### 3.1613 $$\int \frac{1}{(d+e x)^3 (a^2+2 a b x+b^2 x^2)^{5/2}} \, dx$$

Optimal. Leaf size=365 $\frac{5 b e^4 (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^6}+\frac{e^4 (a+b x)}{2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^2 (b d-a e)^5}+\frac{10 b^2 e^3}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^6}-\frac{3 b^2 e^2}{(a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^5}+\frac{15 b^2 e^4 (a+b x) \log (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^7}-\frac{15 b^2 e^4 (a+b x) \log (d+e x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^7}+\frac{b^2 e}{(a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^4}-\frac{b^2}{4 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}$

[Out]

(10*b^2*e^3)/((b*d - a*e)^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - b^2/(4*(b*d - a*e)^3*(a + b*x)^3*Sqrt[a^2 + 2*a*b
*x + b^2*x^2]) + (b^2*e)/((b*d - a*e)^4*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (3*b^2*e^2)/((b*d - a*e)^
5*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (e^4*(a + b*x))/(2*(b*d - a*e)^5*(d + e*x)^2*Sqrt[a^2 + 2*a*b*x +
b^2*x^2]) + (5*b*e^4*(a + b*x))/((b*d - a*e)^6*(d + e*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (15*b^2*e^4*(a + b*
x)*Log[a + b*x])/((b*d - a*e)^7*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (15*b^2*e^4*(a + b*x)*Log[d + e*x])/((b*d - a
*e)^7*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

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Rubi [A]  time = 0.26331, antiderivative size = 365, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 28, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.071, Rules used = {646, 44} $\frac{5 b e^4 (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^6}+\frac{e^4 (a+b x)}{2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^2 (b d-a e)^5}+\frac{10 b^2 e^3}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^6}-\frac{3 b^2 e^2}{(a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^5}+\frac{15 b^2 e^4 (a+b x) \log (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^7}-\frac{15 b^2 e^4 (a+b x) \log (d+e x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^7}+\frac{b^2 e}{(a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^4}-\frac{b^2}{4 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/((d + e*x)^3*(a^2 + 2*a*b*x + b^2*x^2)^(5/2)),x]

[Out]

(10*b^2*e^3)/((b*d - a*e)^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - b^2/(4*(b*d - a*e)^3*(a + b*x)^3*Sqrt[a^2 + 2*a*b
*x + b^2*x^2]) + (b^2*e)/((b*d - a*e)^4*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (3*b^2*e^2)/((b*d - a*e)^
5*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (e^4*(a + b*x))/(2*(b*d - a*e)^5*(d + e*x)^2*Sqrt[a^2 + 2*a*b*x +
b^2*x^2]) + (5*b*e^4*(a + b*x))/((b*d - a*e)^6*(d + e*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (15*b^2*e^4*(a + b*
x)*Log[a + b*x])/((b*d - a*e)^7*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (15*b^2*e^4*(a + b*x)*Log[d + e*x])/((b*d - a
*e)^7*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

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 44

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

Rubi steps

\begin{align*} \int \frac{1}{(d+e x)^3 \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx &=\frac{\left (b^4 \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right )^5 (d+e x)^3} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{\left (b^4 \left (a b+b^2 x\right )\right ) \int \left (\frac{1}{b^2 (b d-a e)^3 (a+b x)^5}-\frac{3 e}{b^2 (b d-a e)^4 (a+b x)^4}+\frac{6 e^2}{b^2 (b d-a e)^5 (a+b x)^3}-\frac{10 e^3}{b^2 (b d-a e)^6 (a+b x)^2}+\frac{15 e^4}{b^2 (b d-a e)^7 (a+b x)}-\frac{e^5}{b^5 (b d-a e)^5 (d+e x)^3}-\frac{5 e^5}{b^4 (b d-a e)^6 (d+e x)^2}-\frac{15 e^5}{b^3 (b d-a e)^7 (d+e x)}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{10 b^2 e^3}{(b d-a e)^6 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{b^2}{4 (b d-a e)^3 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{b^2 e}{(b d-a e)^4 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 b^2 e^2}{(b d-a e)^5 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{e^4 (a+b x)}{2 (b d-a e)^5 (d+e x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{5 b e^4 (a+b x)}{(b d-a e)^6 (d+e x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{15 b^2 e^4 (a+b x) \log (a+b x)}{(b d-a e)^7 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{15 b^2 e^4 (a+b x) \log (d+e x)}{(b d-a e)^7 \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.157949, size = 209, normalized size = 0.57 $\frac{40 b^2 e^3 (a+b x)^2 (b d-a e)-12 b^2 e^2 (a+b x) (b d-a e)^2-60 b^2 e^4 (a+b x)^3 \log (d+e x)-\frac{b^2 (b d-a e)^4}{a+b x}+4 b^2 e (b d-a e)^3+60 b^2 e^4 (a+b x)^3 \log (a+b x)+\frac{20 b e^4 (a+b x)^3 (b d-a e)}{d+e x}+\frac{2 e^4 (a+b x)^3 (b d-a e)^2}{(d+e x)^2}}{4 \left ((a+b x)^2\right )^{3/2} (b d-a e)^7}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((d + e*x)^3*(a^2 + 2*a*b*x + b^2*x^2)^(5/2)),x]

