3.1612 \(\int \frac{1}{(d+e x)^2 (a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=307 \[ \frac{e^4 (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^5}+\frac{4 b e^3}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^5}-\frac{3 b e^2}{2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^4}+\frac{5 b e^4 (a+b x) \log (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^6}-\frac{5 b e^4 (a+b x) \log (d+e x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^6}+\frac{2 b e}{3 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}-\frac{b}{4 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2} \]

[Out]

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

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Rubi [A]  time = 0.210104, antiderivative size = 307, 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} \[ \frac{e^4 (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^5}+\frac{4 b e^3}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^5}-\frac{3 b e^2}{2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^4}+\frac{5 b e^4 (a+b x) \log (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^6}-\frac{5 b e^4 (a+b x) \log (d+e x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^6}+\frac{2 b e}{3 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}-\frac{b}{4 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b*e^3)/((b*d - a*e)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - b/(4*(b*d - a*e)^2*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x +
b^2*x^2]) + (2*b*e)/(3*(b*d - a*e)^3*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (3*b*e^2)/(2*(b*d - a*e)^4*(
a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (e^4*(a + b*x))/((b*d - a*e)^5*(d + e*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^
2]) + (5*b*e^4*(a + b*x)*Log[a + b*x])/((b*d - a*e)^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (5*b*e^4*(a + b*x)*Log[
d + e*x])/((b*d - a*e)^6*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 )^{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)^2} \, 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^3 (b d-a e)^2 (a+b x)^5}-\frac{2 e}{b^3 (b d-a e)^3 (a+b x)^4}+\frac{3 e^2}{b^3 (b d-a e)^4 (a+b x)^3}-\frac{4 e^3}{b^3 (b d-a e)^5 (a+b x)^2}+\frac{5 e^4}{b^3 (b d-a e)^6 (a+b x)}-\frac{e^5}{b^5 (b d-a e)^5 (d+e x)^2}-\frac{5 e^5}{b^4 (b d-a e)^6 (d+e x)}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{4 b e^3}{(b d-a e)^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{b}{4 (b d-a e)^2 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 b e}{3 (b d-a e)^3 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 b e^2}{2 (b d-a e)^4 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{e^4 (a+b x)}{(b d-a e)^5 (d+e x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{5 b e^4 (a+b x) \log (a+b x)}{(b d-a e)^6 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 b e^4 (a+b x) \log (d+e x)}{(b d-a e)^6 \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.119407, size = 167, normalized size = 0.54 \[ \frac{\frac{12 e^4 (a+b x)^3 (b d-a e)}{d+e x}+48 b e^3 (a+b x)^2 (b d-a e)-18 b e^2 (a+b x) (b d-a e)^2-60 b e^4 (a+b x)^3 \log (d+e x)-\frac{3 b (b d-a e)^4}{a+b x}+8 b e (b d-a e)^3+60 b e^4 (a+b x)^3 \log (a+b x)}{12 \left ((a+b x)^2\right )^{3/2} (b d-a e)^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*b*e*(b*d - a*e)^3 - (3*b*(b*d - a*e)^4)/(a + b*x) - 18*b*e^2*(b*d - a*e)^2*(a + b*x) + 48*b*e^3*(b*d - a*e)
*(a + b*x)^2 + (12*e^4*(b*d - a*e)*(a + b*x)^3)/(d + e*x) + 60*b*e^4*(a + b*x)^3*Log[a + b*x] - 60*b*e^4*(a +
b*x)^3*Log[d + e*x])/(12*(b*d - a*e)^6*((a + b*x)^2)^(3/2))

