}{b}^{2}}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \right ) }-{\frac{e{c}^{3}A}{ \left ( be-cd \right ) ^{2}{b}^{2} \left ( cex+be \right ) }\sqrt{ex+d}}+{\frac{e{c}^{2}B}{ \left ( be-cd \right ) ^{2}b \left ( cex+be \right ) }\sqrt{ex+d}}-7\,{\frac{e{c}^{3}A}{ \left ( be-cd \right ) ^{2}{b}^{2}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }+4\,{\frac{{c}^{4}Ad}{ \left ( be-cd \right ) ^{2}{b}^{3}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }+5\,{\frac{e{c}^{2}B}{ \left ( be-cd \right ) ^{2}b\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-2\,{\frac{{c}^{3}Bd}{ \left ( be-cd \right ) ^{2}{b}^{2}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) } \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+b*x)^2,x)

[Out]

-2*e^3/d^2/(b*e-c*d)^2/(e*x+d)^(1/2)*A+2*e^2/d/(b*e-c*d)^2/(e*x+d)^(1/2)*B-1/d^2/b^2*A*(e*x+d)^(1/2)/x+3*e/d^(
5/2)/b^2*arctanh((e*x+d)^(1/2)/d^(1/2))*A+4/d^(3/2)/b^3*arctanh((e*x+d)^(1/2)/d^(1/2))*A*c-2/d^(3/2)/b^2*arcta
nh((e*x+d)^(1/2)/d^(1/2))*B-e*c^3/(b*e-c*d)^2/b^2*(e*x+d)^(1/2)/(c*e*x+b*e)*A+e*c^2/(b*e-c*d)^2/b*(e*x+d)^(1/2
)/(c*e*x+b*e)*B-7*e*c^3/(b*e-c*d)^2/b^2/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*A+4*c^
4/(b*e-c*d)^2/b^3/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*A*d+5*e*c^2/(b*e-c*d)^2/b/((
b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*B-2*c^3/(b*e-c*d)^2/b^2/((b*e-c*d)*c)^(1/2)*arct
an((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*B*d

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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+b*x)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 85.8818, size = 6530, normalized size = 25.71 \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)^(3/2)/(c*x^2+b*x)^2,x, algorithm="fricas")

[Out]

