3.1575 \(\int \frac{(a^2+2 a b x+b^2 x^2)^{5/2}}{d+e x} \, dx\)

Optimal. Leaf size=254 \[ \frac{b x \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^4}{e^5 (a+b x)}-\frac{(a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}{2 e^4}+\frac{(a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}{3 e^3}-\frac{(a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}{4 e^2}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^5 \log (d+e x)}{e^6 (a+b x)}+\frac{(a+b x)^4 \sqrt{a^2+2 a b x+b^2 x^2}}{5 e} \]

[Out]

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

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Rubi [A]  time = 0.125486, antiderivative size = 254, 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, 43} \[ \frac{b x \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^4}{e^5 (a+b x)}-\frac{(a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}{2 e^4}+\frac{(a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}{3 e^3}-\frac{(a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}{4 e^2}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^5 \log (d+e x)}{e^6 (a+b x)}+\frac{(a+b x)^4 \sqrt{a^2+2 a b x+b^2 x^2}}{5 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0963047, size = 185, normalized size = 0.73 \[ \frac{\sqrt{(a+b x)^2} \left (b e x \left (100 a^2 b^2 e^2 \left (6 d^2-3 d e x+2 e^2 x^2\right )+300 a^3 b e^3 (e x-2 d)+300 a^4 e^4+25 a b^3 e \left (6 d^2 e x-12 d^3-4 d e^2 x^2+3 e^3 x^3\right )+b^4 \left (20 d^2 e^2 x^2-30 d^3 e x+60 d^4-15 d e^3 x^3+12 e^4 x^4\right )\right )-60 (b d-a e)^5 \log (d+e x)\right )}{60 e^6 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a + b*x)^2]*(b*e*x*(300*a^4*e^4 + 300*a^3*b*e^3*(-2*d + e*x) + 100*a^2*b^2*e^2*(6*d^2 - 3*d*e*x + 2*e^2
*x^2) + 25*a*b^3*e*(-12*d^3 + 6*d^2*e*x - 4*d*e^2*x^2 + 3*e^3*x^3) + b^4*(60*d^4 - 30*d^3*e*x + 20*d^2*e^2*x^2
- 15*d*e^3*x^3 + 12*e^4*x^4)) - 60*(b*d - a*e)^5*Log[d + e*x]))/(60*e^6*(a + b*x))

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Maple [A]  time = 0.197, size = 318, normalized size = 1.3 \begin{align*}{\frac{12\,{x}^{5}{b}^{5}{e}^{5}+75\,{x}^{4}a{b}^{4}{e}^{5}-15\,{x}^{4}{b}^{5}d{e}^{4}+200\,{x}^{3}{a}^{2}{b}^{3}{e}^{5}-100\,{x}^{3}a{b}^{4}d{e}^{4}+20\,{x}^{3}{b}^{5}{d}^{2}{e}^{3}+300\,{x}^{2}{a}^{3}{b}^{2}{e}^{5}-300\,{x}^{2}{a}^{2}{b}^{3}d{e}^{4}+150\,{x}^{2}a{b}^{4}{d}^{2}{e}^{3}-30\,{x}^{2}{b}^{5}{d}^{3}{e}^{2}+60\,\ln \left ( ex+d \right ){a}^{5}{e}^{5}-300\,\ln \left ( ex+d \right ){a}^{4}bd{e}^{4}+600\,\ln \left ( ex+d \right ){a}^{3}{b}^{2}{d}^{2}{e}^{3}-600\,\ln \left ( ex+d \right ){a}^{2}{b}^{3}{d}^{3}{e}^{2}+300\,\ln \left ( ex+d \right ) a{b}^{4}{d}^{4}e-60\,\ln \left ( ex+d \right ){b}^{5}{d}^{5}+300\,x{a}^{4}b{e}^{5}-600\,x{a}^{3}{b}^{2}d{e}^{4}+600\,x{a}^{2}{b}^{3}{d}^{2}{e}^{3}-300\,xa{b}^{4}{d}^{3}{e}^{2}+60\,x{b}^{5}{d}^{4}e}{60\, \left ( bx+a \right ) ^{5}{e}^{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((b^2*x^2+2*a*b*x+a^2)^(5/2)/(e*x+d),x)

[Out]

1/60*((b*x+a)^2)^(5/2)*(12*x^5*b^5*e^5+75*x^4*a*b^4*e^5-15*x^4*b^5*d*e^4+200*x^3*a^2*b^3*e^5-100*x^3*a*b^4*d*e
^4+20*x^3*b^5*d^2*e^3+300*x^2*a^3*b^2*e^5-300*x^2*a^2*b^3*d*e^4+150*x^2*a*b^4*d^2*e^3-30*x^2*b^5*d^3*e^2+60*ln
(e*x+d)*a^5*e^5-300*ln(e*x+d)*a^4*b*d*e^4+600*ln(e*x+d)*a^3*b^2*d^2*e^3-600*ln(e*x+d)*a^2*b^3*d^3*e^2+300*ln(e
*x+d)*a*b^4*d^4*e-60*ln(e*x+d)*b^5*d^5+300*x*a^4*b*e^5-600*x*a^3*b^2*d*e^4+600*x*a^2*b^3*d^2*e^3-300*x*a*b^4*d
^3*e^2+60*x*b^5*d^4*e)/(b*x+a)^5/e^6

