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

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

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Rubi [A]  time = 0.13, antiderivative size = 254, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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} \end {gather*}

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 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])

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]

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*}

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Mathematica [A]  time = 0.09, size = 185, normalized size = 0.73 \begin {gather*} \frac {\sqrt {(a+b x)^2} \left (b e x \left (300 a^4 e^4+300 a^3 b e^3 (e x-2 d)+100 a^2 b^2 e^2 \left (6 d^2-3 d e x+2 e^2 x^2\right )+25 a b^3 e \left (-12 d^3+6 d^2 e x-4 d e^2 x^2+3 e^3 x^3\right )+b^4 \left (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\right )\right )-60 (b d-a e)^5 \log (d+e x)\right )}{60 e^6 (a+b x)} \end {gather*}

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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IntegrateAlgebraic [F]  time = 2.47, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{d+e x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.39, size = 259, normalized size = 1.02 \begin {gather*} \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 {gather*}

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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giac [B]  time = 0.19, size = 385, normalized size = 1.52 \begin {gather*} -{\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 {gather*}

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)

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maple [A]  time = 0.06, size = 318, normalized size = 1.25 \begin {gather*} \frac {\left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} \left (12 b^{5} e^{5} x^{5}+75 a \,b^{4} e^{5} x^{4}-15 b^{5} d \,e^{4} x^{4}+200 a^{2} b^{3} e^{5} x^{3}-100 a \,b^{4} d \,e^{4} x^{3}+20 b^{5} d^{2} e^{3} x^{3}+300 a^{3} b^{2} e^{5} x^{2}-300 a^{2} b^{3} d \,e^{4} x^{2}+150 a \,b^{4} d^{2} e^{3} x^{2}-30 b^{5} d^{3} e^{2} x^{2}+60 a^{5} e^{5} \ln \left (e x +d \right )-300 a^{4} b d \,e^{4} \ln \left (e x +d \right )+300 a^{4} b \,e^{5} x +600 a^{3} b^{2} d^{2} e^{3} \ln \left (e x +d \right )-600 a^{3} b^{2} d \,e^{4} x -600 a^{2} b^{3} d^{3} e^{2} \ln \left (e x +d \right )+600 a^{2} b^{3} d^{2} e^{3} x +300 a \,b^{4} d^{4} e \ln \left (e x +d \right )-300 a \,b^{4} d^{3} e^{2} x -60 b^{5} d^{5} \ln \left (e x +d \right )+60 b^{5} d^{4} e x \right )}{60 \left (b x +a \right )^{5} e^{6}} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}}{d+e\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}{d + e x}\, dx \end {gather*}

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]

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

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