[Out]

(4*b^2*e*(b*d - a*e)^3 - (b^2*(b*d - a*e)^4)/(a + b*x) - 12*b^2*e^2*(b*d - a*e)^2*(a + b*x) + 40*b^2*e^3*(b*d
- a*e)*(a + b*x)^2 + (2*e^4*(b*d - a*e)^2*(a + b*x)^3)/(d + e*x)^2 + (20*b*e^4*(b*d - a*e)*(a + b*x)^3)/(d + e
*x) + 60*b^2*e^4*(a + b*x)^3*Log[a + b*x] - 60*b^2*e^4*(a + b*x)^3*Log[d + e*x])/(4*(b*d - a*e)^7*((a + b*x)^2
)^(3/2))

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Maple [B]  time = 0.235, size = 982, normalized size = 2.7 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

1/4*(24*a^5*b*d*e^5-2*a^6*e^6-60*ln(b*x+a)*x^6*b^6*e^6+5*x^2*b^6*d^4*e^2+12*x*a^5*b*e^6-2*x*b^6*d^5*e+260*x^3*
a^3*b^3*e^6-20*x^3*b^6*d^3*e^3+125*x^2*a^4*b^2*e^6+60*x^5*a*b^5*e^6-60*x^5*b^6*d*e^5+210*x^4*a^2*b^4*e^6-90*x^
4*b^6*d^2*e^4+60*ln(e*x+d)*x^6*b^6*e^6+240*ln(e*x+d)*x^5*a*b^5*e^6+120*ln(e*x+d)*x^5*b^6*d*e^5-240*ln(b*x+a)*x
^5*a*b^5*e^6-120*ln(b*x+a)*x^5*b^6*d*e^5+360*ln(e*x+d)*x^4*a^2*b^4*e^6+60*ln(e*x+d)*x^4*b^6*d^2*e^4+240*ln(e*x
+d)*x^3*a^3*b^3*e^6-480*ln(b*x+a)*x^4*a*b^5*d*e^5-8*a*b^5*d^5*e-330*x^2*a^2*b^4*d^2*e^4-80*x^2*a*b^5*d^3*e^3-1
00*x*a^3*b^3*d^2*e^4-120*x*a^2*b^4*d^3*e^3+20*x*a*b^5*d^4*e^2-300*x^3*a*b^5*d^2*e^4-360*ln(b*x+a)*x^4*a^2*b^4*
e^6-60*ln(b*x+a)*x^4*b^6*d^2*e^4-240*ln(b*x+a)*x^3*a^3*b^3*e^6-60*ln(b*x+a)*x^2*a^4*b^2*e^6+d^6*b^6+480*ln(e*x
+d)*x^2*a^3*b^3*d*e^5+360*ln(e*x+d)*x^2*a^2*b^4*d^2*e^4-60*ln(b*x+a)*a^4*b^2*d^2*e^4+60*x^3*a^2*b^4*d*e^5+280*
x^2*a^3*b^3*d*e^5+190*x*a^4*b^2*d*e^5-120*x^4*a*b^5*d*e^5+30*a^2*b^4*d^4*e^2+480*ln(e*x+d)*x^4*a*b^5*d*e^5+720
*ln(e*x+d)*x^3*a^2*b^4*d*e^5+240*ln(e*x+d)*x^3*a*b^5*d^2*e^4+35*d^2*e^4*a^4*b^2-80*b^3*a^3*d^3*e^3-120*ln(b*x+
a)*x*a^4*b^2*d*e^5-240*ln(b*x+a)*x*a^3*b^3*d^2*e^4-720*ln(b*x+a)*x^3*a^2*b^4*d*e^5-240*ln(b*x+a)*x^3*a*b^5*d^2
*e^4-480*ln(b*x+a)*x^2*a^3*b^3*d*e^5-360*ln(b*x+a)*x^2*a^2*b^4*d^2*e^4+60*ln(e*x+d)*a^4*b^2*d^2*e^4+60*ln(e*x+
d)*x^2*a^4*b^2*e^6+120*ln(e*x+d)*x*a^4*b^2*d*e^5+240*ln(e*x+d)*x*a^3*b^3*d^2*e^4)*(b*x+a)/(e*x+d)^2/(a*e-b*d)^
7/((b*x+a)^2)^(5/2)

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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)^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.13331, size = 3136, normalized size = 8.59 \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)^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

