3.1868 \(\int \frac{(A+B x) (d+e x)^{3/2}}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=280 \[ -\frac{(d+e x)^{5/2} (A b-a B)}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac{(d+e x)^{3/2} (-5 a B e+A b e+4 b B d)}{4 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}+\frac{3 e (a+b x) \sqrt{d+e x} (-5 a B e+A b e+4 b B d)}{4 b^3 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac{3 e (a+b x) (-5 a B e+A b e+4 b B d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 b^{7/2} \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{b d-a e}} \]

[Out]

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

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Rubi [A]  time = 0.229144, antiderivative size = 280, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {770, 78, 47, 50, 63, 208} \[ -\frac{(d+e x)^{5/2} (A b-a B)}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac{(d+e x)^{3/2} (-5 a B e+A b e+4 b B d)}{4 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}+\frac{3 e (a+b x) \sqrt{d+e x} (-5 a B e+A b e+4 b B d)}{4 b^3 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac{3 e (a+b x) (-5 a B e+A b e+4 b B d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 b^{7/2} \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{b d-a e}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac{\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac{(A+B x) (d+e x)^{3/2}}{\left (a b+b^2 x\right )^3} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{(A b-a B) (d+e x)^{5/2}}{2 b (b d-a e) (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left ((4 b B d+A b e-5 a B e) \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^{3/2}}{\left (a b+b^2 x\right )^2} \, dx}{4 (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{(4 b B d+A b e-5 a B e) (d+e x)^{3/2}}{4 b^2 (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(A b-a B) (d+e x)^{5/2}}{2 b (b d-a e) (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (3 e (4 b B d+A b e-5 a B e) \left (a b+b^2 x\right )\right ) \int \frac{\sqrt{d+e x}}{a b+b^2 x} \, dx}{8 b^2 (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{3 e (4 b B d+A b e-5 a B e) (a+b x) \sqrt{d+e x}}{4 b^3 (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(4 b B d+A b e-5 a B e) (d+e x)^{3/2}}{4 b^2 (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(A b-a B) (d+e x)^{5/2}}{2 b (b d-a e) (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (3 e \left (b^2 d-a b e\right ) (4 b B d+A b e-5 a B e) \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right ) \sqrt{d+e x}} \, dx}{8 b^4 (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{3 e (4 b B d+A b e-5 a B e) (a+b x) \sqrt{d+e x}}{4 b^3 (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(4 b B d+A b e-5 a B e) (d+e x)^{3/2}}{4 b^2 (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(A b-a B) (d+e x)^{5/2}}{2 b (b d-a e) (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (3 \left (b^2 d-a b e\right ) (4 b B d+A b e-5 a B e) \left (a b+b^2 x\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a b-\frac{b^2 d}{e}+\frac{b^2 x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{4 b^4 (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{3 e (4 b B d+A b e-5 a B e) (a+b x) \sqrt{d+e x}}{4 b^3 (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(4 b B d+A b e-5 a B e) (d+e x)^{3/2}}{4 b^2 (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(A b-a B) (d+e x)^{5/2}}{2 b (b d-a e) (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 e (4 b B d+A b e-5 a B e) (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 b^{7/2} \sqrt{b d-a e} \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0961638, size = 110, normalized size = 0.39 \[ \frac{(a+b x) (d+e x)^{5/2} \left (\frac{e (a+b x)^2 (-5 a B e+A b e+4 b B d) \, _2F_1\left (2,\frac{5}{2};\frac{7}{2};\frac{b (d+e x)}{b d-a e}\right )}{(b d-a e)^2}+5 a B-5 A b\right )}{10 b \left ((a+b x)^2\right )^{3/2} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)*(d + e*x)^(5/2)*(-5*A*b + 5*a*B + (e*(4*b*B*d + A*b*e - 5*a*B*e)*(a + b*x)^2*Hypergeometric2F1[2, 5
/2, 7/2, (b*(d + e*x))/(b*d - a*e)])/(b*d - a*e)^2))/(10*b*(b*d - a*e)*((a + b*x)^2)^(3/2))

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Maple [B]  time = 0.019, size = 608, normalized size = 2.2 \begin{align*} -{\frac{bx+a}{4\,{b}^{3}e} \left ( -3\,A\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){x}^{2}{b}^{3}{e}^{3}+15\,B\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){x}^{2}a{b}^{2}{e}^{3}-12\,B\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){x}^{2}{b}^{3}d{e}^{2}-6\,A\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) xa{b}^{2}{e}^{3}-8\,B\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{x}^{2}{b}^{2}{e}^{2}+30\,B\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) x{a}^{2}b{e}^{3}-24\,B\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) xa{b}^{2}d{e}^{2}+5\,A\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}{b}^{2}e-3\,A\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){a}^{2}b{e}^{3}-9\,B\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}abe+4\,B\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}{b}^{2}d-16\,B\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}xab{e}^{2}+15\,B\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){a}^{3}{e}^{3}-12\,B\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){a}^{2}bd{e}^{2}+3\,A\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}ab{e}^{2}-3\,A\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{b}^{2}de-15\,B\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{a}^{2}{e}^{2}+11\,B\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}abde-4\,B\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{b}^{2}{d}^{2} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

