3.13.80 \(\int \frac {(a^2+2 a b x+b^2 x^2)^{5/2}}{(d+e x)^4} \, dx\)

Optimal. Leaf size=292 \[ \frac {10 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}{e^6 (a+b x) (d+e x)}-\frac {5 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}{2 e^6 (a+b x) (d+e x)^2}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^5}{3 e^6 (a+b x) (d+e x)^3}+\frac {b^5 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^4 (a+b x)}-\frac {b^4 x \sqrt {a^2+2 a b x+b^2 x^2} (4 b d-5 a e)}{e^5 (a+b x)}+\frac {10 b^3 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2 \log (d+e x)}{e^6 (a+b x)} \]

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Rubi [A]  time = 0.16, antiderivative size = 292, 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^4 x \sqrt {a^2+2 a b x+b^2 x^2} (4 b d-5 a e)}{e^5 (a+b x)}+\frac {10 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}{e^6 (a+b x) (d+e x)}-\frac {5 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}{2 e^6 (a+b x) (d+e x)^2}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^5}{3 e^6 (a+b x) (d+e x)^3}+\frac {10 b^3 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2 \log (d+e x)}{e^6 (a+b x)}+\frac {b^5 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^4 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/(d + e*x)^4,x]

[Out]

-((b^4*(4*b*d - 5*a*e)*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^5*(a + b*x))) + (b^5*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x
^2])/(2*e^4*(a + b*x)) + ((b*d - a*e)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^6*(a + b*x)*(d + e*x)^3) - (5*b*(b
*d - a*e)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*e^6*(a + b*x)*(d + e*x)^2) + (10*b^2*(b*d - a*e)^3*Sqrt[a^2 + 2*
a*b*x + b^2*x^2])/(e^6*(a + b*x)*(d + e*x)) + (10*b^3*(b*d - a*e)^2*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)^4} \, 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)^4} \, 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^9 (4 b d-5 a e)}{e^5}+\frac {b^{10} x}{e^4}-\frac {b^5 (b d-a e)^5}{e^5 (d+e x)^4}+\frac {5 b^6 (b d-a e)^4}{e^5 (d+e x)^3}-\frac {10 b^7 (b d-a e)^3}{e^5 (d+e x)^2}+\frac {10 b^8 (b d-a e)^2}{e^5 (d+e x)}\right ) \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=-\frac {b^4 (4 b d-5 a e) x \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x)}+\frac {b^5 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^4 (a+b x)}+\frac {(b d-a e)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^6 (a+b x) (d+e x)^3}-\frac {5 b (b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^6 (a+b x) (d+e x)^2}+\frac {10 b^2 (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{e^6 (a+b x) (d+e x)}+\frac {10 b^3 (b d-a e)^2 \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 = 0.11, size = 247, normalized size = 0.85 \begin {gather*} \frac {\sqrt {(a+b x)^2} \left (-2 a^5 e^5-5 a^4 b e^4 (d+3 e x)-20 a^3 b^2 e^3 \left (d^2+3 d e x+3 e^2 x^2\right )+10 a^2 b^3 d e^2 \left (11 d^2+27 d e x+18 e^2 x^2\right )+10 a b^4 e \left (-13 d^4-27 d^3 e x-9 d^2 e^2 x^2+9 d e^3 x^3+3 e^4 x^4\right )+60 b^3 (d+e x)^3 (b d-a e)^2 \log (d+e x)+b^5 \left (47 d^5+81 d^4 e x-9 d^3 e^2 x^2-63 d^2 e^3 x^3-15 d e^4 x^4+3 e^5 x^5\right )\right )}{6 e^6 (a+b x) (d+e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/(d + e*x)^4,x]

[Out]

