3.1245 \(\int \frac{A+B x}{(d+e x)^{5/2} (b x+c x^2)^2} \, dx\)

Optimal. Leaf size=344 \[ -\frac{e \left (-b^2 c d e (6 B d-11 A e)+b^3 e^2 (2 B d-5 A e)-b c^2 d^2 (3 A e+B d)+2 A c^3 d^3\right )}{b^2 d^3 \sqrt{d+e x} (c d-b e)^3}-\frac{e \left (b^2 (-e) (2 B d-5 A e)-3 b c d (2 A e+B d)+6 A c^2 d^2\right )}{3 b^2 d^2 (d+e x)^{3/2} (c d-b e)^2}+\frac{c^{5/2} \left (9 A b c e-4 A c^2 d-7 b^2 B e+2 b B c d\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b^3 (c d-b e)^{7/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right ) (-5 A b e-4 A c d+2 b B d)}{b^3 d^{7/2}}-\frac{c x (2 A c d-b (A e+B d))+A b (c d-b e)}{b^2 d \left (b x+c x^2\right ) (d+e x)^{3/2} (c d-b e)} \]

[Out]

-(e*(6*A*c^2*d^2 - b^2*e*(2*B*d - 5*A*e) - 3*b*c*d*(B*d + 2*A*e)))/(3*b^2*d^2*(c*d - b*e)^2*(d + e*x)^(3/2)) -
(e*(2*A*c^3*d^3 - b^2*c*d*e*(6*B*d - 11*A*e) + b^3*e^2*(2*B*d - 5*A*e) - b*c^2*d^2*(B*d + 3*A*e)))/(b^2*d^3*(
c*d - b*e)^3*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)*(d + e*x)^(
3/2)*(b*x + c*x^2)) - ((2*b*B*d - 4*A*c*d - 5*A*b*e)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(b^3*d^(7/2)) + (c^(5/2)*
(2*b*B*c*d - 4*A*c^2*d - 7*b^2*B*e + 9*A*b*c*e)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(b^3*(c*d -
b*e)^(7/2))

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Rubi [A]  time = 0.917087, antiderivative size = 344, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ -\frac{e \left (-b^2 c d e (6 B d-11 A e)+b^3 e^2 (2 B d-5 A e)-b c^2 d^2 (3 A e+B d)+2 A c^3 d^3\right )}{b^2 d^3 \sqrt{d+e x} (c d-b e)^3}-\frac{e \left (b^2 (-e) (2 B d-5 A e)-3 b c d (2 A e+B d)+6 A c^2 d^2\right )}{3 b^2 d^2 (d+e x)^{3/2} (c d-b e)^2}+\frac{c^{5/2} \left (9 A b c e-4 A c^2 d-7 b^2 B e+2 b B c d\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b^3 (c d-b e)^{7/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right ) (-5 A b e-4 A c d+2 b B d)}{b^3 d^{7/2}}-\frac{c x (2 A c d-b (A e+B d))+A b (c d-b e)}{b^2 d \left (b x+c x^2\right ) (d+e x)^{3/2} (c d-b e)} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/((d + e*x)^(5/2)*(b*x + c*x^2)^2),x]

[Out]

-(e*(6*A*c^2*d^2 - b^2*e*(2*B*d - 5*A*e) - 3*b*c*d*(B*d + 2*A*e)))/(3*b^2*d^2*(c*d - b*e)^2*(d + e*x)^(3/2)) -
(e*(2*A*c^3*d^3 - b^2*c*d*e*(6*B*d - 11*A*e) + b^3*e^2*(2*B*d - 5*A*e) - b*c^2*d^2*(B*d + 3*A*e)))/(b^2*d^3*(
c*d - b*e)^3*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)*(d + e*x)^(
3/2)*(b*x + c*x^2)) - ((2*b*B*d - 4*A*c*d - 5*A*b*e)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(b^3*d^(7/2)) + (c^(5/2)*
(2*b*B*c*d - 4*A*c^2*d - 7*b^2*B*e + 9*A*b*c*e)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(b^3*(c*d -
b*e)^(7/2))

