### 3.1656 $$\int \frac{(d+e x)^{7/2}}{(a^2+2 a b x+b^2 x^2)^2} \, dx$$

Optimal. Leaf size=145 $-\frac{35 e^2 (d+e x)^{3/2}}{24 b^3 (a+b x)}-\frac{35 e^3 \sqrt{b d-a e} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 b^{9/2}}-\frac{7 e (d+e x)^{5/2}}{12 b^2 (a+b x)^2}-\frac{(d+e x)^{7/2}}{3 b (a+b x)^3}+\frac{35 e^3 \sqrt{d+e x}}{8 b^4}$

[Out]

(35*e^3*Sqrt[d + e*x])/(8*b^4) - (35*e^2*(d + e*x)^(3/2))/(24*b^3*(a + b*x)) - (7*e*(d + e*x)^(5/2))/(12*b^2*(
a + b*x)^2) - (d + e*x)^(7/2)/(3*b*(a + b*x)^3) - (35*e^3*Sqrt[b*d - a*e]*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt
[b*d - a*e]])/(8*b^(9/2))

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Rubi [A]  time = 0.0692103, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 28, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.179, Rules used = {27, 47, 50, 63, 208} $-\frac{35 e^2 (d+e x)^{3/2}}{24 b^3 (a+b x)}-\frac{35 e^3 \sqrt{b d-a e} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 b^{9/2}}-\frac{7 e (d+e x)^{5/2}}{12 b^2 (a+b x)^2}-\frac{(d+e x)^{7/2}}{3 b (a+b x)^3}+\frac{35 e^3 \sqrt{d+e x}}{8 b^4}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(35*e^3*Sqrt[d + e*x])/(8*b^4) - (35*e^2*(d + e*x)^(3/2))/(24*b^3*(a + b*x)) - (7*e*(d + e*x)^(5/2))/(12*b^2*(
a + b*x)^2) - (d + e*x)^(7/2)/(3*b*(a + b*x)^3) - (35*e^3*Sqrt[b*d - a*e]*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt
[b*d - a*e]])/(8*b^(9/2))

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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

Mathematica [C]  time = 0.0186881, size = 52, normalized size = 0.36 $\frac{2 e^3 (d+e x)^{9/2} \, _2F_1\left (4,\frac{9}{2};\frac{11}{2};-\frac{b (d+e x)}{a e-b d}\right )}{9 (a e-b d)^4}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*e^3*(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)

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Maple [B]  time = 0.208, size = 352, normalized size = 2.4 \begin{align*} 2\,{\frac{{e}^{3}\sqrt{ex+d}}{{b}^{4}}}+{\frac{29\,{e}^{4}a}{8\,{b}^{2} \left ( bxe+ae \right ) ^{3}} \left ( ex+d \right ) ^{{\frac{5}{2}}}}-{\frac{29\,{e}^{3}d}{8\,b \left ( bxe+ae \right ) ^{3}} \left ( ex+d \right ) ^{{\frac{5}{2}}}}+{\frac{17\,{a}^{2}{e}^{5}}{3\,{b}^{3} \left ( bxe+ae \right ) ^{3}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{34\,{e}^{4}ad}{3\,{b}^{2} \left ( bxe+ae \right ) ^{3}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+{\frac{17\,{e}^{3}{d}^{2}}{3\,b \left ( bxe+ae \right ) ^{3}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+{\frac{19\,{a}^{3}{e}^{6}}{8\,{b}^{4} \left ( bxe+ae \right ) ^{3}}\sqrt{ex+d}}-{\frac{57\,{e}^{5}d{a}^{2}}{8\,{b}^{3} \left ( bxe+ae \right ) ^{3}}\sqrt{ex+d}}+{\frac{57\,{e}^{4}a{d}^{2}}{8\,{b}^{2} \left ( bxe+ae \right ) ^{3}}\sqrt{ex+d}}-{\frac{19\,{e}^{3}{d}^{3}}{8\,b \left ( bxe+ae \right ) ^{3}}\sqrt{ex+d}}-{\frac{35\,{e}^{4}a}{8\,{b}^{4}}\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}}+{\frac{35\,{e}^{3}d}{8\,{b}^{3}}\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*e^3*(e*x+d)^(1/2)/b^4+29/8*e^4/b^2/(b*e*x+a*e)^3*(e*x+d)^(5/2)*a-29/8*e^3/b/(b*e*x+a*e)^3*(e*x+d)^(5/2)*d+17
/3*e^5/b^3/(b*e*x+a*e)^3*(e*x+d)^(3/2)*a^2-34/3*e^4/b^2/(b*e*x+a*e)^3*(e*x+d)^(3/2)*a*d+17/3*e^3/b/(b*e*x+a*e)
^3*(e*x+d)^(3/2)*d^2+19/8*e^6/b^4/(b*e*x+a*e)^3*(e*x+d)^(1/2)*a^3-57/8*e^5/b^3/(b*e*x+a*e)^3*(e*x+d)^(1/2)*d*a
^2+57/8*e^4/b^2/(b*e*x+a*e)^3*(e*x+d)^(1/2)*a*d^2-19/8*e^3/b/(b*e*x+a*e)^3*(e*x+d)^(1/2)*d^3-35/8*e^4/b^4/((a*
e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a+35/8*e^3/b^3/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d
)^(1/2)/((a*e-b*d)*b)^(1/2))*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((e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.94319, size = 1065, normalized size = 7.34 \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((e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2)^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((e*x+d)**(7/2)/(b**2*x**2+2*a*b*x+a**2)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.24008, size = 335, normalized size = 2.31 \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}} + \frac{2 \, \sqrt{x e + d} e^{3}}{b^{4}} - \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}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2)^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) + 2*sqrt(x*e +
d)*e^3/b^4 - 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 + 272*(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)