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Maple [B]  time = 0.211, size = 651, normalized size = 2.1 \begin{align*} -{\frac{ \left ( 240\,\ln \left ( ex+d \right ) x{a}^{3}{b}^{2}d{e}^{4}+360\,\ln \left ( ex+d \right ){x}^{2}{a}^{2}{b}^{3}d{e}^{4}+240\,\ln \left ( ex+d \right ){x}^{2}{a}^{3}{b}^{2}{e}^{5}+12\,{a}^{5}{e}^{5}+3\,{b}^{5}{d}^{5}+65\,d{e}^{4}{a}^{4}b+60\,\ln \left ( ex+d \right ){x}^{5}{b}^{5}{e}^{5}-5\,x{b}^{5}{d}^{4}e+60\,{x}^{4}a{b}^{4}{e}^{5}-60\,{x}^{4}{b}^{5}d{e}^{4}+210\,{x}^{3}{a}^{2}{b}^{3}{e}^{5}-30\,{x}^{3}{b}^{5}{d}^{2}{e}^{3}+260\,{x}^{2}{a}^{3}{b}^{2}{e}^{5}+10\,{x}^{2}{b}^{5}{d}^{3}{e}^{2}+125\,x{a}^{4}b{e}^{5}-240\,\ln \left ( bx+a \right ){x}^{3}a{b}^{4}d{e}^{4}+60\,\ln \left ( ex+d \right ) x{a}^{4}b{e}^{5}-120\,{x}^{2}a{b}^{4}{d}^{2}{e}^{3}-150\,{x}^{2}{a}^{2}{b}^{3}d{e}^{4}-180\,{x}^{3}a{b}^{4}d{e}^{4}+60\,\ln \left ( ex+d \right ){a}^{4}bd{e}^{4}+60\,{a}^{2}{b}^{3}{d}^{3}{e}^{2}-120\,{a}^{3}{b}^{2}{d}^{2}{e}^{3}+40\,xa{b}^{4}{d}^{3}{e}^{2}+20\,x{a}^{3}{b}^{2}d{e}^{4}-180\,x{a}^{2}{b}^{3}{d}^{2}{e}^{3}-60\,\ln \left ( bx+a \right ){x}^{5}{b}^{5}{e}^{5}+360\,\ln \left ( ex+d \right ){x}^{3}{a}^{2}{b}^{3}{e}^{5}-240\,\ln \left ( bx+a \right ){x}^{2}{a}^{3}{b}^{2}{e}^{5}-60\,\ln \left ( bx+a \right ) x{a}^{4}b{e}^{5}-60\,\ln \left ( bx+a \right ){a}^{4}bd{e}^{4}+240\,\ln \left ( ex+d \right ){x}^{4}a{b}^{4}{e}^{5}+60\,\ln \left ( ex+d \right ){x}^{4}{b}^{5}d{e}^{4}-240\,\ln \left ( bx+a \right ){x}^{4}a{b}^{4}{e}^{5}-60\,\ln \left ( bx+a \right ){x}^{4}{b}^{5}d{e}^{4}-360\,\ln \left ( bx+a \right ){x}^{3}{a}^{2}{b}^{3}{e}^{5}-360\,\ln \left ( bx+a \right ){x}^{2}{a}^{2}{b}^{3}d{e}^{4}+240\,\ln \left ( ex+d \right ){x}^{3}a{b}^{4}d{e}^{4}-240\,\ln \left ( bx+a \right ) x{a}^{3}{b}^{2}d{e}^{4}-20\,a{b}^{4}{d}^{4}e \right ) \left ( bx+a \right ) }{ \left ( 12\,ex+12\,d \right ) \left ( ae-bd \right ) ^{6}} \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)^2/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

-1/12*(240*ln(e*x+d)*x*a^3*b^2*d*e^4+360*ln(e*x+d)*x^2*a^2*b^3*d*e^4+240*ln(e*x+d)*x^2*a^3*b^2*e^5+12*a^5*e^5+
3*b^5*d^5+65*d*e^4*a^4*b+60*ln(e*x+d)*x^5*b^5*e^5-5*x*b^5*d^4*e+60*x^4*a*b^4*e^5-60*x^4*b^5*d*e^4+210*x^3*a^2*
b^3*e^5-30*x^3*b^5*d^2*e^3+260*x^2*a^3*b^2*e^5+10*x^2*b^5*d^3*e^2+125*x*a^4*b*e^5-240*ln(b*x+a)*x^3*a*b^4*d*e^
4+60*ln(e*x+d)*x*a^4*b*e^5-120*x^2*a*b^4*d^2*e^3-150*x^2*a^2*b^3*d*e^4-180*x^3*a*b^4*d*e^4+60*ln(e*x+d)*a^4*b*
d*e^4+60*a^2*b^3*d^3*e^2-120*a^3*b^2*d^2*e^3+40*x*a*b^4*d^3*e^2+20*x*a^3*b^2*d*e^4-180*x*a^2*b^3*d^2*e^3-60*ln
(b*x+a)*x^5*b^5*e^5+360*ln(e*x+d)*x^3*a^2*b^3*e^5-240*ln(b*x+a)*x^2*a^3*b^2*e^5-60*ln(b*x+a)*x*a^4*b*e^5-60*ln
(b*x+a)*a^4*b*d*e^4+240*ln(e*x+d)*x^4*a*b^4*e^5+60*ln(e*x+d)*x^4*b^5*d*e^4-240*ln(b*x+a)*x^4*a*b^4*e^5-60*ln(b
*x+a)*x^4*b^5*d*e^4-360*ln(b*x+a)*x^3*a^2*b^3*e^5-360*ln(b*x+a)*x^2*a^2*b^3*d*e^4+240*ln(e*x+d)*x^3*a*b^4*d*e^
4-240*ln(b*x+a)*x*a^3*b^2*d*e^4-20*a*b^4*d^4*e)*(b*x+a)/(e*x+d)/(a*e-b*d)^6/((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)^2/(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 = 1.91157, size = 2196, normalized size = 7.15 \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)^2/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

