### 3.1611 $$\int \frac{1}{(d+e x) (a^2+2 a b x+b^2 x^2)^{5/2}} \, dx$$

Optimal. Leaf size=253 $\frac{e^3}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^4}-\frac{e^2}{2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}+\frac{e^4 (a+b x) \log (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^5}-\frac{e^4 (a+b x) \log (d+e x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^5}+\frac{e}{3 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac{1}{4 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}$

[Out]

e^3/((b*d - a*e)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - 1/(4*(b*d - a*e)*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]
) + e/(3*(b*d - a*e)^2*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - e^2/(2*(b*d - a*e)^3*(a + b*x)*Sqrt[a^2 +
2*a*b*x + b^2*x^2]) + (e^4*(a + b*x)*Log[a + b*x])/((b*d - a*e)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (e^4*(a + b
*x)*Log[d + e*x])/((b*d - a*e)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

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Rubi [A]  time = 0.157273, antiderivative size = 253, 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} $\frac{e^3}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^4}-\frac{e^2}{2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}+\frac{e^4 (a+b x) \log (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^5}-\frac{e^4 (a+b x) \log (d+e x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^5}+\frac{e}{3 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac{1}{4 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

e^3/((b*d - a*e)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - 1/(4*(b*d - a*e)*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]
) + e/(3*(b*d - a*e)^2*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - e^2/(2*(b*d - a*e)^3*(a + b*x)*Sqrt[a^2 +
2*a*b*x + b^2*x^2]) + (e^4*(a + b*x)*Log[a + b*x])/((b*d - a*e)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (e^4*(a + b
*x)*Log[d + e*x])/((b*d - a*e)^5*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) \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)} \, 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^4 (b d-a e) (a+b x)^5}-\frac{e}{b^4 (b d-a e)^2 (a+b x)^4}+\frac{e^2}{b^4 (b d-a e)^3 (a+b x)^3}-\frac{e^3}{b^4 (b d-a e)^4 (a+b x)^2}+\frac{e^4}{b^4 (b d-a e)^5 (a+b x)}-\frac{e^5}{b^5 (b d-a e)^5 (d+e x)}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{e^3}{(b d-a e)^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{1}{4 (b d-a e) (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{e}{3 (b d-a e)^2 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{e^2}{2 (b d-a e)^3 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{e^4 (a+b x) \log (a+b x)}{(b d-a e)^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{e^4 (a+b x) \log (d+e x)}{(b d-a e)^5 \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.123977, size = 163, normalized size = 0.64 $\frac{-(b d-a e) \left (a^2 b e^2 (23 d-52 e x)-25 a^3 e^3+a b^2 e \left (-13 d^2+20 d e x-42 e^2 x^2\right )+b^3 \left (-4 d^2 e x+3 d^3+6 d e^2 x^2-12 e^3 x^3\right )\right )-12 e^4 (a+b x)^4 \log (d+e x)+12 e^4 (a+b x)^4 \log (a+b x)}{12 (a+b x)^3 \sqrt{(a+b x)^2} (b d-a e)^5}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-((b*d - a*e)*(-25*a^3*e^3 + a^2*b*e^2*(23*d - 52*e*x) + a*b^2*e*(-13*d^2 + 20*d*e*x - 42*e^2*x^2) + b^3*(3*d
^3 - 4*d^2*e*x + 6*d*e^2*x^2 - 12*e^3*x^3))) + 12*e^4*(a + b*x)^4*Log[a + b*x] - 12*e^4*(a + b*x)^4*Log[d + e*
x])/(12*(b*d - a*e)^5*(a + b*x)^3*Sqrt[(a + b*x)^2])

