### 3.1646 $$\int \frac{(d+e x)^{7/2}}{a^2+2 a b x+b^2 x^2} \, dx$$

Optimal. Leaf size=137 $\frac{7 e (d+e x)^{3/2} (b d-a e)}{3 b^3}+\frac{7 e \sqrt{d+e x} (b d-a e)^2}{b^4}-\frac{7 e (b d-a e)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{9/2}}-\frac{(d+e x)^{7/2}}{b (a+b x)}+\frac{7 e (d+e x)^{5/2}}{5 b^2}$

[Out]

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

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Rubi [A]  time = 0.0757445, antiderivative size = 137, 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} $\frac{7 e (d+e x)^{3/2} (b d-a e)}{3 b^3}+\frac{7 e \sqrt{d+e x} (b d-a e)^2}{b^4}-\frac{7 e (b d-a e)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{9/2}}-\frac{(d+e x)^{7/2}}{b (a+b x)}+\frac{7 e (d+e x)^{5/2}}{5 b^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(7*e*(b*d - a*e)^2*Sqrt[d + e*x])/b^4 + (7*e*(b*d - a*e)*(d + e*x)^(3/2))/(3*b^3) + (7*e*(d + e*x)^(5/2))/(5*b
^2) - (d + e*x)^(7/2)/(b*(a + b*x)) - (7*e*(b*d - a*e)^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])
/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}}{a^2+2 a b x+b^2 x^2} \, dx &=\int \frac{(d+e x)^{7/2}}{(a+b x)^2} \, dx\\ &=-\frac{(d+e x)^{7/2}}{b (a+b x)}+\frac{(7 e) \int \frac{(d+e x)^{5/2}}{a+b x} \, dx}{2 b}\\ &=\frac{7 e (d+e x)^{5/2}}{5 b^2}-\frac{(d+e x)^{7/2}}{b (a+b x)}+\frac{(7 e (b d-a e)) \int \frac{(d+e x)^{3/2}}{a+b x} \, dx}{2 b^2}\\ &=\frac{7 e (b d-a e) (d+e x)^{3/2}}{3 b^3}+\frac{7 e (d+e x)^{5/2}}{5 b^2}-\frac{(d+e x)^{7/2}}{b (a+b x)}+\frac{\left (7 e (b d-a e)^2\right ) \int \frac{\sqrt{d+e x}}{a+b x} \, dx}{2 b^3}\\ &=\frac{7 e (b d-a e)^2 \sqrt{d+e x}}{b^4}+\frac{7 e (b d-a e) (d+e x)^{3/2}}{3 b^3}+\frac{7 e (d+e x)^{5/2}}{5 b^2}-\frac{(d+e x)^{7/2}}{b (a+b x)}+\frac{\left (7 e (b d-a e)^3\right ) \int \frac{1}{(a+b x) \sqrt{d+e x}} \, dx}{2 b^4}\\ &=\frac{7 e (b d-a e)^2 \sqrt{d+e x}}{b^4}+\frac{7 e (b d-a e) (d+e x)^{3/2}}{3 b^3}+\frac{7 e (d+e x)^{5/2}}{5 b^2}-\frac{(d+e x)^{7/2}}{b (a+b x)}+\frac{\left (7 (b d-a e)^3\right ) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b d}{e}+\frac{b x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{b^4}\\ &=\frac{7 e (b d-a e)^2 \sqrt{d+e x}}{b^4}+\frac{7 e (b d-a e) (d+e x)^{3/2}}{3 b^3}+\frac{7 e (d+e x)^{5/2}}{5 b^2}-\frac{(d+e x)^{7/2}}{b (a+b x)}-\frac{7 e (b d-a e)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{9/2}}\\ \end{align*}