-1/12*(3*b^5*d^5 - 20*a*b^4*d^4*e + 60*a^2*b^3*d^3*e^2 - 120*a^3*b^2*d^2*e^3 + 65*a^4*b*d*e^4 + 12*a^5*e^5 - 6
0*(b^5*d*e^4 - a*b^4*e^5)*x^4 - 30*(b^5*d^2*e^3 + 6*a*b^4*d*e^4 - 7*a^2*b^3*e^5)*x^3 + 10*(b^5*d^3*e^2 - 12*a*
b^4*d^2*e^3 - 15*a^2*b^3*d*e^4 + 26*a^3*b^2*e^5)*x^2 - 5*(b^5*d^4*e - 8*a*b^4*d^3*e^2 + 36*a^2*b^3*d^2*e^3 - 4
*a^3*b^2*d*e^4 - 25*a^4*b*e^5)*x - 60*(b^5*e^5*x^5 + a^4*b*d*e^4 + (b^5*d*e^4 + 4*a*b^4*e^5)*x^4 + 2*(2*a*b^4*
d*e^4 + 3*a^2*b^3*e^5)*x^3 + 2*(3*a^2*b^3*d*e^4 + 2*a^3*b^2*e^5)*x^2 + (4*a^3*b^2*d*e^4 + a^4*b*e^5)*x)*log(b*
x + a) + 60*(b^5*e^5*x^5 + a^4*b*d*e^4 + (b^5*d*e^4 + 4*a*b^4*e^5)*x^4 + 2*(2*a*b^4*d*e^4 + 3*a^2*b^3*e^5)*x^3
+ 2*(3*a^2*b^3*d*e^4 + 2*a^3*b^2*e^5)*x^2 + (4*a^3*b^2*d*e^4 + a^4*b*e^5)*x)*log(e*x + d))/(a^4*b^6*d^7 - 6*a
^5*b^5*d^6*e + 15*a^6*b^4*d^5*e^2 - 20*a^7*b^3*d^4*e^3 + 15*a^8*b^2*d^3*e^4 - 6*a^9*b*d^2*e^5 + a^10*d*e^6 + (
b^10*d^6*e - 6*a*b^9*d^5*e^2 + 15*a^2*b^8*d^4*e^3 - 20*a^3*b^7*d^3*e^4 + 15*a^4*b^6*d^2*e^5 - 6*a^5*b^5*d*e^6
+ a^6*b^4*e^7)*x^5 + (b^10*d^7 - 2*a*b^9*d^6*e - 9*a^2*b^8*d^5*e^2 + 40*a^3*b^7*d^4*e^3 - 65*a^4*b^6*d^3*e^4 +
54*a^5*b^5*d^2*e^5 - 23*a^6*b^4*d*e^6 + 4*a^7*b^3*e^7)*x^4 + 2*(2*a*b^9*d^7 - 9*a^2*b^8*d^6*e + 12*a^3*b^7*d^
5*e^2 + 5*a^4*b^6*d^4*e^3 - 30*a^5*b^5*d^3*e^4 + 33*a^6*b^4*d^2*e^5 - 16*a^7*b^3*d*e^6 + 3*a^8*b^2*e^7)*x^3 +
2*(3*a^2*b^8*d^7 - 16*a^3*b^7*d^6*e + 33*a^4*b^6*d^5*e^2 - 30*a^5*b^5*d^4*e^3 + 5*a^6*b^4*d^3*e^4 + 12*a^7*b^3
*d^2*e^5 - 9*a^8*b^2*d*e^6 + 2*a^9*b*e^7)*x^2 + (4*a^3*b^7*d^7 - 23*a^4*b^6*d^6*e + 54*a^5*b^5*d^5*e^2 - 65*a^
6*b^4*d^4*e^3 + 40*a^7*b^3*d^3*e^4 - 9*a^8*b^2*d^2*e^5 - 2*a^9*b*d*e^6 + a^10*e^7)*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)**2/(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*} \int \frac{1}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{5}{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)^(5/2),x, algorithm="giac")

[Out]

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