-1/4*(-3*A*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*x^2*b^3*e^3+15*B*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(
1/2))*x^2*a*b^2*e^3-12*B*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*x^2*b^3*d*e^2-6*A*arctan((e*x+d)^(1/2)*b/
((a*e-b*d)*b)^(1/2))*x*a*b^2*e^3-8*B*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*x^2*b^2*e^2+30*B*arctan((e*x+d)^(1/2)*b
/((a*e-b*d)*b)^(1/2))*x*a^2*b*e^3-24*B*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*x*a*b^2*d*e^2+5*A*((a*e-b*d
)*b)^(1/2)*(e*x+d)^(3/2)*b^2*e-3*A*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*a^2*b*e^3-9*B*((a*e-b*d)*b)^(1/
2)*(e*x+d)^(3/2)*a*b*e+4*B*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*b^2*d-16*B*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*x*a*
b*e^2+15*B*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*a^3*e^3-12*B*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2)
)*a^2*b*d*e^2+3*A*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a*b*e^2-3*A*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*b^2*d*e-15*B
*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a^2*e^2+11*B*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a*b*d*e-4*B*((a*e-b*d)*b)^(1
/2)*(e*x+d)^(1/2)*b^2*d^2)/e*(b*x+a)/((a*e-b*d)*b)^(1/2)/b^3/((b*x+a)^2)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x + A\right )}{\left (e x + d\right )}^{\frac{3}{2}}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\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)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((B*x + A)*(e*x + d)^(3/2)/(b^2*x^2 + 2*a*b*x + a^2)^(3/2), x)

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Fricas [A]  time = 1.66612, size = 1432, normalized size = 5.11 \begin{align*} \left [\frac{3 \,{\left (4 \, B a^{2} b d e -{\left (5 \, B a^{3} - A a^{2} b\right )} e^{2} +{\left (4 \, B b^{3} d e -{\left (5 \, B a b^{2} - A b^{3}\right )} e^{2}\right )} x^{2} + 2 \,{\left (4 \, B a b^{2} d e -{\left (5 \, B a^{2} b - A a b^{2}\right )} e^{2}\right )} x\right )} \sqrt{b^{2} d - a b e} \log \left (\frac{b e x + 2 \, b d - a e - 2 \, \sqrt{b^{2} d - a b e} \sqrt{e x + d}}{b x + a}\right ) - 2 \,{\left (2 \,{\left (B a b^{3} + A b^{4}\right )} d^{2} -{\left (17 \, B a^{2} b^{2} - A a b^{3}\right )} d e + 3 \,{\left (5 \, B a^{3} b - A a^{2} b^{2}\right )} e^{2} - 8 \,{\left (B b^{4} d e - B a b^{3} e^{2}\right )} x^{2} +{\left (4 \, B b^{4} d^{2} -{\left (29 \, B a b^{3} - 5 \, A b^{4}\right )} d e + 5 \,{\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} e^{2}\right )} x\right )} \sqrt{e x + d}}{8 \,{\left (a^{2} b^{5} d - a^{3} b^{4} e +{\left (b^{7} d - a b^{6} e\right )} x^{2} + 2 \,{\left (a b^{6} d - a^{2} b^{5} e\right )} x\right )}}, \frac{3 \,{\left (4 \, B a^{2} b d e -{\left (5 \, B a^{3} - A a^{2} b\right )} e^{2} +{\left (4 \, B b^{3} d e -{\left (5 \, B a b^{2} - A b^{3}\right )} e^{2}\right )} x^{2} + 2 \,{\left (4 \, B a b^{2} d e -{\left (5 \, B a^{2} b - A a b^{2}\right )} e^{2}\right )} x\right )} \sqrt{-b^{2} d + a b e} \arctan \left (\frac{\sqrt{-b^{2} d + a b e} \sqrt{e x + d}}{b e x + b d}\right ) -{\left (2 \,{\left (B a b^{3} + A b^{4}\right )} d^{2} -{\left (17 \, B a^{2} b^{2} - A a b^{3}\right )} d e + 3 \,{\left (5 \, B a^{3} b - A a^{2} b^{2}\right )} e^{2} - 8 \,{\left (B b^{4} d e - B a b^{3} e^{2}\right )} x^{2} +{\left (4 \, B b^{4} d^{2} -{\left (29 \, B a b^{3} - 5 \, A b^{4}\right )} d e + 5 \,{\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} e^{2}\right )} x\right )} \sqrt{e x + d}}{4 \,{\left (a^{2} b^{5} d - a^{3} b^{4} e +{\left (b^{7} d - a b^{6} e\right )} x^{2} + 2 \,{\left (a b^{6} d - a^{2} b^{5} e\right )} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

