3.2138 \(\int \frac{(a+b x) (d+e x)^{7/2}}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=241 \[ -\frac{35 e^2 (d+e x)^{3/2}}{24 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{35 e^3 (a+b x) \sqrt{d+e x}}{8 b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{35 e^3 (a+b x) \sqrt{b d-a e} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 b^{9/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{7/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac{7 e (d+e x)^{5/2}}{12 b^2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

-(d + e*x)^(7/2)/(3*b*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)) + (35*e^3*(a + b*x)*Sqrt[d + e*x])/(8*b^4*Sqrt[a^2 + 2*
a*b*x + b^2*x^2]) - (35*e^2*(d + e*x)^(3/2))/(24*b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (7*e*(d + e*x)^(5/2))/(1
2*b^2*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (35*e^3*Sqrt[b*d - a*e]*(a + b*x)*ArcTanh[(Sqrt[b]*Sqrt[d + e
*x])/Sqrt[b*d - a*e]])/(8*b^(9/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

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Rubi [A]  time = 0.137094, antiderivative size = 241, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {768, 646, 47, 50, 63, 208} \[ -\frac{35 e^2 (d+e x)^{3/2}}{24 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{35 e^3 (a+b x) \sqrt{d+e x}}{8 b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{35 e^3 (a+b x) \sqrt{b d-a e} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 b^{9/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{7/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac{7 e (d+e x)^{5/2}}{12 b^2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*(d + e*x)^(7/2))/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

-(d + e*x)^(7/2)/(3*b*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)) + (35*e^3*(a + b*x)*Sqrt[d + e*x])/(8*b^4*Sqrt[a^2 + 2*
a*b*x + b^2*x^2]) - (35*e^2*(d + e*x)^(3/2))/(24*b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (7*e*(d + e*x)^(5/2))/(1
2*b^2*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (35*e^3*Sqrt[b*d - a*e]*(a + b*x)*ArcTanh[(Sqrt[b]*Sqrt[d + e
*x])/Sqrt[b*d - a*e]])/(8*b^(9/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

Rule 768

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

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

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)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx &=-\frac{(d+e x)^{7/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac{(7 e) \int \frac{(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx}{6 b}\\ &=-\frac{(d+e x)^{7/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac{\left (7 b e \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^{5/2}}{\left (a b+b^2 x\right )^3} \, dx}{6 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{(d+e x)^{7/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac{7 e (d+e x)^{5/2}}{12 b^2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (35 e^2 \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^{3/2}}{\left (a b+b^2 x\right )^2} \, dx}{24 b \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{(d+e x)^{7/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac{35 e^2 (d+e x)^{3/2}}{24 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{7 e (d+e x)^{5/2}}{12 b^2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (35 e^3 \left (a b+b^2 x\right )\right ) \int \frac{\sqrt{d+e x}}{a b+b^2 x} \, dx}{16 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{(d+e x)^{7/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac{35 e^3 (a+b x) \sqrt{d+e x}}{8 b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{35 e^2 (d+e x)^{3/2}}{24 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{7 e (d+e x)^{5/2}}{12 b^2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (35 e^3 \left (b^2 d-a b e\right ) \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right ) \sqrt{d+e x}} \, dx}{16 b^5 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{(d+e x)^{7/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac{35 e^3 (a+b x) \sqrt{d+e x}}{8 b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{35 e^2 (d+e x)^{3/2}}{24 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{7 e (d+e x)^{5/2}}{12 b^2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (35 e^2 \left (b^2 d-a b e\right ) \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 )}{8 b^5 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{(d+e x)^{7/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac{35 e^3 (a+b x) \sqrt{d+e x}}{8 b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{35 e^2 (d+e x)^{3/2}}{24 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{7 e (d+e x)^{5/2}}{12 b^2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{35 e^3 \sqrt{b d-a e} (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 b^{9/2} \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0446623, size = 68, normalized size = 0.28 \[ \frac{2 e^3 (a+b x) (d+e x)^{9/2} \, _2F_1\left (4,\frac{9}{2};\frac{11}{2};-\frac{b (d+e x)}{a e-b d}\right )}{9 \sqrt{(a+b x)^2} (a e-b d)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*(d + e*x)^(7/2))/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

(2*e^3*(a + b*x)*(d + e*x)^(9/2)*Hypergeometric2F1[4, 9/2, 11/2, -((b*(d + e*x))/(-(b*d) + a*e))])/(9*(-(b*d)
+ a*e)^4*Sqrt[(a + b*x)^2])