[1/2*(((2*(B*b*c^3 - 2*A*c^4)*d^4*e - (5*B*b^2*c^2 - 7*A*b*c^3)*d^3*e^2)*x^3 + (2*(B*b*c^3 - 2*A*c^4)*d^5 - 3*
(B*b^2*c^2 - A*b*c^3)*d^4*e - (5*B*b^3*c - 7*A*b^2*c^2)*d^3*e^2)*x^2 + (2*(B*b^2*c^2 - 2*A*b*c^3)*d^5 - (5*B*b
^3*c - 7*A*b^2*c^2)*d^4*e)*x)*sqrt(c/(c*d - b*e))*log((c*e*x + 2*c*d - b*e + 2*(c*d - b*e)*sqrt(e*x + d)*sqrt(
c/(c*d - b*e)))/(c*x + b)) + ((3*A*b^3*c*e^4 - 2*(B*b*c^3 - 2*A*c^4)*d^3*e + (4*B*b^2*c^2 - 5*A*b*c^3)*d^2*e^2
- 2*(B*b^3*c + A*b^2*c^2)*d*e^3)*x^3 + (3*A*b^4*e^4 - 2*(B*b*c^3 - 2*A*c^4)*d^4 + (2*B*b^2*c^2 - A*b*c^3)*d^3
*e + (2*B*b^3*c - 7*A*b^2*c^2)*d^2*e^2 - (2*B*b^4 - A*b^3*c)*d*e^3)*x^2 + (3*A*b^4*d*e^3 - 2*(B*b^2*c^2 - 2*A*
b*c^3)*d^4 + (4*B*b^3*c - 5*A*b^2*c^2)*d^3*e - 2*(B*b^4 + A*b^3*c)*d^2*e^2)*x)*sqrt(d)*log((e*x + 2*sqrt(e*x +
d)*sqrt(d) + 2*d)/x) - 2*(A*b^2*c^2*d^4 - 2*A*b^3*c*d^3*e + A*b^4*d^2*e^2 + (3*A*b^3*c*d*e^3 - (B*b^2*c^2 - 2
*A*b*c^3)*d^3*e - 2*(B*b^3*c + A*b^2*c^2)*d^2*e^2)*x^2 - (A*b^2*c^2*d^3*e - 3*A*b^4*d*e^3 + (B*b^2*c^2 - 2*A*b
*c^3)*d^4 + (2*B*b^4 + A*b^3*c)*d^2*e^2)*x)*sqrt(e*x + d))/((b^3*c^3*d^5*e - 2*b^4*c^2*d^4*e^2 + b^5*c*d^3*e^3
)*x^3 + (b^3*c^3*d^6 - b^4*c^2*d^5*e - b^5*c*d^4*e^2 + b^6*d^3*e^3)*x^2 + (b^4*c^2*d^6 - 2*b^5*c*d^5*e + b^6*d
^4*e^2)*x), 1/2*(2*((2*(B*b*c^3 - 2*A*c^4)*d^4*e - (5*B*b^2*c^2 - 7*A*b*c^3)*d^3*e^2)*x^3 + (2*(B*b*c^3 - 2*A*
c^4)*d^5 - 3*(B*b^2*c^2 - A*b*c^3)*d^4*e - (5*B*b^3*c - 7*A*b^2*c^2)*d^3*e^2)*x^2 + (2*(B*b^2*c^2 - 2*A*b*c^3)
*d^5 - (5*B*b^3*c - 7*A*b^2*c^2)*d^4*e)*x)*sqrt(-c/(c*d - b*e))*arctan(-(c*d - b*e)*sqrt(e*x + d)*sqrt(-c/(c*d
- b*e))/(c*e*x + c*d)) + ((3*A*b^3*c*e^4 - 2*(B*b*c^3 - 2*A*c^4)*d^3*e + (4*B*b^2*c^2 - 5*A*b*c^3)*d^2*e^2 -
2*(B*b^3*c + A*b^2*c^2)*d*e^3)*x^3 + (3*A*b^4*e^4 - 2*(B*b*c^3 - 2*A*c^4)*d^4 + (2*B*b^2*c^2 - A*b*c^3)*d^3*e
+ (2*B*b^3*c - 7*A*b^2*c^2)*d^2*e^2 - (2*B*b^4 - A*b^3*c)*d*e^3)*x^2 + (3*A*b^4*d*e^3 - 2*(B*b^2*c^2 - 2*A*b*c
^3)*d^4 + (4*B*b^3*c - 5*A*b^2*c^2)*d^3*e - 2*(B*b^4 + A*b^3*c)*d^2*e^2)*x)*sqrt(d)*log((e*x + 2*sqrt(e*x + d)
*sqrt(d) + 2*d)/x) - 2*(A*b^2*c^2*d^4 - 2*A*b^3*c*d^3*e + A*b^4*d^2*e^2 + (3*A*b^3*c*d*e^3 - (B*b^2*c^2 - 2*A*
b*c^3)*d^3*e - 2*(B*b^3*c + A*b^2*c^2)*d^2*e^2)*x^2 - (A*b^2*c^2*d^3*e - 3*A*b^4*d*e^3 + (B*b^2*c^2 - 2*A*b*c^
3)*d^4 + (2*B*b^4 + A*b^3*c)*d^2*e^2)*x)*sqrt(e*x + d))/((b^3*c^3*d^5*e - 2*b^4*c^2*d^4*e^2 + b^5*c*d^3*e^3)*x
^3 + (b^3*c^3*d^6 - b^4*c^2*d^5*e - b^5*c*d^4*e^2 + b^6*d^3*e^3)*x^2 + (b^4*c^2*d^6 - 2*b^5*c*d^5*e + b^6*d^4*
e^2)*x), -1/2*(2*((3*A*b^3*c*e^4 - 2*(B*b*c^3 - 2*A*c^4)*d^3*e + (4*B*b^2*c^2 - 5*A*b*c^3)*d^2*e^2 - 2*(B*b^3*
c + A*b^2*c^2)*d*e^3)*x^3 + (3*A*b^4*e^4 - 2*(B*b*c^3 - 2*A*c^4)*d^4 + (2*B*b^2*c^2 - A*b*c^3)*d^3*e + (2*B*b^
3*c - 7*A*b^2*c^2)*d^2*e^2 - (2*B*b^4 - A*b^3*c)*d*e^3)*x^2 + (3*A*b^4*d*e^3 - 2*(B*b^2*c^2 - 2*A*b*c^3)*d^4 +
(4*B*b^3*c - 5*A*b^2*c^2)*d^3*e - 2*(B*b^4 + A*b^3*c)*d^2*e^2)*x)*sqrt(-d)*arctan(sqrt(e*x + d)*sqrt(-d)/d) -
((2*(B*b*c^3 - 2*A*c^4)*d^4*e - (5*B*b^2*c^2 - 7*A*b*c^3)*d^3*e^2)*x^3 + (2*(B*b*c^3 - 2*A*c^4)*d^5 - 3*(B*b^
2*c^2 - A*b*c^3)*d^4*e - (5*B*b^3*c - 7*A*b^2*c^2)*d^3*e^2)*x^2 + (2*(B*b^2*c^2 - 2*A*b*c^3)*d^5 - (5*B*b^3*c
- 7*A*b^2*c^2)*d^4*e)*x)*sqrt(c/(c*d - b*e))*log((c*e*x + 2*c*d - b*e + 2*(c*d - b*e)*sqrt(e*x + d)*sqrt(c/(c*
d - b*e)))/(c*x + b)) + 2*(A*b^2*c^2*d^4 - 2*A*b^3*c*d^3*e + A*b^4*d^2*e^2 + (3*A*b^3*c*d*e^3 - (B*b^2*c^2 - 2
*A*b*c^3)*d^3*e - 2*(B*b^3*c + A*b^2*c^2)*d^2*e^2)*x^2 - (A*b^2*c^2*d^3*e - 3*A*b^4*d*e^3 + (B*b^2*c^2 - 2*A*b
*c^3)*d^4 + (2*B*b^4 + A*b^3*c)*d^2*e^2)*x)*sqrt(e*x + d))/((b^3*c^3*d^5*e - 2*b^4*c^2*d^4*e^2 + b^5*c*d^3*e^3
)*x^3 + (b^3*c^3*d^6 - b^4*c^2*d^5*e - b^5*c*d^4*e^2 + b^6*d^3*e^3)*x^2 + (b^4*c^2*d^6 - 2*b^5*c*d^5*e + b^6*d
^4*e^2)*x), (((2*(B*b*c^3 - 2*A*c^4)*d^4*e - (5*B*b^2*c^2 - 7*A*b*c^3)*d^3*e^2)*x^3 + (2*(B*b*c^3 - 2*A*c^4)*d
^5 - 3*(B*b^2*c^2 - A*b*c^3)*d^4*e - (5*B*b^3*c - 7*A*b^2*c^2)*d^3*e^2)*x^2 + (2*(B*b^2*c^2 - 2*A*b*c^3)*d^5 -
(5*B*b^3*c - 7*A*b^2*c^2)*d^4*e)*x)*sqrt(-c/(c*d - b*e))*arctan(-(c*d - b*e)*sqrt(e*x + d)*sqrt(-c/(c*d - b*e
))/(c*e*x + c*d)) - ((3*A*b^3*c*e^4 - 2*(B*b*c^3 - 2*A*c^4)*d^3*e + (4*B*b^2*c^2 - 5*A*b*c^3)*d^2*e^2 - 2*(B*b
^3*c + A*b^2*c^2)*d*e^3)*x^3 + (3*A*b^4*e^4 - 2*(B*b*c^3 - 2*A*c^4)*d^4 + (2*B*b^2*c^2 - A*b*c^3)*d^3*e + (2*B
*b^3*c - 7*A*b^2*c^2)*d^2*e^2 - (2*B*b^4 - A*b^3*c)*d*e^3)*x^2 + (3*A*b^4*d*e^3 - 2*(B*b^2*c^2 - 2*A*b*c^3)*d^
4 + (4*B*b^3*c - 5*A*b^2*c^2)*d^3*e - 2*(B*b^4 + A*b^3*c)*d^2*e^2)*x)*sqrt(-d)*arctan(sqrt(e*x + d)*sqrt(-d)/d
) - (A*b^2*c^2*d^4 - 2*A*b^3*c*d^3*e + A*b^4*d^2*e^2 + (3*A*b^3*c*d*e^3 - (B*b^2*c^2 - 2*A*b*c^3)*d^3*e - 2*(B
*b^3*c + A*b^2*c^2)*d^2*e^2)*x^2 - (A*b^2*c^2*d^3*e - 3*A*b^4*d*e^3 + (B*b^2*c^2 - 2*A*b*c^3)*d^4 + (2*B*b^4 +
A*b^3*c)*d^2*e^2)*x)*sqrt(e*x + d))/((b^3*c^3*d^5*e - 2*b^4*c^2*d^4*e^2 + b^5*c*d^3*e^3)*x^3 + (b^3*c^3*d^6 -
b^4*c^2*d^5*e - b^5*c*d^4*e^2 + b^6*d^3*e^3)*x^2 + (b^4*c^2*d^6 - 2*b^5*c*d^5*e + b^6*d^4*e^2)*x)]

