3.1240 \(\int \frac{(A+B x) (d+e x)^{5/2}}{(b x+c x^2)^2} \, dx\)

Optimal. Leaf size=225 \[ -\frac{(d+e x)^{3/2} \left (x \left (-b c (A e+B d)+2 A c^2 d+b^2 B e\right )+A b c d\right )}{b^2 c \left (b x+c x^2\right )}+\frac{e \sqrt{d+e x} \left (-b c (A e+B d)+2 A c^2 d+3 b^2 B e\right )}{b^2 c^2}-\frac{(c d-b e)^{3/2} \left (-b c (2 B d-A e)+4 A c^2 d-3 b^2 B e\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b^3 c^{5/2}}-\frac{d^{3/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right ) (5 A b e-4 A c d+2 b B d)}{b^3} \]

[Out]

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

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Rubi [A]  time = 0.582095, antiderivative size = 225, 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 = {818, 824, 826, 1166, 208} \[ -\frac{(d+e x)^{3/2} \left (x \left (-b c (A e+B d)+2 A c^2 d+b^2 B e\right )+A b c d\right )}{b^2 c \left (b x+c x^2\right )}+\frac{e \sqrt{d+e x} \left (-b c (A e+B d)+2 A c^2 d+3 b^2 B e\right )}{b^2 c^2}-\frac{(c d-b e)^{3/2} \left (-b c (2 B d-A e)+4 A c^2 d-3 b^2 B e\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b^3 c^{5/2}}-\frac{d^{3/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right ) (5 A b e-4 A c d+2 b B d)}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 818

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

Rule 824

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

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

Mathematica [A]  time = 1.39871, size = 302, normalized size = 1.34 \[ -\frac{\frac{\frac{2 d \left (b c (2 B d-A e)-4 A c^2 d+3 b^2 B e\right ) \left (\sqrt{c} \sqrt{d+e x} \left (15 b^2 e^2-5 b c e (7 d+e x)+c^2 \left (23 d^2+11 d e x+3 e^2 x^2\right )\right )-15 (c d-b e)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )\right )}{c^{5/2} (c d-b e)}+2 \left (15 d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )-\sqrt{d+e x} \left (23 d^2+11 d e x+3 e^2 x^2\right )\right ) (5 A b e-4 A c d+2 b B d)}{30 b^2}+\frac{c (d+e x)^{7/2} (A b e-2 A c d+b B d)}{b (b+c x) (b e-c d)}+\frac{A (d+e x)^{7/2}}{x (b+c x)}}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((c*(b*B*d - 2*A*c*d + A*b*e)*(d + e*x)^(7/2))/(b*(-(c*d) + b*e)*(b + c*x)) + (A*(d + e*x)^(7/2))/(x*(b + c*
x)) + (2*(2*b*B*d - 4*A*c*d + 5*A*b*e)*(-(Sqrt[d + e*x]*(23*d^2 + 11*d*e*x + 3*e^2*x^2)) + 15*d^(5/2)*ArcTanh[
Sqrt[d + e*x]/Sqrt[d]]) + (2*d*(-4*A*c^2*d + 3*b^2*B*e + b*c*(2*B*d - A*e))*(Sqrt[c]*Sqrt[d + e*x]*(15*b^2*e^2
- 5*b*c*e*(7*d + e*x) + c^2*(23*d^2 + 11*d*e*x + 3*e^2*x^2)) - 15*(c*d - b*e)^(5/2)*ArcTanh[(Sqrt[c]*Sqrt[d +
e*x])/Sqrt[c*d - b*e]]))/(c^(5/2)*(c*d - b*e)))/(30*b^2))/(b*d))

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Maple [B]  time = 0.023, size = 614, normalized size = 2.7 \begin{align*} 2\,{\frac{{e}^{2}B\sqrt{ex+d}}{{c}^{2}}}-{\frac{{d}^{2}A}{{b}^{2}x}\sqrt{ex+d}}-5\,{\frac{e{d}^{3/2}A}{{b}^{2}}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \right ) }+4\,{\frac{{d}^{5/2}Ac}{{b}^{3}}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \right ) }-2\,{\frac{{d}^{5/2}B}{{b}^{2}}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \right ) }-{\frac{{e}^{3}A}{c \left ( cex+be \right ) }\sqrt{ex+d}}+2\,{\frac{{e}^{2}\sqrt{ex+d}Ad}{b \left ( cex+be \right ) }}-{\frac{Ac{d}^{2}e}{{b}^{2} \left ( cex+be \right ) }\sqrt{ex+d}}+{\frac{B{e}^{3}b}{{c}^{2} \left ( cex+be \right ) }\sqrt{ex+d}}-2\,{\frac{{e}^{2}B\sqrt{ex+d}d}{c \left ( cex+be \right ) }}+{\frac{Be{d}^{2}}{b \left ( cex+be \right ) }\sqrt{ex+d}}+{\frac{{e}^{3}A}{c}\arctan \left ({c\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( be-cd \right ) c}}}} \right ){\frac{1}{\sqrt{ \left ( be-cd \right ) c}}}}+2\,{\frac{{e}^{2}Ad}{b\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-7\,{\frac{Ac{d}^{2}e}{{b}^{2}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }+4\,{\frac{A{d}^{3}{c}^{2}}{{b}^{3}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-3\,{\frac{B{e}^{3}b}{{c}^{2}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }+4\,{\frac{{e}^{2}Bd}{c\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }+{\frac{Be{d}^{2}}{b}\arctan \left ({c\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( be-cd \right ) c}}}} \right ){\frac{1}{\sqrt{ \left ( be-cd \right ) c}}}}-2\,{\frac{Bc{d}^{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*e^2*B/c^2*(e*x+d)^(1/2)-d^2/b^2*A*(e*x+d)^(1/2)/x-5*e*d^(3/2)/b^2*arctanh((e*x+d)^(1/2)/d^(1/2))*A+4*d^(5/2)
/b^3*arctanh((e*x+d)^(1/2)/d^(1/2))*A*c-2*d^(5/2)/b^2*arctanh((e*x+d)^(1/2)/d^(1/2))*B-e^3/c*(e*x+d)^(1/2)/(c*
e*x+b*e)*A+2*e^2/b*(e*x+d)^(1/2)/(c*e*x+b*e)*A*d-e*c/b^2*(e*x+d)^(1/2)/(c*e*x+b*e)*A*d^2+e^3/c^2*b*(e*x+d)^(1/
2)/(c*e*x+b*e)*B-2*e^2/c*(e*x+d)^(1/2)/(c*e*x+b*e)*B*d+e/b*(e*x+d)^(1/2)/(c*e*x+b*e)*B*d^2+e^3/c/((b*e-c*d)*c)
^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*A+2*e^2/b/((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/b^2/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*A*d^2+4*c^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^3-3*e^3/c^2*b/((b*e-c*d)*c)^(1/2)*arctan((e
*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*B+4*e^2/c/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*B
*d+e/b/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*B*d^2-2*c/b^2/((b*e-c*d)*c)^(1/2)*arcta
n((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*B*d^3