[1/8*(3*(4*B*a^2*b*d*e - (5*B*a^3 - A*a^2*b)*e^2 + (4*B*b^3*d*e - (5*B*a*b^2 - A*b^3)*e^2)*x^2 + 2*(4*B*a*b^2*
d*e - (5*B*a^2*b - A*a*b^2)*e^2)*x)*sqrt(b^2*d - a*b*e)*log((b*e*x + 2*b*d - a*e - 2*sqrt(b^2*d - a*b*e)*sqrt(
e*x + d))/(b*x + a)) - 2*(2*(B*a*b^3 + A*b^4)*d^2 - (17*B*a^2*b^2 - A*a*b^3)*d*e + 3*(5*B*a^3*b - A*a^2*b^2)*e
^2 - 8*(B*b^4*d*e - B*a*b^3*e^2)*x^2 + (4*B*b^4*d^2 - (29*B*a*b^3 - 5*A*b^4)*d*e + 5*(5*B*a^2*b^2 - A*a*b^3)*e
^2)*x)*sqrt(e*x + d))/(a^2*b^5*d - a^3*b^4*e + (b^7*d - a*b^6*e)*x^2 + 2*(a*b^6*d - a^2*b^5*e)*x), 1/4*(3*(4*B
*a^2*b*d*e - (5*B*a^3 - A*a^2*b)*e^2 + (4*B*b^3*d*e - (5*B*a*b^2 - A*b^3)*e^2)*x^2 + 2*(4*B*a*b^2*d*e - (5*B*a
^2*b - A*a*b^2)*e^2)*x)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) - (2*(B*
a*b^3 + A*b^4)*d^2 - (17*B*a^2*b^2 - A*a*b^3)*d*e + 3*(5*B*a^3*b - A*a^2*b^2)*e^2 - 8*(B*b^4*d*e - B*a*b^3*e^2
)*x^2 + (4*B*b^4*d^2 - (29*B*a*b^3 - 5*A*b^4)*d*e + 5*(5*B*a^2*b^2 - A*a*b^3)*e^2)*x)*sqrt(e*x + d))/(a^2*b^5*
d - a^3*b^4*e + (b^7*d - a*b^6*e)*x^2 + 2*(a*b^6*d - a^2*b^5*e)*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)/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.22162, size = 421, normalized size = 1.5 \begin{align*} \frac{2 \, \sqrt{x e + d} B e}{b^{3} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} + \frac{3 \,{\left (4 \, B b d e^{2} - 5 \, B a e^{3} + A b e^{3}\right )} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right ) e^{\left (-1\right )}}{4 \, \sqrt{-b^{2} d + a b e} b^{3} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} - \frac{{\left (4 \,{\left (x e + d\right )}^{\frac{3}{2}} B b^{2} d e^{2} - 4 \, \sqrt{x e + d} B b^{2} d^{2} e^{2} - 9 \,{\left (x e + d\right )}^{\frac{3}{2}} B a b e^{3} + 5 \,{\left (x e + d\right )}^{\frac{3}{2}} A b^{2} e^{3} + 11 \, \sqrt{x e + d} B a b d e^{3} - 3 \, \sqrt{x e + d} A b^{2} d e^{3} - 7 \, \sqrt{x e + d} B a^{2} e^{4} + 3 \, \sqrt{x e + d} A a b e^{4}\right )} e^{\left (-1\right )}}{4 \,{\left ({\left (x e + d\right )} b - b d + a e\right )}^{2} b^{3} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(x*e + d)*B*e/(b^3*sgn((x*e + d)*b*e - b*d*e + a*e^2)) + 3/4*(4*B*b*d*e^2 - 5*B*a*e^3 + A*b*e^3)*arctan(
sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^(-1)/(sqrt(-b^2*d + a*b*e)*b^3*sgn((x*e + d)*b*e - b*d*e + a*e^2)) - 1
/4*(4*(x*e + d)^(3/2)*B*b^2*d*e^2 - 4*sqrt(x*e + d)*B*b^2*d^2*e^2 - 9*(x*e + d)^(3/2)*B*a*b*e^3 + 5*(x*e + d)^
(3/2)*A*b^2*e^3 + 11*sqrt(x*e + d)*B*a*b*d*e^3 - 3*sqrt(x*e + d)*A*b^2*d*e^3 - 7*sqrt(x*e + d)*B*a^2*e^4 + 3*s
qrt(x*e + d)*A*a*b*e^4)*e^(-1)/(((x*e + d)*b - b*d + a*e)^2*b^3*sgn((x*e + d)*b*e - b*d*e + a*e^2))