Mathematica [C]  time = 0.0173275, size = 50, normalized size = 0.36 $\frac{2 e (d+e x)^{9/2} \, _2F_1\left (2,\frac{9}{2};\frac{11}{2};-\frac{b (d+e x)}{a e-b d}\right )}{9 (a e-b d)^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [B]  time = 0.202, size = 387, normalized size = 2.8 \begin{align*}{\frac{2\,e}{5\,{b}^{2}} \left ( ex+d \right ) ^{{\frac{5}{2}}}}-{\frac{4\,a{e}^{2}}{3\,{b}^{3}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+{\frac{4\,de}{3\,{b}^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+6\,{\frac{{a}^{2}{e}^{3}\sqrt{ex+d}}{{b}^{4}}}-12\,{\frac{ad{e}^{2}\sqrt{ex+d}}{{b}^{3}}}+6\,{\frac{e{d}^{2}\sqrt{ex+d}}{{b}^{2}}}+{\frac{{a}^{3}{e}^{4}}{{b}^{4} \left ( bxe+ae \right ) }\sqrt{ex+d}}-3\,{\frac{\sqrt{ex+d}d{e}^{3}{a}^{2}}{{b}^{3} \left ( bxe+ae \right ) }}+3\,{\frac{\sqrt{ex+d}a{d}^{2}{e}^{2}}{{b}^{2} \left ( bxe+ae \right ) }}-{\frac{e{d}^{3}}{b \left ( bxe+ae \right ) }\sqrt{ex+d}}-7\,{\frac{{a}^{3}{e}^{4}}{{b}^{4}\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) }+21\,{\frac{d{e}^{3}{a}^{2}}{{b}^{3}\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) }-21\,{\frac{a{d}^{2}{e}^{2}}{{b}^{2}\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) }+7\,{\frac{e{d}^{3}}{b\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) } \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),x)

[Out]

2/5*e*(e*x+d)^(5/2)/b^2-4/3/b^3*(e*x+d)^(3/2)*a*e^2+4/3*e/b^2*(e*x+d)^(3/2)*d+6/b^4*a^2*e^3*(e*x+d)^(1/2)-12/b
^3*a*d*e^2*(e*x+d)^(1/2)+6*e/b^2*d^2*(e*x+d)^(1/2)+1/b^4*(e*x+d)^(1/2)/(b*e*x+a*e)*a^3*e^4-3/b^3*(e*x+d)^(1/2)
/(b*e*x+a*e)*d*e^3*a^2+3/b^2*(e*x+d)^(1/2)/(b*e*x+a*e)*a*d^2*e^2-e/b*(e*x+d)^(1/2)/(b*e*x+a*e)*d^3-7/b^4/((a*e
-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a^3*e^4+21/b^3/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)
^(1/2)/((a*e-b*d)*b)^(1/2))*d*e^3*a^2-21/b^2/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a
*d^2*e^2+7*e/b/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*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((e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.92009, size = 1049, normalized size = 7.66 \begin{align*} \left [\frac{105 \,{\left (a b^{2} d^{2} e - 2 \, a^{2} b d e^{2} + a^{3} e^{3} +{\left (b^{3} d^{2} e - 2 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x\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 (6 \, b^{3} e^{3} x^{3} - 15 \, b^{3} d^{3} + 161 \, a b^{2} d^{2} e - 245 \, a^{2} b d e^{2} + 105 \, a^{3} e^{3} + 2 \,{\left (16 \, b^{3} d e^{2} - 7 \, a b^{2} e^{3}\right )} x^{2} + 2 \,{\left (58 \, b^{3} d^{2} e - 84 \, a b^{2} d e^{2} + 35 \, a^{2} b e^{3}\right )} x\right )} \sqrt{e x + d}}{30 \,{\left (b^{5} x + a b^{4}\right )}}, -\frac{105 \,{\left (a b^{2} d^{2} e - 2 \, a^{2} b d e^{2} + a^{3} e^{3} +{\left (b^{3} d^{2} e - 2 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x\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 (6 \, b^{3} e^{3} x^{3} - 15 \, b^{3} d^{3} + 161 \, a b^{2} d^{2} e - 245 \, a^{2} b d e^{2} + 105 \, a^{3} e^{3} + 2 \,{\left (16 \, b^{3} d e^{2} - 7 \, a b^{2} e^{3}\right )} x^{2} + 2 \,{\left (58 \, b^{3} d^{2} e - 84 \, a b^{2} d e^{2} + 35 \, a^{2} b e^{3}\right )} x\right )} \sqrt{e x + d}}{15 \,{\left (b^{5} x + a 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),x, algorithm="fricas")

