### 3.1602 $$\int \frac{1}{(d+e x)^2 (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx$$

Optimal. Leaf size=217 $\frac{e^2 (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^3}+\frac{3 b e^2 (a+b x) \log (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^4}-\frac{3 b e^2 (a+b x) \log (d+e x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^4}+\frac{2 b e}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}-\frac{b}{2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}$

[Out]

(2*b*e)/((b*d - a*e)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - b/(2*(b*d - a*e)^2*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*
x^2]) + (e^2*(a + b*x))/((b*d - a*e)^3*(d + e*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (3*b*e^2*(a + b*x)*Log[a + b
*x])/((b*d - a*e)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (3*b*e^2*(a + b*x)*Log[d + e*x])/((b*d - a*e)^4*Sqrt[a^2
+ 2*a*b*x + b^2*x^2])

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Rubi [A]  time = 0.126441, antiderivative size = 217, 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} $\frac{e^2 (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^3}+\frac{3 b e^2 (a+b x) \log (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^4}-\frac{3 b e^2 (a+b x) \log (d+e x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^4}+\frac{2 b e}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}-\frac{b}{2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*b*e)/((b*d - a*e)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - b/(2*(b*d - a*e)^2*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*
x^2]) + (e^2*(a + b*x))/((b*d - a*e)^3*(d + e*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (3*b*e^2*(a + b*x)*Log[a + b
*x])/((b*d - a*e)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (3*b*e^2*(a + b*x)*Log[d + e*x])/((b*d - a*e)^4*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)^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac{\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right )^3 (d+e x)^2} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{\left (b^2 \left (a b+b^2 x\right )\right ) \int \left (\frac{1}{b (b d-a e)^2 (a+b x)^3}-\frac{2 e}{b (b d-a e)^3 (a+b x)^2}+\frac{3 e^2}{b (b d-a e)^4 (a+b x)}-\frac{e^3}{b^3 (b d-a e)^3 (d+e x)^2}-\frac{3 e^3}{b^2 (b d-a e)^4 (d+e x)}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{2 b e}{(b d-a e)^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{b}{2 (b d-a e)^2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{e^2 (a+b x)}{(b d-a e)^3 (d+e x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{3 b e^2 (a+b x) \log (a+b x)}{(b d-a e)^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 b e^2 (a+b x) \log (d+e x)}{(b d-a e)^4 \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0992049, size = 141, normalized size = 0.65 $\frac{-(b d-a e) \left (-2 a^2 e^2-a b e (5 d+9 e x)+b^2 \left (d^2-3 d e x-6 e^2 x^2\right )\right )+6 b e^2 (a+b x)^2 (d+e x) \log (a+b x)-6 b e^2 (a+b x)^2 (d+e x) \log (d+e x)}{2 (a+b x) \sqrt{(a+b x)^2} (d+e x) (b d-a e)^4}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-((b*d - a*e)*(-2*a^2*e^2 - a*b*e*(5*d + 9*e*x) + b^2*(d^2 - 3*d*e*x - 6*e^2*x^2))) + 6*b*e^2*(a + b*x)^2*(d
+ e*x)*Log[a + b*x] - 6*b*e^2*(a + b*x)^2*(d + e*x)*Log[d + e*x])/(2*(b*d - a*e)^4*(a + b*x)*Sqrt[(a + b*x)^2]
*(d + e*x))

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Maple [B]  time = 0.204, size = 329, normalized size = 1.5 \begin{align*} -{\frac{ \left ( 6\,\ln \left ( ex+d \right ){x}^{3}{b}^{3}{e}^{3}-6\,\ln \left ( bx+a \right ){x}^{3}{b}^{3}{e}^{3}+12\,\ln \left ( ex+d \right ){x}^{2}a{b}^{2}{e}^{3}+6\,\ln \left ( ex+d \right ){x}^{2}{b}^{3}d{e}^{2}-12\,\ln \left ( bx+a \right ){x}^{2}a{b}^{2}{e}^{3}-6\,\ln \left ( bx+a \right ){x}^{2}{b}^{3}d{e}^{2}+6\,\ln \left ( ex+d \right ) x{a}^{2}b{e}^{3}+12\,\ln \left ( ex+d \right ) xa{b}^{2}d{e}^{2}-6\,\ln \left ( bx+a \right ) x{a}^{2}b{e}^{3}-12\,\ln \left ( bx+a \right ) xa{b}^{2}d{e}^{2}+6\,{x}^{2}a{b}^{2}{e}^{3}-6\,{x}^{2}{b}^{3}d{e}^{2}+6\,\ln \left ( ex+d \right ){a}^{2}bd{e}^{2}-6\,\ln \left ( bx+a \right ){a}^{2}bd{e}^{2}+9\,x{a}^{2}b{e}^{3}-6\,xa{b}^{2}d{e}^{2}-3\,x{b}^{3}{d}^{2}e+2\,{a}^{3}{e}^{3}+3\,d{e}^{2}{a}^{2}b-6\,a{b}^{2}{d}^{2}e+{b}^{3}{d}^{3} \right ) \left ( bx+a \right ) }{ \left ( 2\,ex+2\,d \right ) \left ( ae-bd \right ) ^{4}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(6*ln(e*x+d)*x^3*b^3*e^3-6*ln(b*x+a)*x^3*b^3*e^3+12*ln(e*x+d)*x^2*a*b^2*e^3+6*ln(e*x+d)*x^2*b^3*d*e^2-12*
ln(b*x+a)*x^2*a*b^2*e^3-6*ln(b*x+a)*x^2*b^3*d*e^2+6*ln(e*x+d)*x*a^2*b*e^3+12*ln(e*x+d)*x*a*b^2*d*e^2-6*ln(b*x+
a)*x*a^2*b*e^3-12*ln(b*x+a)*x*a*b^2*d*e^2+6*x^2*a*b^2*e^3-6*x^2*b^3*d*e^2+6*ln(e*x+d)*a^2*b*d*e^2-6*ln(b*x+a)*
a^2*b*d*e^2+9*x*a^2*b*e^3-6*x*a*b^2*d*e^2-3*x*b^3*d^2*e+2*a^3*e^3+3*d*e^2*a^2*b-6*a*b^2*d^2*e+b^3*d^3)*(b*x+a)
/(e*x+d)/(a*e-b*d)^4/((b*x+a)^2)^(3/2)