(Sqrt[(a + b*x)^2]*(-2*a^5*e^5 - 5*a^4*b*e^4*(d + 3*e*x) - 20*a^3*b^2*e^3*(d^2 + 3*d*e*x + 3*e^2*x^2) + 10*a^2
*b^3*d*e^2*(11*d^2 + 27*d*e*x + 18*e^2*x^2) + 10*a*b^4*e*(-13*d^4 - 27*d^3*e*x - 9*d^2*e^2*x^2 + 9*d*e^3*x^3 +
3*e^4*x^4) + b^5*(47*d^5 + 81*d^4*e*x - 9*d^3*e^2*x^2 - 63*d^2*e^3*x^3 - 15*d*e^4*x^4 + 3*e^5*x^5) + 60*b^3*(
b*d - a*e)^2*(d + e*x)^3*Log[d + e*x]))/(6*e^6*(a + b*x)*(d + e*x)^3)

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IntegrateAlgebraic [F]  time = 5.54, 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)^4} \, 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)^4,x]

[Out]

Defer[IntegrateAlgebraic][(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/(d + e*x)^4, x]

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fricas [A]  time = 0.40, size = 425, normalized size = 1.46 \begin {gather*} \frac {3 \, b^{5} e^{5} x^{5} + 47 \, b^{5} d^{5} - 130 \, a b^{4} d^{4} e + 110 \, a^{2} b^{3} d^{3} e^{2} - 20 \, a^{3} b^{2} d^{2} e^{3} - 5 \, a^{4} b d e^{4} - 2 \, a^{5} e^{5} - 15 \, {\left (b^{5} d e^{4} - 2 \, a b^{4} e^{5}\right )} x^{4} - 9 \, {\left (7 \, b^{5} d^{2} e^{3} - 10 \, a b^{4} d e^{4}\right )} x^{3} - 3 \, {\left (3 \, b^{5} d^{3} e^{2} + 30 \, a b^{4} d^{2} e^{3} - 60 \, a^{2} b^{3} d e^{4} + 20 \, a^{3} b^{2} e^{5}\right )} x^{2} + 3 \, {\left (27 \, b^{5} d^{4} e - 90 \, a b^{4} d^{3} e^{2} + 90 \, a^{2} b^{3} d^{2} e^{3} - 20 \, a^{3} b^{2} d e^{4} - 5 \, a^{4} b e^{5}\right )} x + 60 \, {\left (b^{5} d^{5} - 2 \, a b^{4} d^{4} e + a^{2} b^{3} d^{3} e^{2} + {\left (b^{5} d^{2} e^{3} - 2 \, a b^{4} d e^{4} + a^{2} b^{3} e^{5}\right )} x^{3} + 3 \, {\left (b^{5} d^{3} e^{2} - 2 \, a b^{4} d^{2} e^{3} + a^{2} b^{3} d e^{4}\right )} x^{2} + 3 \, {\left (b^{5} d^{4} e - 2 \, a b^{4} d^{3} e^{2} + a^{2} b^{3} d^{2} e^{3}\right )} x\right )} \log \left (e x + d\right )}{6 \, {\left (e^{9} x^{3} + 3 \, d e^{8} x^{2} + 3 \, d^{2} e^{7} x + d^{3} e^{6}\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)^4,x, algorithm="fricas")

[Out]

1/6*(3*b^5*e^5*x^5 + 47*b^5*d^5 - 130*a*b^4*d^4*e + 110*a^2*b^3*d^3*e^2 - 20*a^3*b^2*d^2*e^3 - 5*a^4*b*d*e^4 -
2*a^5*e^5 - 15*(b^5*d*e^4 - 2*a*b^4*e^5)*x^4 - 9*(7*b^5*d^2*e^3 - 10*a*b^4*d*e^4)*x^3 - 3*(3*b^5*d^3*e^2 + 30
*a*b^4*d^2*e^3 - 60*a^2*b^3*d*e^4 + 20*a^3*b^2*e^5)*x^2 + 3*(27*b^5*d^4*e - 90*a*b^4*d^3*e^2 + 90*a^2*b^3*d^2*
e^3 - 20*a^3*b^2*d*e^4 - 5*a^4*b*e^5)*x + 60*(b^5*d^5 - 2*a*b^4*d^4*e + a^2*b^3*d^3*e^2 + (b^5*d^2*e^3 - 2*a*b
^4*d*e^4 + a^2*b^3*e^5)*x^3 + 3*(b^5*d^3*e^2 - 2*a*b^4*d^2*e^3 + a^2*b^3*d*e^4)*x^2 + 3*(b^5*d^4*e - 2*a*b^4*d
^3*e^2 + a^2*b^3*d^2*e^3)*x)*log(e*x + d))/(e^9*x^3 + 3*d*e^8*x^2 + 3*d^2*e^7*x + d^3*e^6)

