### 3.1653 $$\int \frac{1}{(d+e x)^{7/2} (a^2+2 a b x+b^2 x^2)} \, dx$$

Optimal. Leaf size=151 $-\frac{7 b^2 e}{\sqrt{d+e x} (b d-a e)^4}+\frac{7 b^{5/2} e \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{(b d-a e)^{9/2}}-\frac{7 b e}{3 (d+e x)^{3/2} (b d-a e)^3}-\frac{1}{(a+b x) (d+e x)^{5/2} (b d-a e)}-\frac{7 e}{5 (d+e x)^{5/2} (b d-a e)^2}$

[Out]

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

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Rubi [A]  time = 0.0996385, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 28, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.143, Rules used = {27, 51, 63, 208} $-\frac{7 b^2 e}{\sqrt{d+e x} (b d-a e)^4}+\frac{7 b^{5/2} e \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{(b d-a e)^{9/2}}-\frac{7 b e}{3 (d+e x)^{3/2} (b d-a e)^3}-\frac{1}{(a+b x) (d+e x)^{5/2} (b d-a e)}-\frac{7 e}{5 (d+e x)^{5/2} (b d-a e)^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/((d + e*x)^(7/2)*(a^2 + 2*a*b*x + b^2*x^2)),x]

[Out]

(-7*e)/(5*(b*d - a*e)^2*(d + e*x)^(5/2)) - 1/((b*d - a*e)*(a + b*x)*(d + e*x)^(5/2)) - (7*b*e)/(3*(b*d - a*e)^
3*(d + e*x)^(3/2)) - (7*b^2*e)/((b*d - a*e)^4*Sqrt[d + e*x]) + (7*b^(5/2)*e*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sq
rt[b*d - a*e]])/(b*d - a*e)^(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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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{1}{(d+e x)^{7/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx &=\int \frac{1}{(a+b x)^2 (d+e x)^{7/2}} \, dx\\ &=-\frac{1}{(b d-a e) (a+b x) (d+e x)^{5/2}}-\frac{(7 e) \int \frac{1}{(a+b x) (d+e x)^{7/2}} \, dx}{2 (b d-a e)}\\ &=-\frac{7 e}{5 (b d-a e)^2 (d+e x)^{5/2}}-\frac{1}{(b d-a e) (a+b x) (d+e x)^{5/2}}-\frac{(7 b e) \int \frac{1}{(a+b x) (d+e x)^{5/2}} \, dx}{2 (b d-a e)^2}\\ &=-\frac{7 e}{5 (b d-a e)^2 (d+e x)^{5/2}}-\frac{1}{(b d-a e) (a+b x) (d+e x)^{5/2}}-\frac{7 b e}{3 (b d-a e)^3 (d+e x)^{3/2}}-\frac{\left (7 b^2 e\right ) \int \frac{1}{(a+b x) (d+e x)^{3/2}} \, dx}{2 (b d-a e)^3}\\ &=-\frac{7 e}{5 (b d-a e)^2 (d+e x)^{5/2}}-\frac{1}{(b d-a e) (a+b x) (d+e x)^{5/2}}-\frac{7 b e}{3 (b d-a e)^3 (d+e x)^{3/2}}-\frac{7 b^2 e}{(b d-a e)^4 \sqrt{d+e x}}-\frac{\left (7 b^3 e\right ) \int \frac{1}{(a+b x) \sqrt{d+e x}} \, dx}{2 (b d-a e)^4}\\ &=-\frac{7 e}{5 (b d-a e)^2 (d+e x)^{5/2}}-\frac{1}{(b d-a e) (a+b x) (d+e x)^{5/2}}-\frac{7 b e}{3 (b d-a e)^3 (d+e x)^{3/2}}-\frac{7 b^2 e}{(b d-a e)^4 \sqrt{d+e x}}-\frac{\left (7 b^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 d-a e)^4}\\ &=-\frac{7 e}{5 (b d-a e)^2 (d+e x)^{5/2}}-\frac{1}{(b d-a e) (a+b x) (d+e x)^{5/2}}-\frac{7 b e}{3 (b d-a e)^3 (d+e x)^{3/2}}-\frac{7 b^2 e}{(b d-a e)^4 \sqrt{d+e x}}+\frac{7 b^{5/2} e \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{(b d-a e)^{9/2}}\\ \end{align*}