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.70269, size = 991, normalized size = 4.57 \begin{align*} -\frac{b^{3} d^{3} - 6 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} + 2 \, a^{3} e^{3} - 6 \,{\left (b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} - 3 \,{\left (b^{3} d^{2} e + 2 \, a b^{2} d e^{2} - 3 \, a^{2} b e^{3}\right )} x - 6 \,{\left (b^{3} e^{3} x^{3} + a^{2} b d e^{2} +{\left (b^{3} d e^{2} + 2 \, a b^{2} e^{3}\right )} x^{2} +{\left (2 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x\right )} \log \left (b x + a\right ) + 6 \,{\left (b^{3} e^{3} x^{3} + a^{2} b d e^{2} +{\left (b^{3} d e^{2} + 2 \, a b^{2} e^{3}\right )} x^{2} +{\left (2 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x\right )} \log \left (e x + d\right )}{2 \,{\left (a^{2} b^{4} d^{5} - 4 \, a^{3} b^{3} d^{4} e + 6 \, a^{4} b^{2} d^{3} e^{2} - 4 \, a^{5} b d^{2} e^{3} + a^{6} d e^{4} +{\left (b^{6} d^{4} e - 4 \, a b^{5} d^{3} e^{2} + 6 \, a^{2} b^{4} d^{2} e^{3} - 4 \, a^{3} b^{3} d e^{4} + a^{4} b^{2} e^{5}\right )} x^{3} +{\left (b^{6} d^{5} - 2 \, a b^{5} d^{4} e - 2 \, a^{2} b^{4} d^{3} e^{2} + 8 \, a^{3} b^{3} d^{2} e^{3} - 7 \, a^{4} b^{2} d e^{4} + 2 \, a^{5} b e^{5}\right )} x^{2} +{\left (2 \, a b^{5} d^{5} - 7 \, a^{2} b^{4} d^{4} e + 8 \, a^{3} b^{3} d^{3} e^{2} - 2 \, a^{4} b^{2} d^{2} e^{3} - 2 \, a^{5} b d e^{4} + a^{6} e^{5}\right )} x\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b^3*d^3 - 6*a*b^2*d^2*e + 3*a^2*b*d*e^2 + 2*a^3*e^3 - 6*(b^3*d*e^2 - a*b^2*e^3)*x^2 - 3*(b^3*d^2*e + 2*a
*b^2*d*e^2 - 3*a^2*b*e^3)*x - 6*(b^3*e^3*x^3 + a^2*b*d*e^2 + (b^3*d*e^2 + 2*a*b^2*e^3)*x^2 + (2*a*b^2*d*e^2 +
a^2*b*e^3)*x)*log(b*x + a) + 6*(b^3*e^3*x^3 + a^2*b*d*e^2 + (b^3*d*e^2 + 2*a*b^2*e^3)*x^2 + (2*a*b^2*d*e^2 + a
^2*b*e^3)*x)*log(e*x + d))/(a^2*b^4*d^5 - 4*a^3*b^3*d^4*e + 6*a^4*b^2*d^3*e^2 - 4*a^5*b*d^2*e^3 + a^6*d*e^4 +
(b^6*d^4*e - 4*a*b^5*d^3*e^2 + 6*a^2*b^4*d^2*e^3 - 4*a^3*b^3*d*e^4 + a^4*b^2*e^5)*x^3 + (b^6*d^5 - 2*a*b^5*d^4
*e - 2*a^2*b^4*d^3*e^2 + 8*a^3*b^3*d^2*e^3 - 7*a^4*b^2*d*e^4 + 2*a^5*b*e^5)*x^2 + (2*a*b^5*d^5 - 7*a^2*b^4*d^4
*e + 8*a^3*b^3*d^3*e^2 - 2*a^4*b^2*d^2*e^3 - 2*a^5*b*d*e^4 + a^6*e^5)*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (d + e x\right )^{2} \left (\left (a + b x\right )^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)**2/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Integral(1/((d + e*x)**2*((a + b*x)**2)**(3/2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{3}{2}}{\left (e x + d\right )}^{2}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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