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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+b*x)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.39892, size = 675, normalized size = 2.66 \begin{align*} -\frac{{\left (2 \, B b c^{3} d - 4 \, A c^{4} d - 5 \, B b^{2} c^{2} e + 7 \, A b c^{3} e\right )} \arctan \left (\frac{\sqrt{x e + d} c}{\sqrt{-c^{2} d + b c e}}\right )}{{\left (b^{3} c^{2} d^{2} - 2 \, b^{4} c d e + b^{5} e^{2}\right )} \sqrt{-c^{2} d + b c e}} + \frac{{\left (x e + d\right )}^{2} B b c^{2} d^{2} e - 2 \,{\left (x e + d\right )}^{2} A c^{3} d^{2} e -{\left (x e + d\right )} B b c^{2} d^{3} e + 2 \,{\left (x e + d\right )} A c^{3} d^{3} e + 2 \,{\left (x e + d\right )}^{2} B b^{2} c d e^{2} + 2 \,{\left (x e + d\right )}^{2} A b c^{2} d e^{2} - 4 \,{\left (x e + d\right )} B b^{2} c d^{2} e^{2} - 3 \,{\left (x e + d\right )} A b c^{2} d^{2} e^{2} + 2 \, B b^{2} c d^{3} e^{2} - 3 \,{\left (x e + d\right )}^{2} A b^{2} c e^{3} + 2 \,{\left (x e + d\right )} B b^{3} d e^{3} + 7 \,{\left (x e + d\right )} A b^{2} c d e^{3} - 2 \, B b^{3} d^{2} e^{3} - 2 \, A b^{2} c d^{2} e^{3} - 3 \,{\left (x e + d\right )} A b^{3} e^{4} + 2 \, A b^{3} d e^{4}}{{\left (b^{2} c^{2} d^{4} - 2 \, b^{3} c d^{3} e + b^{4} d^{2} e^{2}\right )}{\left ({\left (x e + d\right )}^{\frac{5}{2}} c - 2 \,{\left (x e + d\right )}^{\frac{3}{2}} c d + \sqrt{x e + d} c d^{2} +{\left (x e + d\right )}^{\frac{3}{2}} b e - \sqrt{x e + d} b d e\right )}} + \frac{{\left (2 \, B b d - 4 \, A c d - 3 \, A b e\right )} \arctan \left (\frac{\sqrt{x e + d}}{\sqrt{-d}}\right )}{b^{3} \sqrt{-d} d^{2}} \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+b*x)^2,x, algorithm="giac")