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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 [A]  time = 71.4824, size = 3272, normalized size = 14.54 \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)^(5/2)/(c*x^2+b*x)^2,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.29958, size = 632, normalized size = 2.81 \begin{align*} \frac{2 \, \sqrt{x e + d} B e^{2}}{c^{2}} + \frac{{\left (2 \, B b d^{3} - 4 \, A c d^{3} + 5 \, A b d^{2} e\right )} \arctan \left (\frac{\sqrt{x e + d}}{\sqrt{-d}}\right )}{b^{3} \sqrt{-d}} - \frac{{\left (2 \, B b c^{3} d^{3} - 4 \, A c^{4} d^{3} - B b^{2} c^{2} d^{2} e + 7 \, A b c^{3} d^{2} e - 4 \, B b^{3} c d e^{2} - 2 \, A b^{2} c^{2} d e^{2} + 3 \, B b^{4} e^{3} - A b^{3} c e^{3}\right )} \arctan \left (\frac{\sqrt{x e + d} c}{\sqrt{-c^{2} d + b c e}}\right )}{\sqrt{-c^{2} d + b c e} b^{3} c^{2}} + \frac{{\left (x e + d\right )}^{\frac{3}{2}} B b c^{2} d^{2} e - 2 \,{\left (x e + d\right )}^{\frac{3}{2}} A c^{3} d^{2} e - \sqrt{x e + d} B b c^{2} d^{3} e + 2 \, \sqrt{x e + d} A c^{3} d^{3} e - 2 \,{\left (x e + d\right )}^{\frac{3}{2}} B b^{2} c d e^{2} + 2 \,{\left (x e + d\right )}^{\frac{3}{2}} A b c^{2} d e^{2} + 2 \, \sqrt{x e + d} B b^{2} c d^{2} e^{2} - 3 \, \sqrt{x e + d} A b c^{2} d^{2} e^{2} +{\left (x e + d\right )}^{\frac{3}{2}} B b^{3} e^{3} -{\left (x e + d\right )}^{\frac{3}{2}} A b^{2} c e^{3} - \sqrt{x e + d} B b^{3} d e^{3} + \sqrt{x e + d} A b^{2} c d e^{3}}{{\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 )} b^{2} c^{2}} \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*sqrt(x*e + d)*B*e^2/c^2 + (2*B*b*d^3 - 4*A*c*d^3 + 5*A*b*d^2*e)*arctan(sqrt(x*e + d)/sqrt(-d))/(b^3*sqrt(-d)
) - (2*B*b*c^3*d^3 - 4*A*c^4*d^3 - B*b^2*c^2*d^2*e + 7*A*b*c^3*d^2*e - 4*B*b^3*c*d*e^2 - 2*A*b^2*c^2*d*e^2 + 3
*B*b^4*e^3 - A*b^3*c*e^3)*arctan(sqrt(x*e + d)*c/sqrt(-c^2*d + b*c*e))/(sqrt(-c^2*d + b*c*e)*b^3*c^2) + ((x*e
+ d)^(3/2)*B*b*c^2*d^2*e - 2*(x*e + d)^(3/2)*A*c^3*d^2*e - sqrt(x*e + d)*B*b*c^2*d^3*e + 2*sqrt(x*e + d)*A*c^3
*d^3*e - 2*(x*e + d)^(3/2)*B*b^2*c*d*e^2 + 2*(x*e + d)^(3/2)*A*b*c^2*d*e^2 + 2*sqrt(x*e + d)*B*b^2*c*d^2*e^2 -
3*sqrt(x*e + d)*A*b*c^2*d^2*e^2 + (x*e + d)^(3/2)*B*b^3*e^3 - (x*e + d)^(3/2)*A*b^2*c*e^3 - sqrt(x*e + d)*B*b
^3*d*e^3 + sqrt(x*e + d)*A*b^2*c*d*e^3)/(((x*e + d)^2*c - 2*(x*e + d)*c*d + c*d^2 + (x*e + d)*b*e - b*d*e)*b^2
*c^2)