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Maple [A]  time = 0.204, size = 359, normalized size = 1.4 \begin{align*}{\frac{ \left ( 12\,\ln \left ( ex+d \right ){x}^{4}{b}^{4}{e}^{4}-12\,\ln \left ( bx+a \right ){x}^{4}{b}^{4}{e}^{4}+48\,\ln \left ( ex+d \right ){x}^{3}a{b}^{3}{e}^{4}-48\,\ln \left ( bx+a \right ){x}^{3}a{b}^{3}{e}^{4}+72\,\ln \left ( ex+d \right ){x}^{2}{a}^{2}{b}^{2}{e}^{4}-72\,\ln \left ( bx+a \right ){x}^{2}{a}^{2}{b}^{2}{e}^{4}+12\,{x}^{3}a{b}^{3}{e}^{4}-12\,{x}^{3}{b}^{4}d{e}^{3}+48\,\ln \left ( ex+d \right ) x{a}^{3}b{e}^{4}-48\,\ln \left ( bx+a \right ) x{a}^{3}b{e}^{4}+42\,{x}^{2}{a}^{2}{b}^{2}{e}^{4}-48\,{x}^{2}a{b}^{3}d{e}^{3}+6\,{x}^{2}{b}^{4}{d}^{2}{e}^{2}+12\,\ln \left ( ex+d \right ){a}^{4}{e}^{4}-12\,\ln \left ( bx+a \right ){a}^{4}{e}^{4}+52\,x{a}^{3}b{e}^{4}-72\,x{a}^{2}{b}^{2}d{e}^{3}+24\,xa{b}^{3}{d}^{2}{e}^{2}-4\,x{b}^{4}{d}^{3}e+25\,{a}^{4}{e}^{4}-48\,{a}^{3}bd{e}^{3}+36\,{d}^{2}{e}^{2}{a}^{2}{b}^{2}-16\,a{b}^{3}{d}^{3}e+3\,{b}^{4}{d}^{4} \right ) \left ( bx+a \right ) }{12\, \left ( ae-bd \right ) ^{5}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(12*ln(e*x+d)*x^4*b^4*e^4-12*ln(b*x+a)*x^4*b^4*e^4+48*ln(e*x+d)*x^3*a*b^3*e^4-48*ln(b*x+a)*x^3*a*b^3*e^4+
72*ln(e*x+d)*x^2*a^2*b^2*e^4-72*ln(b*x+a)*x^2*a^2*b^2*e^4+12*x^3*a*b^3*e^4-12*x^3*b^4*d*e^3+48*ln(e*x+d)*x*a^3
*b*e^4-48*ln(b*x+a)*x*a^3*b*e^4+42*x^2*a^2*b^2*e^4-48*x^2*a*b^3*d*e^3+6*x^2*b^4*d^2*e^2+12*ln(e*x+d)*a^4*e^4-1
2*ln(b*x+a)*a^4*e^4+52*x*a^3*b*e^4-72*x*a^2*b^2*d*e^3+24*x*a*b^3*d^2*e^2-4*x*b^4*d^3*e+25*a^4*e^4-48*a^3*b*d*e
^3+36*d^2*e^2*a^2*b^2-16*a*b^3*d^3*e+3*b^4*d^4)*(b*x+a)/(a*e-b*d)^5/((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)/(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 = 1.69628, size = 1320, normalized size = 5.22 \begin{align*} -\frac{3 \, b^{4} d^{4} - 16 \, a b^{3} d^{3} e + 36 \, a^{2} b^{2} d^{2} e^{2} - 48 \, a^{3} b d e^{3} + 25 \, a^{4} e^{4} - 12 \,{\left (b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} + 6 \,{\left (b^{4} d^{2} e^{2} - 8 \, a b^{3} d e^{3} + 7 \, a^{2} b^{2} e^{4}\right )} x^{2} - 4 \,{\left (b^{4} d^{3} e - 6 \, a b^{3} d^{2} e^{2} + 18 \, a^{2} b^{2} d e^{3} - 13 \, a^{3} b e^{4}\right )} x - 12 \,{\left (b^{4} e^{4} x^{4} + 4 \, a b^{3} e^{4} x^{3} + 6 \, a^{2} b^{2} e^{4} x^{2} + 4 \, a^{3} b e^{4} x + a^{4} e^{4}\right )} \log \left (b x + a\right ) + 12 \,{\left (b^{4} e^{4} x^{4} + 4 \, a b^{3} e^{4} x^{3} + 6 \, a^{2} b^{2} e^{4} x^{2} + 4 \, a^{3} b e^{4} x + a^{4} e^{4}\right )} \log \left (e x + d\right )}{12 \,{\left (a^{4} b^{5} d^{5} - 5 \, a^{5} b^{4} d^{4} e + 10 \, a^{6} b^{3} d^{3} e^{2} - 10 \, a^{7} b^{2} d^{2} e^{3} + 5 \, a^{8} b d e^{4} - a^{9} e^{5} +{\left (b^{9} d^{5} - 5 \, a b^{8} d^{4} e + 10 \, a^{2} b^{7} d^{3} e^{2} - 10 \, a^{3} b^{6} d^{2} e^{3} + 5 \, a^{4} b^{5} d e^{4} - a^{5} b^{4} e^{5}\right )} x^{4} + 4 \,{\left (a b^{8} d^{5} - 5 \, a^{2} b^{7} d^{4} e + 10 \, a^{3} b^{6} d^{3} e^{2} - 10 \, a^{4} b^{5} d^{2} e^{3} + 5 \, a^{5} b^{4} d e^{4} - a^{6} b^{3} e^{5}\right )} x^{3} + 6 \,{\left (a^{2} b^{7} d^{5} - 5 \, a^{3} b^{6} d^{4} e + 10 \, a^{4} b^{5} d^{3} e^{2} - 10 \, a^{5} b^{4} d^{2} e^{3} + 5 \, a^{6} b^{3} d e^{4} - a^{7} b^{2} e^{5}\right )} x^{2} + 4 \,{\left (a^{3} b^{6} d^{5} - 5 \, a^{4} b^{5} d^{4} e + 10 \, a^{5} b^{4} d^{3} e^{2} - 10 \, a^{6} b^{3} d^{2} e^{3} + 5 \, a^{7} b^{2} d e^{4} - a^{8} b e^{5}\right )} x\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*b^4*d^4 - 16*a*b^3*d^3*e + 36*a^2*b^2*d^2*e^2 - 48*a^3*b*d*e^3 + 25*a^4*e^4 - 12*(b^4*d*e^3 - a*b^3*e
^4)*x^3 + 6*(b^4*d^2*e^2 - 8*a*b^3*d*e^3 + 7*a^2*b^2*e^4)*x^2 - 4*(b^4*d^3*e - 6*a*b^3*d^2*e^2 + 18*a^2*b^2*d*
e^3 - 13*a^3*b*e^4)*x - 12*(b^4*e^4*x^4 + 4*a*b^3*e^4*x^3 + 6*a^2*b^2*e^4*x^2 + 4*a^3*b*e^4*x + a^4*e^4)*log(b
*x + a) + 12*(b^4*e^4*x^4 + 4*a*b^3*e^4*x^3 + 6*a^2*b^2*e^4*x^2 + 4*a^3*b*e^4*x + a^4*e^4)*log(e*x + d))/(a^4*
b^5*d^5 - 5*a^5*b^4*d^4*e + 10*a^6*b^3*d^3*e^2 - 10*a^7*b^2*d^2*e^3 + 5*a^8*b*d*e^4 - a^9*e^5 + (b^9*d^5 - 5*a
*b^8*d^4*e + 10*a^2*b^7*d^3*e^2 - 10*a^3*b^6*d^2*e^3 + 5*a^4*b^5*d*e^4 - a^5*b^4*e^5)*x^4 + 4*(a*b^8*d^5 - 5*a
^2*b^7*d^4*e + 10*a^3*b^6*d^3*e^2 - 10*a^4*b^5*d^2*e^3 + 5*a^5*b^4*d*e^4 - a^6*b^3*e^5)*x^3 + 6*(a^2*b^7*d^5 -
5*a^3*b^6*d^4*e + 10*a^4*b^5*d^3*e^2 - 10*a^5*b^4*d^2*e^3 + 5*a^6*b^3*d*e^4 - a^7*b^2*e^5)*x^2 + 4*(a^3*b^6*d
^5 - 5*a^4*b^5*d^4*e + 10*a^5*b^4*d^3*e^2 - 10*a^6*b^3*d^2*e^3 + 5*a^7*b^2*d*e^4 - a^8*b*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 ) \left (\left (a + b x\right )^{2}\right )^{\frac{5}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Integral(1/((d + e*x)*((a + b*x)**2)**(5/2)), x)

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

[Out]

sage0*x