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giac [A]  time = 0.19, size = 375, normalized size = 1.28 \begin {gather*} 10 \, {\left (b^{5} d^{2} \mathrm {sgn}\left (b x + a\right ) - 2 \, a b^{4} d e \mathrm {sgn}\left (b x + a\right ) + a^{2} b^{3} e^{2} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-6\right )} \log \left ({\left | x e + d \right |}\right ) + \frac {1}{2} \, {\left (b^{5} x^{2} e^{4} \mathrm {sgn}\left (b x + a\right ) - 8 \, b^{5} d x e^{3} \mathrm {sgn}\left (b x + a\right ) + 10 \, a b^{4} x e^{4} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-8\right )} + \frac {{\left (47 \, b^{5} d^{5} \mathrm {sgn}\left (b x + a\right ) - 130 \, a b^{4} d^{4} e \mathrm {sgn}\left (b x + a\right ) + 110 \, a^{2} b^{3} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) - 20 \, 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 ) - 2 \, a^{5} e^{5} \mathrm {sgn}\left (b x + a\right ) + 60 \, {\left (b^{5} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) - 3 \, a b^{4} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b^{3} d e^{4} \mathrm {sgn}\left (b x + a\right ) - a^{3} b^{2} e^{5} \mathrm {sgn}\left (b x + a\right )\right )} x^{2} + 15 \, {\left (7 \, b^{5} d^{4} e \mathrm {sgn}\left (b x + a\right ) - 20 \, a b^{4} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) + 18 \, a^{2} b^{3} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) - 4 \, a^{3} b^{2} d e^{4} \mathrm {sgn}\left (b x + a\right ) - a^{4} b e^{5} \mathrm {sgn}\left (b x + a\right )\right )} x\right )} e^{\left (-6\right )}}{6 \, {\left (x e + d\right )}^{3}} \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)^4,x, algorithm="giac")

[Out]

10*(b^5*d^2*sgn(b*x + a) - 2*a*b^4*d*e*sgn(b*x + a) + a^2*b^3*e^2*sgn(b*x + a))*e^(-6)*log(abs(x*e + d)) + 1/2
*(b^5*x^2*e^4*sgn(b*x + a) - 8*b^5*d*x*e^3*sgn(b*x + a) + 10*a*b^4*x*e^4*sgn(b*x + a))*e^(-8) + 1/6*(47*b^5*d^
5*sgn(b*x + a) - 130*a*b^4*d^4*e*sgn(b*x + a) + 110*a^2*b^3*d^3*e^2*sgn(b*x + a) - 20*a^3*b^2*d^2*e^3*sgn(b*x
+ a) - 5*a^4*b*d*e^4*sgn(b*x + a) - 2*a^5*e^5*sgn(b*x + a) + 60*(b^5*d^3*e^2*sgn(b*x + a) - 3*a*b^4*d^2*e^3*sg
n(b*x + a) + 3*a^2*b^3*d*e^4*sgn(b*x + a) - a^3*b^2*e^5*sgn(b*x + a))*x^2 + 15*(7*b^5*d^4*e*sgn(b*x + a) - 20*
a*b^4*d^3*e^2*sgn(b*x + a) + 18*a^2*b^3*d^2*e^3*sgn(b*x + a) - 4*a^3*b^2*d*e^4*sgn(b*x + a) - a^4*b*e^5*sgn(b*
x + a))*x)*e^(-6)/(x*e + d)^3