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)^{5/2} \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) (d+e x)^{3/2} \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-5 A b e)-\frac{5}{2} c e (b B d-2 A c d+A b e) x}{(d+e x)^{5/2} \left (b x+c x^2\right )} \, dx}{b^2 d (c d-b e)}\\ &=-\frac{e \left (6 A c^2 d^2-b^2 e (2 B d-5 A e)-3 b c d (B d+2 A e)\right )}{3 b^2 d^2 (c d-b e)^2 (d+e x)^{3/2}}-\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) (d+e x)^{3/2} \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-5 A b e)+\frac{1}{2} c e \left (6 A c^2 d^2-b^2 e (2 B d-5 A e)-3 b c d (B d+2 A e)\right ) x}{(d+e x)^{3/2} \left (b x+c x^2\right )} \, dx}{b^2 d^2 (c d-b e)^2}\\ &=-\frac{e \left (6 A c^2 d^2-b^2 e (2 B d-5 A e)-3 b c d (B d+2 A e)\right )}{3 b^2 d^2 (c d-b e)^2 (d+e x)^{3/2}}-\frac{e \left (2 A c^3 d^3-b^2 c d e (6 B d-11 A e)+b^3 e^2 (2 B d-5 A e)-b c^2 d^2 (B d+3 A e)\right )}{b^2 d^3 (c d-b e)^3 \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) (d+e x)^{3/2} \left (b x+c x^2\right )}-\frac{\int \frac{-\frac{1}{2} (c d-b e)^3 (2 b B d-4 A c d-5 A b e)+\frac{1}{2} c e \left (2 A c^3 d^3-b^2 c d e (6 B d-11 A e)+b^3 e^2 (2 B d-5 A e)-b c^2 d^2 (B d+3 A e)\right ) x}{\sqrt{d+e x} \left (b x+c x^2\right )} \, dx}{b^2 d^3 (c d-b e)^3}\\ &=-\frac{e \left (6 A c^2 d^2-b^2 e (2 B d-5 A e)-3 b c d (B d+2 A e)\right )}{3 b^2 d^2 (c d-b e)^2 (d+e x)^{3/2}}-\frac{e \left (2 A c^3 d^3-b^2 c d e (6 B d-11 A e)+b^3 e^2 (2 B d-5 A e)-b c^2 d^2 (B d+3 A e)\right )}{b^2 d^3 (c d-b e)^3 \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) (d+e x)^{3/2} \left (b x+c x^2\right )}-\frac{2 \operatorname{Subst}\left (\int \frac{-\frac{1}{2} e (c d-b e)^3 (2 b B d-4 A c d-5 A b e)-\frac{1}{2} c d e \left (2 A c^3 d^3-b^2 c d e (6 B d-11 A e)+b^3 e^2 (2 B d-5 A e)-b c^2 d^2 (B d+3 A e)\right )+\frac{1}{2} c e \left (2 A c^3 d^3-b^2 c d e (6 B d-11 A e)+b^3 e^2 (2 B d-5 A e)-b c^2 d^2 (B d+3 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^3 (c d-b e)^3}\\ &=-\frac{e \left (6 A c^2 d^2-b^2 e (2 B d-5 A e)-3 b c d (B d+2 A e)\right )}{3 b^2 d^2 (c d-b e)^2 (d+e x)^{3/2}}-\frac{e \left (2 A c^3 d^3-b^2 c d e (6 B d-11 A e)+b^3 e^2 (2 B d-5 A e)-b c^2 d^2 (B d+3 A e)\right )}{b^2 d^3 (c d-b e)^3 \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) (d+e x)^{3/2} \left (b x+c x^2\right )}+\frac{(c (2 b B d-4 A c d-5 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^3}-\frac{\left (c^3 \left (2 b B c d-4 A c^2 d-7 b^2 B e+9 A b c 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^3 (c d-b e)^3}\\ &=-\frac{e \left (6 A c^2 d^2-b^2 e (2 B d-5 A e)-3 b c d (B d+2 A e)\right )}{3 b^2 d^2 (c d-b e)^2 (d+e x)^{3/2}}-\frac{e \left (2 A c^3 d^3-b^2 c d e (6 B d-11 A e)+b^3 e^2 (2 B d-5 A e)-b c^2 d^2 (B d+3 A e)\right )}{b^2 d^3 (c d-b e)^3 \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) (d+e x)^{3/2} \left (b x+c x^2\right )}-\frac{(2 b B d-4 A c d-5 A b e) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b^3 d^{7/2}}-\frac{c^{5/2} \left (4 A c^2 d+7 b^2 B e-b c (2 B d+9 A e)\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b^3 (c d-b e)^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.1792, size = 194, normalized size = 0.56 \[ \frac{-x (b+c x) \left (c d^2 \left (b c (9 A e+2 B d)-4 A c^2 d-7 b^2 B e\right ) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{c (d+e x)}{c d-b e}\right )+(c d-b e)^2 \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{e x}{d}+1\right ) (5 A b e+4 A c d-2 b B d)\right )-3 A b^2 d (c d-b e)^2-3 b c d x (b e-c d) (A b e-2 A c d+b B d)}{3 b^3 d^2 x (b+c x) (d+e x)^{3/2} (c d-b e)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/((d + e*x)^(5/2)*(b*x + c*x^2)^2),x]

[Out]