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Maple [B]  time = 0.019, size = 638, normalized size = 2.7 \begin{align*}{\frac{ \left ( bx+a \right ) ^{2}}{24\,{b}^{4}} \left ( -105\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){x}^{3}a{b}^{3}{e}^{4}+105\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){x}^{3}{b}^{4}d{e}^{3}+48\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{x}^{3}{b}^{3}{e}^{3}-315\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){x}^{2}{a}^{2}{b}^{2}{e}^{4}+315\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){x}^{2}a{b}^{3}d{e}^{3}+87\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{5/2}a{b}^{2}e-87\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{5/2}{b}^{3}d+144\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{x}^{2}a{b}^{2}{e}^{3}-315\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) x{a}^{3}b{e}^{4}+315\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) x{a}^{2}{b}^{2}d{e}^{3}+136\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}{a}^{2}b{e}^{2}-272\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}a{b}^{2}de+136\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}{b}^{3}{d}^{2}+144\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}x{a}^{2}b{e}^{3}-105\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){a}^{4}{e}^{4}+105\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){a}^{3}bd{e}^{3}+105\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{a}^{3}{e}^{3}-171\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{a}^{2}bd{e}^{2}+171\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}a{b}^{2}{d}^{2}e-57\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{b}^{3}{d}^{3} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*(e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

1/24*(-105*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*x^3*a*b^3*e^4+105*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^
(1/2))*x^3*b^4*d*e^3+48*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*x^3*b^3*e^3-315*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)
^(1/2))*x^2*a^2*b^2*e^4+315*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*x^2*a*b^3*d*e^3+87*((a*e-b*d)*b)^(1/2)
*(e*x+d)^(5/2)*a*b^2*e-87*((a*e-b*d)*b)^(1/2)*(e*x+d)^(5/2)*b^3*d+144*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*x^2*a*
b^2*e^3-315*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*x*a^3*b*e^4+315*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(
1/2))*x*a^2*b^2*d*e^3+136*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*a^2*b*e^2-272*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*a*
b^2*d*e+136*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*b^3*d^2+144*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*x*a^2*b*e^3-105*ar
ctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*a^4*e^4+105*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*a^3*b*d*e^3+
105*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a^3*e^3-171*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a^2*b*d*e^2+171*((a*e-b*d)
*b)^(1/2)*(e*x+d)^(1/2)*a*b^2*d^2*e-57*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*b^3*d^3)*(b*x+a)^2/((a*e-b*d)*b)^(1/2
)/b^4/((b*x+a)^2)^(5/2)

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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{7}{2}}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")

[Out]

integrate((b*x + a)*(e*x + d)^(7/2)/(b^2*x^2 + 2*a*b*x + a^2)^(5/2), x)

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Fricas [A]  time = 1.05066, size = 1065, normalized size = 4.42 \begin{align*} \left [\frac{105 \,{\left (b^{3} e^{3} x^{3} + 3 \, a b^{2} e^{3} x^{2} + 3 \, a^{2} b e^{3} x + a^{3} e^{3}\right )} \sqrt{\frac{b d - a e}{b}} \log \left (\frac{b e x + 2 \, b d - a e - 2 \, \sqrt{e x + d} b \sqrt{\frac{b d - a e}{b}}}{b x + a}\right ) + 2 \,{\left (48 \, b^{3} e^{3} x^{3} - 8 \, b^{3} d^{3} - 14 \, a b^{2} d^{2} e - 35 \, a^{2} b d e^{2} + 105 \, a^{3} e^{3} - 3 \,{\left (29 \, b^{3} d e^{2} - 77 \, a b^{2} e^{3}\right )} x^{2} - 2 \,{\left (19 \, b^{3} d^{2} e + 49 \, a b^{2} d e^{2} - 140 \, a^{2} b e^{3}\right )} x\right )} \sqrt{e x + d}}{48 \,{\left (b^{7} x^{3} + 3 \, a b^{6} x^{2} + 3 \, a^{2} b^{5} x + a^{3} b^{4}\right )}}, -\frac{105 \,{\left (b^{3} e^{3} x^{3} + 3 \, a b^{2} e^{3} x^{2} + 3 \, a^{2} b e^{3} x + a^{3} e^{3}\right )} \sqrt{-\frac{b d - a e}{b}} \arctan \left (-\frac{\sqrt{e x + d} b \sqrt{-\frac{b d - a e}{b}}}{b d - a e}\right ) -{\left (48 \, b^{3} e^{3} x^{3} - 8 \, b^{3} d^{3} - 14 \, a b^{2} d^{2} e - 35 \, a^{2} b d e^{2} + 105 \, a^{3} e^{3} - 3 \,{\left (29 \, b^{3} d e^{2} - 77 \, a b^{2} e^{3}\right )} x^{2} - 2 \,{\left (19 \, b^{3} d^{2} e + 49 \, a b^{2} d e^{2} - 140 \, a^{2} b e^{3}\right )} x\right )} \sqrt{e x + d}}{24 \,{\left (b^{7} x^{3} + 3 \, a b^{6} x^{2} + 3 \, a^{2} b^{5} x + a^{3} b^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