Mathematica [C]  time = 0.0171591, size = 50, normalized size = 0.33 $-\frac{2 e \, _2F_1\left (-\frac{5}{2},2;-\frac{3}{2};-\frac{b (d+e x)}{a e-b d}\right )}{5 (d+e x)^{5/2} (a e-b d)^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((d + e*x)^(7/2)*(a^2 + 2*a*b*x + b^2*x^2)),x]

[Out]

(-2*e*Hypergeometric2F1[-5/2, 2, -3/2, -((b*(d + e*x))/(-(b*d) + a*e))])/(5*(-(b*d) + a*e)^2*(d + e*x)^(5/2))

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Maple [A]  time = 0.207, size = 149, normalized size = 1. \begin{align*} -{\frac{2\,e}{5\, \left ( ae-bd \right ) ^{2}} \left ( ex+d \right ) ^{-{\frac{5}{2}}}}-6\,{\frac{{b}^{2}e}{ \left ( ae-bd \right ) ^{4}\sqrt{ex+d}}}+{\frac{4\,be}{3\, \left ( ae-bd \right ) ^{3}} \left ( ex+d \right ) ^{-{\frac{3}{2}}}}-{\frac{e{b}^{3}}{ \left ( ae-bd \right ) ^{4} \left ( bxe+ae \right ) }\sqrt{ex+d}}-7\,{\frac{e{b}^{3}}{ \left ( ae-bd \right ) ^{4}\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(1/(e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2),x)

[Out]

-2/5*e/(a*e-b*d)^2/(e*x+d)^(5/2)-6*e/(a*e-b*d)^4*b^2/(e*x+d)^(1/2)+4/3*e/(a*e-b*d)^3*b/(e*x+d)^(3/2)-e*b^3/(a*
e-b*d)^4*(e*x+d)^(1/2)/(b*e*x+a*e)-7*e*b^3/(a*e-b*d)^4/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b
)^(1/2))

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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(1/(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 = 2.13549, size = 2471, normalized size = 16.36 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="fricas")

[Out]