(-3*A*b^2*d*(c*d - b*e)^2 - 3*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 - 7*b^2*B*e + b*c*(2*B*d + 9*A*e))*Hypergeometric2F1[-3/2, 1, -1/2, (c*(d + e*x))/(c*d - b*e)] + (c*d - b*e)
^2*(-2*b*B*d + 4*A*c*d + 5*A*b*e)*Hypergeometric2F1[-3/2, 1, -1/2, 1 + (e*x)/d]))/(3*b^3*d^2*(c*d - b*e)^2*x*(
b + c*x)*(d + e*x)^(3/2))

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Maple [A]  time = 0.031, size = 535, normalized size = 1.6 \begin{align*} -{\frac{2\,{e}^{3}A}{3\,{d}^{2} \left ( be-cd \right ) ^{2}} \left ( ex+d \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,{e}^{2}B}{3\,d \left ( be-cd \right ) ^{2}} \left ( ex+d \right ) ^{-{\frac{3}{2}}}}-4\,{\frac{{e}^{4}Ab}{{d}^{3} \left ( be-cd \right ) ^{3}\sqrt{ex+d}}}+8\,{\frac{{e}^{3}Ac}{{d}^{2} \left ( be-cd \right ) ^{3}\sqrt{ex+d}}}+2\,{\frac{B{e}^{3}b}{{d}^{2} \left ( be-cd \right ) ^{3}\sqrt{ex+d}}}-6\,{\frac{{e}^{2}Bc}{d \left ( be-cd \right ) ^{3}\sqrt{ex+d}}}-{\frac{A}{{b}^{2}{d}^{3}x}\sqrt{ex+d}}+5\,{\frac{Ae}{{b}^{2}{d}^{7/2}}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \right ) }+4\,{\frac{Ac}{{b}^{3}{d}^{5/2}}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \right ) }-2\,{\frac{B}{{b}^{2}{d}^{5/2}}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \right ) }+{\frac{e{c}^{4}A}{ \left ( be-cd \right ) ^{3}{b}^{2} \left ( cex+be \right ) }\sqrt{ex+d}}-{\frac{e{c}^{3}B}{ \left ( be-cd \right ) ^{3}b \left ( cex+be \right ) }\sqrt{ex+d}}+9\,{\frac{e{c}^{4}A}{ \left ( be-cd \right ) ^{3}{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}^{5}Ad}{ \left ( be-cd \right ) ^{3}{b}^{3}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-7\,{\frac{e{c}^{3}B}{ \left ( be-cd \right ) ^{3}b\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }+2\,{\frac{{c}^{4}Bd}{ \left ( be-cd \right ) ^{3}{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)^(5/2)/(c*x^2+b*x)^2,x)

[Out]