[1/48*(105*(b^3*e^3*x^3 + 3*a*b^2*e^3*x^2 + 3*a^2*b*e^3*x + a^3*e^3)*sqrt((b*d - a*e)/b)*log((b*e*x + 2*b*d -
a*e - 2*sqrt(e*x + d)*b*sqrt((b*d - a*e)/b))/(b*x + a)) + 2*(48*b^3*e^3*x^3 - 8*b^3*d^3 - 14*a*b^2*d^2*e - 35*
a^2*b*d*e^2 + 105*a^3*e^3 - 3*(29*b^3*d*e^2 - 77*a*b^2*e^3)*x^2 - 2*(19*b^3*d^2*e + 49*a*b^2*d*e^2 - 140*a^2*b
*e^3)*x)*sqrt(e*x + d))/(b^7*x^3 + 3*a*b^6*x^2 + 3*a^2*b^5*x + a^3*b^4), -1/24*(105*(b^3*e^3*x^3 + 3*a*b^2*e^3
*x^2 + 3*a^2*b*e^3*x + a^3*e^3)*sqrt(-(b*d - a*e)/b)*arctan(-sqrt(e*x + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e))
- (48*b^3*e^3*x^3 - 8*b^3*d^3 - 14*a*b^2*d^2*e - 35*a^2*b*d*e^2 + 105*a^3*e^3 - 3*(29*b^3*d*e^2 - 77*a*b^2*e^
3)*x^2 - 2*(19*b^3*d^2*e + 49*a*b^2*d*e^2 - 140*a^2*b*e^3)*x)*sqrt(e*x + d))/(b^7*x^3 + 3*a*b^6*x^2 + 3*a^2*b^
5*x + a^3*b^4)]

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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)**(7/2)/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.26938, size = 432, normalized size = 1.79 \begin{align*} \frac{35 \,{\left (b d e^{3} - a e^{4}\right )} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right )}{8 \, \sqrt{-b^{2} d + a b e} b^{4} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} + \frac{2 \, \sqrt{x e + d} e^{3}}{b^{4} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} - \frac{87 \,{\left (x e + d\right )}^{\frac{5}{2}} b^{3} d e^{3} - 136 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{3} d^{2} e^{3} + 57 \, \sqrt{x e + d} b^{3} d^{3} e^{3} - 87 \,{\left (x e + d\right )}^{\frac{5}{2}} a b^{2} e^{4} + 272 \,{\left (x e + d\right )}^{\frac{3}{2}} a b^{2} d e^{4} - 171 \, \sqrt{x e + d} a b^{2} d^{2} e^{4} - 136 \,{\left (x e + d\right )}^{\frac{3}{2}} a^{2} b e^{5} + 171 \, \sqrt{x e + d} a^{2} b d e^{5} - 57 \, \sqrt{x e + d} a^{3} e^{6}}{24 \,{\left ({\left (x e + d\right )} b - b d + a e\right )}^{3} b^{4} \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)^(7/2)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

35/8*(b*d*e^3 - a*e^4)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*b^4*sgn((x*e + d)*b*
e - b*d*e + a*e^2)) + 2*sqrt(x*e + d)*e^3/(b^4*sgn((x*e + d)*b*e - b*d*e + a*e^2)) - 1/24*(87*(x*e + d)^(5/2)*
b^3*d*e^3 - 136*(x*e + d)^(3/2)*b^3*d^2*e^3 + 57*sqrt(x*e + d)*b^3*d^3*e^3 - 87*(x*e + d)^(5/2)*a*b^2*e^4 + 27
2*(x*e + d)^(3/2)*a*b^2*d*e^4 - 171*sqrt(x*e + d)*a*b^2*d^2*e^4 - 136*(x*e + d)^(3/2)*a^2*b*e^5 + 171*sqrt(x*e
+ d)*a^2*b*d*e^5 - 57*sqrt(x*e + d)*a^3*e^6)/(((x*e + d)*b - b*d + a*e)^3*b^4*sgn((x*e + d)*b*e - b*d*e + a*e
^2))