-1/4*(b^6*d^6 - 8*a*b^5*d^5*e + 30*a^2*b^4*d^4*e^2 - 80*a^3*b^3*d^3*e^3 + 35*a^4*b^2*d^2*e^4 + 24*a^5*b*d*e^5
- 2*a^6*e^6 - 60*(b^6*d*e^5 - a*b^5*e^6)*x^5 - 30*(3*b^6*d^2*e^4 + 4*a*b^5*d*e^5 - 7*a^2*b^4*e^6)*x^4 - 20*(b^
6*d^3*e^3 + 15*a*b^5*d^2*e^4 - 3*a^2*b^4*d*e^5 - 13*a^3*b^3*e^6)*x^3 + 5*(b^6*d^4*e^2 - 16*a*b^5*d^3*e^3 - 66*
a^2*b^4*d^2*e^4 + 56*a^3*b^3*d*e^5 + 25*a^4*b^2*e^6)*x^2 - 2*(b^6*d^5*e - 10*a*b^5*d^4*e^2 + 60*a^2*b^4*d^3*e^
3 + 50*a^3*b^3*d^2*e^4 - 95*a^4*b^2*d*e^5 - 6*a^5*b*e^6)*x - 60*(b^6*e^6*x^6 + a^4*b^2*d^2*e^4 + 2*(b^6*d*e^5
+ 2*a*b^5*e^6)*x^5 + (b^6*d^2*e^4 + 8*a*b^5*d*e^5 + 6*a^2*b^4*e^6)*x^4 + 4*(a*b^5*d^2*e^4 + 3*a^2*b^4*d*e^5 +
a^3*b^3*e^6)*x^3 + (6*a^2*b^4*d^2*e^4 + 8*a^3*b^3*d*e^5 + a^4*b^2*e^6)*x^2 + 2*(2*a^3*b^3*d^2*e^4 + a^4*b^2*d*
e^5)*x)*log(b*x + a) + 60*(b^6*e^6*x^6 + a^4*b^2*d^2*e^4 + 2*(b^6*d*e^5 + 2*a*b^5*e^6)*x^5 + (b^6*d^2*e^4 + 8*
a*b^5*d*e^5 + 6*a^2*b^4*e^6)*x^4 + 4*(a*b^5*d^2*e^4 + 3*a^2*b^4*d*e^5 + a^3*b^3*e^6)*x^3 + (6*a^2*b^4*d^2*e^4
+ 8*a^3*b^3*d*e^5 + a^4*b^2*e^6)*x^2 + 2*(2*a^3*b^3*d^2*e^4 + a^4*b^2*d*e^5)*x)*log(e*x + d))/(a^4*b^7*d^9 - 7
*a^5*b^6*d^8*e + 21*a^6*b^5*d^7*e^2 - 35*a^7*b^4*d^6*e^3 + 35*a^8*b^3*d^5*e^4 - 21*a^9*b^2*d^4*e^5 + 7*a^10*b*
d^3*e^6 - a^11*d^2*e^7 + (b^11*d^7*e^2 - 7*a*b^10*d^6*e^3 + 21*a^2*b^9*d^5*e^4 - 35*a^3*b^8*d^4*e^5 + 35*a^4*b
^7*d^3*e^6 - 21*a^5*b^6*d^2*e^7 + 7*a^6*b^5*d*e^8 - a^7*b^4*e^9)*x^6 + 2*(b^11*d^8*e - 5*a*b^10*d^7*e^2 + 7*a^
2*b^9*d^6*e^3 + 7*a^3*b^8*d^5*e^4 - 35*a^4*b^7*d^4*e^5 + 49*a^5*b^6*d^3*e^6 - 35*a^6*b^5*d^2*e^7 + 13*a^7*b^4*
d*e^8 - 2*a^8*b^3*e^9)*x^5 + (b^11*d^9 + a*b^10*d^8*e - 29*a^2*b^9*d^7*e^2 + 91*a^3*b^8*d^6*e^3 - 119*a^4*b^7*
d^5*e^4 + 49*a^5*b^6*d^4*e^5 + 49*a^6*b^5*d^3*e^6 - 71*a^7*b^4*d^2*e^7 + 34*a^8*b^3*d*e^8 - 6*a^9*b^2*e^9)*x^4
+ 4*(a*b^10*d^9 - 4*a^2*b^9*d^8*e + a^3*b^8*d^7*e^2 + 21*a^4*b^7*d^6*e^3 - 49*a^5*b^6*d^5*e^4 + 49*a^6*b^5*d^
4*e^5 - 21*a^7*b^4*d^3*e^6 - a^8*b^3*d^2*e^7 + 4*a^9*b^2*d*e^8 - a^10*b*e^9)*x^3 + (6*a^2*b^9*d^9 - 34*a^3*b^8
*d^8*e + 71*a^4*b^7*d^7*e^2 - 49*a^5*b^6*d^6*e^3 - 49*a^6*b^5*d^5*e^4 + 119*a^7*b^4*d^4*e^5 - 91*a^8*b^3*d^3*e
^6 + 29*a^9*b^2*d^2*e^7 - a^10*b*d*e^8 - a^11*e^9)*x^2 + 2*(2*a^3*b^8*d^9 - 13*a^4*b^7*d^8*e + 35*a^5*b^6*d^7*
e^2 - 49*a^6*b^5*d^6*e^3 + 35*a^7*b^4*d^5*e^4 - 7*a^8*b^3*d^4*e^5 - 7*a^9*b^2*d^3*e^6 + 5*a^10*b*d^2*e^7 - a^1
1*d*e^8)*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(1/(e*x+d)**3/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

sage0*x