-2/3*e^3/d^2/(b*e-c*d)^2/(e*x+d)^(3/2)*A+2/3*e^2/d/(b*e-c*d)^2/(e*x+d)^(3/2)*B-4*e^4/d^3/(b*e-c*d)^3/(e*x+d)^(
1/2)*A*b+8*e^3/d^2/(b*e-c*d)^3/(e*x+d)^(1/2)*A*c+2*e^3/d^2/(b*e-c*d)^3/(e*x+d)^(1/2)*B*b-6*e^2/d/(b*e-c*d)^3/(
e*x+d)^(1/2)*B*c-1/b^2/d^3*A*(e*x+d)^(1/2)/x+5*e/b^2/d^(7/2)*arctanh((e*x+d)^(1/2)/d^(1/2))*A+4/b^3/d^(5/2)*ar
ctanh((e*x+d)^(1/2)/d^(1/2))*A*c-2/b^2/d^(5/2)*arctanh((e*x+d)^(1/2)/d^(1/2))*B+e*c^4/(b*e-c*d)^3/b^2*(e*x+d)^
(1/2)/(c*e*x+b*e)*A-e*c^3/(b*e-c*d)^3/b*(e*x+d)^(1/2)/(c*e*x+b*e)*B+9*e*c^4/(b*e-c*d)^3/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^5/(b*e-c*d)^3/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-7*e*c^3/(b*e-c*d)^3/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^4/(b*e-c*d)^3/b^2/((b*e-c*d)*c)^(1/2)*arctan((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)^(5/2)/(c*x^2+b*x)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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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)^(5/2)/(c*x^2+b*x)^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)**(5/2)/(c*x**2+b*x)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.33841, size = 814, normalized size = 2.37 \begin{align*} -\frac{{\left (2 \, B b c^{4} d - 4 \, A c^{5} d - 7 \, B b^{2} c^{3} e + 9 \, A b c^{4} e\right )} \arctan \left (\frac{\sqrt{x e + d} c}{\sqrt{-c^{2} d + b c e}}\right )}{{\left (b^{3} c^{3} d^{3} - 3 \, b^{4} c^{2} d^{2} e + 3 \, b^{5} c d e^{2} - b^{6} e^{3}\right )} \sqrt{-c^{2} d + b c e}} + \frac{{\left (x e + d\right )}^{\frac{3}{2}} B b c^{3} d^{3} e - 2 \,{\left (x e + d\right )}^{\frac{3}{2}} A c^{4} d^{3} e - \sqrt{x e + d} B b c^{3} d^{4} e + 2 \, \sqrt{x e + d} A c^{4} d^{4} e + 3 \,{\left (x e + d\right )}^{\frac{3}{2}} A b c^{3} d^{2} e^{2} - 4 \, \sqrt{x e + d} A b c^{3} d^{3} e^{2} - 3 \,{\left (x e + d\right )}^{\frac{3}{2}} A b^{2} c^{2} d e^{3} + 6 \, \sqrt{x e + d} A b^{2} c^{2} d^{2} e^{3} +{\left (x e + d\right )}^{\frac{3}{2}} A b^{3} c e^{4} - 4 \, \sqrt{x e + d} A b^{3} c d e^{4} + \sqrt{x e + d} A b^{4} e^{5}}{{\left (b^{2} c^{3} d^{6} - 3 \, b^{3} c^{2} d^{5} e + 3 \, b^{4} c d^{4} e^{2} - b^{5} d^{3} e^{3}\right )}{\left ({\left (x e + d\right )}^{2} c - 2 \,{\left (x e + d\right )} c d + c d^{2} +{\left (x e + d\right )} b e - b d e\right )}} + \frac{2 \,{\left (9 \,{\left (x e + d\right )} B c d^{2} e^{2} + B c d^{3} e^{2} - 3 \,{\left (x e + d\right )} B b d e^{3} - 12 \,{\left (x e + d\right )} A c d e^{3} - B b d^{2} e^{3} - A c d^{2} e^{3} + 6 \,{\left (x e + d\right )} A b e^{4} + A b d e^{4}\right )}}{3 \,{\left (c^{3} d^{6} - 3 \, b c^{2} d^{5} e + 3 \, b^{2} c d^{4} e^{2} - b^{3} d^{3} e^{3}\right )}{\left (x e + d\right )}^{\frac{3}{2}}} + \frac{{\left (2 \, B b d - 4 \, A c d - 5 \, A b e\right )} \arctan \left (\frac{\sqrt{x e + d}}{\sqrt{-d}}\right )}{b^{3} \sqrt{-d} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)^(5/2)/(c*x^2+b*x)^2,x, algorithm="giac")

[Out]

-(2*B*b*c^4*d - 4*A*c^5*d - 7*B*b^2*c^3*e + 9*A*b*c^4*e)*arctan(sqrt(x*e + d)*c/sqrt(-c^2*d + b*c*e))/((b^3*c^
3*d^3 - 3*b^4*c^2*d^2*e + 3*b^5*c*d*e^2 - b^6*e^3)*sqrt(-c^2*d + b*c*e)) + ((x*e + d)^(3/2)*B*b*c^3*d^3*e - 2*
(x*e + d)^(3/2)*A*c^4*d^3*e - sqrt(x*e + d)*B*b*c^3*d^4*e + 2*sqrt(x*e + d)*A*c^4*d^4*e + 3*(x*e + d)^(3/2)*A*
b*c^3*d^2*e^2 - 4*sqrt(x*e + d)*A*b*c^3*d^3*e^2 - 3*(x*e + d)^(3/2)*A*b^2*c^2*d*e^3 + 6*sqrt(x*e + d)*A*b^2*c^
2*d^2*e^3 + (x*e + d)^(3/2)*A*b^3*c*e^4 - 4*sqrt(x*e + d)*A*b^3*c*d*e^4 + sqrt(x*e + d)*A*b^4*e^5)/((b^2*c^3*d
^6 - 3*b^3*c^2*d^5*e + 3*b^4*c*d^4*e^2 - b^5*d^3*e^3)*((x*e + d)^2*c - 2*(x*e + d)*c*d + c*d^2 + (x*e + d)*b*e
- b*d*e)) + 2/3*(9*(x*e + d)*B*c*d^2*e^2 + B*c*d^3*e^2 - 3*(x*e + d)*B*b*d*e^3 - 12*(x*e + d)*A*c*d*e^3 - B*b
*d^2*e^3 - A*c*d^2*e^3 + 6*(x*e + d)*A*b*e^4 + A*b*d*e^4)/((c^3*d^6 - 3*b*c^2*d^5*e + 3*b^2*c*d^4*e^2 - b^3*d^
3*e^3)*(x*e + d)^(3/2)) + (2*B*b*d - 4*A*c*d - 5*A*b*e)*arctan(sqrt(x*e + d)/sqrt(-d))/(b^3*sqrt(-d)*d^3)