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maple [B]  time = 0.06, size = 502, normalized size = 1.72 \begin {gather*} \frac {\left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} \left (3 b^{5} e^{5} x^{5}+60 a^{2} b^{3} e^{5} x^{3} \ln \left (e x +d \right )-120 a \,b^{4} d \,e^{4} x^{3} \ln \left (e x +d \right )+30 a \,b^{4} e^{5} x^{4}+60 b^{5} d^{2} e^{3} x^{3} \ln \left (e x +d \right )-15 b^{5} d \,e^{4} x^{4}+180 a^{2} b^{3} d \,e^{4} x^{2} \ln \left (e x +d \right )-360 a \,b^{4} d^{2} e^{3} x^{2} \ln \left (e x +d \right )+90 a \,b^{4} d \,e^{4} x^{3}+180 b^{5} d^{3} e^{2} x^{2} \ln \left (e x +d \right )-63 b^{5} d^{2} e^{3} x^{3}-60 a^{3} b^{2} e^{5} x^{2}+180 a^{2} b^{3} d^{2} e^{3} x \ln \left (e x +d \right )+180 a^{2} b^{3} d \,e^{4} x^{2}-360 a \,b^{4} d^{3} e^{2} x \ln \left (e x +d \right )-90 a \,b^{4} d^{2} e^{3} x^{2}+180 b^{5} d^{4} e x \ln \left (e x +d \right )-9 b^{5} d^{3} e^{2} x^{2}-15 a^{4} b \,e^{5} x -60 a^{3} b^{2} d \,e^{4} x +60 a^{2} b^{3} d^{3} e^{2} \ln \left (e x +d \right )+270 a^{2} b^{3} d^{2} e^{3} x -120 a \,b^{4} d^{4} e \ln \left (e x +d \right )-270 a \,b^{4} d^{3} e^{2} x +60 b^{5} d^{5} \ln \left (e x +d \right )+81 b^{5} d^{4} e x -2 a^{5} e^{5}-5 a^{4} b d \,e^{4}-20 a^{3} b^{2} d^{2} e^{3}+110 a^{2} b^{3} d^{3} e^{2}-130 a \,b^{4} d^{4} e +47 b^{5} d^{5}\right )}{6 \left (b x +a \right )^{5} \left (e x +d \right )^{3} 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)^4,x)

[Out]

1/6*((b*x+a)^2)^(5/2)*(-5*a^4*b*d*e^4-20*a^3*b^2*d^2*e^3+110*a^2*b^3*d^3*e^2-130*a*b^4*d^4*e+60*ln(e*x+d)*x^3*
b^5*d^2*e^3-120*ln(e*x+d)*x^3*a*b^4*d*e^4+180*a^2*b^3*d*e^4*x^2*ln(e*x+d)-360*a*b^4*d^2*e^3*x^2*ln(e*x+d)-360*
a*b^4*d^3*e^2*x*ln(e*x+d)+180*a^2*b^3*d^2*e^3*x*ln(e*x+d)-2*a^5*e^5+60*a^2*b^3*d^3*e^2*ln(e*x+d)+60*ln(e*x+d)*
x^3*a^2*b^3*e^5+47*b^5*d^5-60*a^3*b^2*e^5*x^2-9*b^5*d^3*e^2*x^2-15*a^4*b*e^5*x+81*b^5*d^4*e*x+30*a*b^4*e^5*x^4
-15*b^5*d*e^4*x^4-63*b^5*d^2*e^3*x^3+180*b^5*d^3*e^2*x^2*ln(e*x+d)-120*a*b^4*d^4*e*ln(e*x+d)-60*a^3*b^2*d*e^4*
x+270*a^2*b^3*d^2*e^3*x-270*a*b^4*d^3*e^2*x+3*b^5*e^5*x^5+60*b^5*d^5*ln(e*x+d)+180*a^2*b^3*d*e^4*x^2-90*a*b^4*
d^2*e^3*x^2+90*a*b^4*d*e^4*x^3+180*b^5*d^4*e*x*ln(e*x+d))/(b*x+a)^5/e^6/(e*x+d)^3

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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)^4,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}}{{\left (d+e\,x\right )}^4} \,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)^4,x)

[Out]

int((a^2 + b^2*x^2 + 2*a*b*x)^(5/2)/(d + e*x)^4, 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}}}{\left (d + e x\right )^{4}}\, 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)**4,x)

[Out]

Integral(((a + b*x)**2)**(5/2)/(d + e*x)**4, x)

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