[Out]

-(2*B*b*c^3*d - 4*A*c^4*d - 5*B*b^2*c^2*e + 7*A*b*c^3*e)*arctan(sqrt(x*e + d)*c/sqrt(-c^2*d + b*c*e))/((b^3*c^
2*d^2 - 2*b^4*c*d*e + b^5*e^2)*sqrt(-c^2*d + b*c*e)) + ((x*e + d)^2*B*b*c^2*d^2*e - 2*(x*e + d)^2*A*c^3*d^2*e
- (x*e + d)*B*b*c^2*d^3*e + 2*(x*e + d)*A*c^3*d^3*e + 2*(x*e + d)^2*B*b^2*c*d*e^2 + 2*(x*e + d)^2*A*b*c^2*d*e^
2 - 4*(x*e + d)*B*b^2*c*d^2*e^2 - 3*(x*e + d)*A*b*c^2*d^2*e^2 + 2*B*b^2*c*d^3*e^2 - 3*(x*e + d)^2*A*b^2*c*e^3
+ 2*(x*e + d)*B*b^3*d*e^3 + 7*(x*e + d)*A*b^2*c*d*e^3 - 2*B*b^3*d^2*e^3 - 2*A*b^2*c*d^2*e^3 - 3*(x*e + d)*A*b^
3*e^4 + 2*A*b^3*d*e^4)/((b^2*c^2*d^4 - 2*b^3*c*d^3*e + b^4*d^2*e^2)*((x*e + d)^(5/2)*c - 2*(x*e + d)^(3/2)*c*d
+ sqrt(x*e + d)*c*d^2 + (x*e + d)^(3/2)*b*e - sqrt(x*e + d)*b*d*e)) + (2*B*b*d - 4*A*c*d - 3*A*b*e)*arctan(sq
rt(x*e + d)/sqrt(-d))/(b^3*sqrt(-d)*d^2)