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.69253, size = 537, normalized size = 2.11 \begin{align*} \frac{12 \, b^{5} e^{5} x^{5} - 15 \,{\left (b^{5} d e^{4} - 5 \, a b^{4} e^{5}\right )} x^{4} + 20 \,{\left (b^{5} d^{2} e^{3} - 5 \, a b^{4} d e^{4} + 10 \, a^{2} b^{3} e^{5}\right )} x^{3} - 30 \,{\left (b^{5} d^{3} e^{2} - 5 \, a b^{4} d^{2} e^{3} + 10 \, a^{2} b^{3} d e^{4} - 10 \, a^{3} b^{2} e^{5}\right )} x^{2} + 60 \,{\left (b^{5} d^{4} e - 5 \, a b^{4} d^{3} e^{2} + 10 \, a^{2} b^{3} d^{2} e^{3} - 10 \, a^{3} b^{2} d e^{4} + 5 \, a^{4} b e^{5}\right )} x - 60 \,{\left (b^{5} d^{5} - 5 \, a b^{4} d^{4} e + 10 \, a^{2} b^{3} d^{3} e^{2} - 10 \, a^{3} b^{2} d^{2} e^{3} + 5 \, a^{4} b d e^{4} - a^{5} e^{5}\right )} \log \left (e x + d\right )}{60 \, e^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(12*b^5*e^5*x^5 - 15*(b^5*d*e^4 - 5*a*b^4*e^5)*x^4 + 20*(b^5*d^2*e^3 - 5*a*b^4*d*e^4 + 10*a^2*b^3*e^5)*x^
3 - 30*(b^5*d^3*e^2 - 5*a*b^4*d^2*e^3 + 10*a^2*b^3*d*e^4 - 10*a^3*b^2*e^5)*x^2 + 60*(b^5*d^4*e - 5*a*b^4*d^3*e
^2 + 10*a^2*b^3*d^2*e^3 - 10*a^3*b^2*d*e^4 + 5*a^4*b*e^5)*x - 60*(b^5*d^5 - 5*a*b^4*d^4*e + 10*a^2*b^3*d^3*e^2
- 10*a^3*b^2*d^2*e^3 + 5*a^4*b*d*e^4 - a^5*e^5)*log(e*x + d))/e^6

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

[Out]

Timed out

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Giac [B]  time = 1.24678, size = 520, normalized size = 2.05 \begin{align*} -{\left (b^{5} d^{5} \mathrm{sgn}\left (b x + a\right ) - 5 \, a b^{4} d^{4} e \mathrm{sgn}\left (b x + a\right ) + 10 \, a^{2} b^{3} d^{3} e^{2} \mathrm{sgn}\left (b x + a\right ) - 10 \, a^{3} b^{2} d^{2} e^{3} \mathrm{sgn}\left (b x + a\right ) + 5 \, a^{4} b d e^{4} \mathrm{sgn}\left (b x + a\right ) - a^{5} e^{5} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-6\right )} \log \left ({\left | x e + d \right |}\right ) + \frac{1}{60} \,{\left (12 \, b^{5} x^{5} e^{4} \mathrm{sgn}\left (b x + a\right ) - 15 \, b^{5} d x^{4} e^{3} \mathrm{sgn}\left (b x + a\right ) + 20 \, b^{5} d^{2} x^{3} e^{2} \mathrm{sgn}\left (b x + a\right ) - 30 \, b^{5} d^{3} x^{2} e \mathrm{sgn}\left (b x + a\right ) + 60 \, b^{5} d^{4} x \mathrm{sgn}\left (b x + a\right ) + 75 \, a b^{4} x^{4} e^{4} \mathrm{sgn}\left (b x + a\right ) - 100 \, a b^{4} d x^{3} e^{3} \mathrm{sgn}\left (b x + a\right ) + 150 \, a b^{4} d^{2} x^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) - 300 \, a b^{4} d^{3} x e \mathrm{sgn}\left (b x + a\right ) + 200 \, a^{2} b^{3} x^{3} e^{4} \mathrm{sgn}\left (b x + a\right ) - 300 \, a^{2} b^{3} d x^{2} e^{3} \mathrm{sgn}\left (b x + a\right ) + 600 \, a^{2} b^{3} d^{2} x e^{2} \mathrm{sgn}\left (b x + a\right ) + 300 \, a^{3} b^{2} x^{2} e^{4} \mathrm{sgn}\left (b x + a\right ) - 600 \, a^{3} b^{2} d x e^{3} \mathrm{sgn}\left (b x + a\right ) + 300 \, a^{4} b x e^{4} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-5\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b^5*d^5*sgn(b*x + a) - 5*a*b^4*d^4*e*sgn(b*x + a) + 10*a^2*b^3*d^3*e^2*sgn(b*x + a) - 10*a^3*b^2*d^2*e^3*sgn
(b*x + a) + 5*a^4*b*d*e^4*sgn(b*x + a) - a^5*e^5*sgn(b*x + a))*e^(-6)*log(abs(x*e + d)) + 1/60*(12*b^5*x^5*e^4
*sgn(b*x + a) - 15*b^5*d*x^4*e^3*sgn(b*x + a) + 20*b^5*d^2*x^3*e^2*sgn(b*x + a) - 30*b^5*d^3*x^2*e*sgn(b*x + a
) + 60*b^5*d^4*x*sgn(b*x + a) + 75*a*b^4*x^4*e^4*sgn(b*x + a) - 100*a*b^4*d*x^3*e^3*sgn(b*x + a) + 150*a*b^4*d
^2*x^2*e^2*sgn(b*x + a) - 300*a*b^4*d^3*x*e*sgn(b*x + a) + 200*a^2*b^3*x^3*e^4*sgn(b*x + a) - 300*a^2*b^3*d*x^
2*e^3*sgn(b*x + a) + 600*a^2*b^3*d^2*x*e^2*sgn(b*x + a) + 300*a^3*b^2*x^2*e^4*sgn(b*x + a) - 600*a^3*b^2*d*x*e
^3*sgn(b*x + a) + 300*a^4*b*x*e^4*sgn(b*x + a))*e^(-5)