[1/30*(105*(b^3*e^4*x^4 + a*b^2*d^3*e + (3*b^3*d*e^3 + a*b^2*e^4)*x^3 + 3*(b^3*d^2*e^2 + a*b^2*d*e^3)*x^2 + (b
^3*d^3*e + 3*a*b^2*d^2*e^2)*x)*sqrt(b/(b*d - a*e))*log((b*e*x + 2*b*d - a*e + 2*(b*d - a*e)*sqrt(e*x + d)*sqrt
(b/(b*d - a*e)))/(b*x + a)) - 2*(105*b^3*e^3*x^3 + 15*b^3*d^3 + 116*a*b^2*d^2*e - 32*a^2*b*d*e^2 + 6*a^3*e^3 +
35*(7*b^3*d*e^2 + 2*a*b^2*e^3)*x^2 + 7*(23*b^3*d^2*e + 24*a*b^2*d*e^2 - 2*a^2*b*e^3)*x)*sqrt(e*x + d))/(a*b^4
*d^7 - 4*a^2*b^3*d^6*e + 6*a^3*b^2*d^5*e^2 - 4*a^4*b*d^4*e^3 + a^5*d^3*e^4 + (b^5*d^4*e^3 - 4*a*b^4*d^3*e^4 +
6*a^2*b^3*d^2*e^5 - 4*a^3*b^2*d*e^6 + a^4*b*e^7)*x^4 + (3*b^5*d^5*e^2 - 11*a*b^4*d^4*e^3 + 14*a^2*b^3*d^3*e^4
- 6*a^3*b^2*d^2*e^5 - a^4*b*d*e^6 + a^5*e^7)*x^3 + 3*(b^5*d^6*e - 3*a*b^4*d^5*e^2 + 2*a^2*b^3*d^4*e^3 + 2*a^3*
b^2*d^3*e^4 - 3*a^4*b*d^2*e^5 + a^5*d*e^6)*x^2 + (b^5*d^7 - a*b^4*d^6*e - 6*a^2*b^3*d^5*e^2 + 14*a^3*b^2*d^4*e
^3 - 11*a^4*b*d^3*e^4 + 3*a^5*d^2*e^5)*x), 1/15*(105*(b^3*e^4*x^4 + a*b^2*d^3*e + (3*b^3*d*e^3 + a*b^2*e^4)*x^
3 + 3*(b^3*d^2*e^2 + a*b^2*d*e^3)*x^2 + (b^3*d^3*e + 3*a*b^2*d^2*e^2)*x)*sqrt(-b/(b*d - a*e))*arctan(-(b*d - a
*e)*sqrt(e*x + d)*sqrt(-b/(b*d - a*e))/(b*e*x + b*d)) - (105*b^3*e^3*x^3 + 15*b^3*d^3 + 116*a*b^2*d^2*e - 32*a
^2*b*d*e^2 + 6*a^3*e^3 + 35*(7*b^3*d*e^2 + 2*a*b^2*e^3)*x^2 + 7*(23*b^3*d^2*e + 24*a*b^2*d*e^2 - 2*a^2*b*e^3)*
x)*sqrt(e*x + d))/(a*b^4*d^7 - 4*a^2*b^3*d^6*e + 6*a^3*b^2*d^5*e^2 - 4*a^4*b*d^4*e^3 + a^5*d^3*e^4 + (b^5*d^4*
e^3 - 4*a*b^4*d^3*e^4 + 6*a^2*b^3*d^2*e^5 - 4*a^3*b^2*d*e^6 + a^4*b*e^7)*x^4 + (3*b^5*d^5*e^2 - 11*a*b^4*d^4*e
^3 + 14*a^2*b^3*d^3*e^4 - 6*a^3*b^2*d^2*e^5 - a^4*b*d*e^6 + a^5*e^7)*x^3 + 3*(b^5*d^6*e - 3*a*b^4*d^5*e^2 + 2*
a^2*b^3*d^4*e^3 + 2*a^3*b^2*d^3*e^4 - 3*a^4*b*d^2*e^5 + a^5*d*e^6)*x^2 + (b^5*d^7 - a*b^4*d^6*e - 6*a^2*b^3*d^
5*e^2 + 14*a^3*b^2*d^4*e^3 - 11*a^4*b*d^3*e^4 + 3*a^5*d^2*e^5)*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b x\right )^{2} \left (d + e x\right )^{\frac{7}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)**(7/2)/(b**2*x**2+2*a*b*x+a**2),x)

[Out]

Integral(1/((a + b*x)**2*(d + e*x)**(7/2)), x)

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Giac [B]  time = 1.20258, size = 410, normalized size = 2.72 \begin{align*} -\frac{7 \, b^{3} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right ) e}{{\left (b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}\right )} \sqrt{-b^{2} d + a b e}} - \frac{\sqrt{x e + d} b^{3} e}{{\left (b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}\right )}{\left ({\left (x e + d\right )} b - b d + a e\right )}} - \frac{2 \,{\left (45 \,{\left (x e + d\right )}^{2} b^{2} e + 10 \,{\left (x e + d\right )} b^{2} d e + 3 \, b^{2} d^{2} e - 10 \,{\left (x e + d\right )} a b e^{2} - 6 \, a b d e^{2} + 3 \, a^{2} e^{3}\right )}}{15 \,{\left (b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}\right )}{\left (x e + d\right )}^{\frac{5}{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="giac")

[Out]

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