[Out]

[1/30*(105*(a*b^2*d^2*e - 2*a^2*b*d*e^2 + a^3*e^3 + (b^3*d^2*e - 2*a*b^2*d*e^2 + a^2*b*e^3)*x)*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*(6*b^3*e^3*x^3 - 15*b^3*
d^3 + 161*a*b^2*d^2*e - 245*a^2*b*d*e^2 + 105*a^3*e^3 + 2*(16*b^3*d*e^2 - 7*a*b^2*e^3)*x^2 + 2*(58*b^3*d^2*e -
84*a*b^2*d*e^2 + 35*a^2*b*e^3)*x)*sqrt(e*x + d))/(b^5*x + a*b^4), -1/15*(105*(a*b^2*d^2*e - 2*a^2*b*d*e^2 + a
^3*e^3 + (b^3*d^2*e - 2*a*b^2*d*e^2 + a^2*b*e^3)*x)*sqrt(-(b*d - a*e)/b)*arctan(-sqrt(e*x + d)*b*sqrt(-(b*d -
a*e)/b)/(b*d - a*e)) - (6*b^3*e^3*x^3 - 15*b^3*d^3 + 161*a*b^2*d^2*e - 245*a^2*b*d*e^2 + 105*a^3*e^3 + 2*(16*b
^3*d*e^2 - 7*a*b^2*e^3)*x^2 + 2*(58*b^3*d^2*e - 84*a*b^2*d*e^2 + 35*a^2*b*e^3)*x)*sqrt(e*x + d))/(b^5*x + a*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),x)

[Out]

Timed out

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Giac [B]  time = 1.2795, size = 379, normalized size = 2.77 \begin{align*} \frac{7 \,{\left (b^{3} d^{3} e - 3 \, a b^{2} d^{2} e^{2} + 3 \, a^{2} b d e^{3} - a^{3} e^{4}\right )} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right )}{\sqrt{-b^{2} d + a b e} b^{4}} - \frac{\sqrt{x e + d} b^{3} d^{3} e - 3 \, \sqrt{x e + d} a b^{2} d^{2} e^{2} + 3 \, \sqrt{x e + d} a^{2} b d e^{3} - \sqrt{x e + d} a^{3} e^{4}}{{\left ({\left (x e + d\right )} b - b d + a e\right )} b^{4}} + \frac{2 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} b^{8} e + 10 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{8} d e + 45 \, \sqrt{x e + d} b^{8} d^{2} e - 10 \,{\left (x e + d\right )}^{\frac{3}{2}} a b^{7} e^{2} - 90 \, \sqrt{x e + d} a b^{7} d e^{2} + 45 \, \sqrt{x e + d} a^{2} b^{6} e^{3}\right )}}{15 \, b^{10}} \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),x, algorithm="giac")

[Out]

7*(b^3*d^3*e - 3*a*b^2*d^2*e^2 + 3*a^2*b*d*e^3 - a^3*e^4)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-
b^2*d + a*b*e)*b^4) - (sqrt(x*e + d)*b^3*d^3*e - 3*sqrt(x*e + d)*a*b^2*d^2*e^2 + 3*sqrt(x*e + d)*a^2*b*d*e^3 -
sqrt(x*e + d)*a^3*e^4)/(((x*e + d)*b - b*d + a*e)*b^4) + 2/15*(3*(x*e + d)^(5/2)*b^8*e + 10*(x*e + d)^(3/2)*b
^8*d*e + 45*sqrt(x*e + d)*b^8*d^2*e - 10*(x*e + d)^(3/2)*a*b^7*e^2 - 90*sqrt(x*e + d)*a*b^7*d*e^2 + 45*sqrt(x*
e + d)*a^2*b^6*e^3)/b^10