3.13.79 \(\int \frac {(a^2+2 a b x+b^2 x^2)^{5/2}}{(d+e x)^3} \, dx\)

Optimal. Leaf size=295 \[ -\frac {5 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}{e^6 (a+b x) (d+e x)}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^5}{2 e^6 (a+b x) (d+e x)^2}-\frac {10 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3 \log (d+e x)}{e^6 (a+b x)}+\frac {b^5 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^3}{3 e^6 (a+b x)}-\frac {5 b^4 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^2 (b d-a e)}{2 e^6 (a+b x)}+\frac {10 b^3 x \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^5 (a+b x)} \]

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Rubi [A]  time = 0.19, antiderivative size = 295, 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^5 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^3}{3 e^6 (a+b x)}-\frac {5 b^4 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^2 (b d-a e)}{2 e^6 (a+b x)}+\frac {10 b^3 x \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^5 (a+b x)}-\frac {5 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}{e^6 (a+b x) (d+e x)}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^5}{2 e^6 (a+b x) (d+e x)^2}-\frac {10 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3 \log (d+e x)}{e^6 (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)^3,x]

[Out]

(10*b^3*(b*d - a*e)^2*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^5*(a + b*x)) + ((b*d - a*e)^5*Sqrt[a^2 + 2*a*b*x + b
^2*x^2])/(2*e^6*(a + b*x)*(d + e*x)^2) - (5*b*(b*d - a*e)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^6*(a + b*x)*(d +
e*x)) - (5*b^4*(b*d - a*e)*(d + e*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*e^6*(a + b*x)) + (b^5*(d + e*x)^3*Sq
rt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^6*(a + b*x)) - (10*b^2*(b*d - a*e)^3*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)^3} \, 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)^3} \, 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 {10 b^8 (b d-a e)^2}{e^5}-\frac {b^5 (b d-a e)^5}{e^5 (d+e x)^3}+\frac {5 b^6 (b d-a e)^4}{e^5 (d+e x)^2}-\frac {10 b^7 (b d-a e)^3}{e^5 (d+e x)}-\frac {5 b^9 (b d-a e) (d+e x)}{e^5}+\frac {b^{10} (d+e x)^2}{e^5}\right ) \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=\frac {10 b^3 (b d-a e)^2 x \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x)}+\frac {(b d-a e)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^6 (a+b x) (d+e x)^2}-\frac {5 b (b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{e^6 (a+b x) (d+e x)}-\frac {5 b^4 (b d-a e) (d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^6 (a+b x)}+\frac {b^5 (d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^6 (a+b x)}-\frac {10 b^2 (b d-a e)^3 \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 = 248, normalized size = 0.84 \begin {gather*} \frac {\sqrt {(a+b x)^2} \left (-3 a^5 e^5-15 a^4 b e^4 (d+2 e x)+30 a^3 b^2 d e^3 (3 d+4 e x)+30 a^2 b^3 e^2 \left (-5 d^3-4 d^2 e x+4 d e^2 x^2+2 e^3 x^3\right )+15 a b^4 e \left (7 d^4+2 d^3 e x-11 d^2 e^2 x^2-4 d e^3 x^3+e^4 x^4\right )-60 b^2 (d+e x)^2 (b d-a e)^3 \log (d+e x)+b^5 \left (-27 d^5+6 d^4 e x+63 d^3 e^2 x^2+20 d^2 e^3 x^3-5 d e^4 x^4+2 e^5 x^5\right )\right )}{6 e^6 (a+b x) (d+e x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/(d + e*x)^3,x]

[Out]

(Sqrt[(a + b*x)^2]*(-3*a^5*e^5 - 15*a^4*b*e^4*(d + 2*e*x) + 30*a^3*b^2*d*e^3*(3*d + 4*e*x) + 30*a^2*b^3*e^2*(-
5*d^3 - 4*d^2*e*x + 4*d*e^2*x^2 + 2*e^3*x^3) + 15*a*b^4*e*(7*d^4 + 2*d^3*e*x - 11*d^2*e^2*x^2 - 4*d*e^3*x^3 +
e^4*x^4) + b^5*(-27*d^5 + 6*d^4*e*x + 63*d^3*e^2*x^2 + 20*d^2*e^3*x^3 - 5*d*e^4*x^4 + 2*e^5*x^5) - 60*b^2*(b*d
- a*e)^3*(d + e*x)^2*Log[d + e*x]))/(6*e^6*(a + b*x)*(d + e*x)^2)

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IntegrateAlgebraic [F]  time = 4.60, 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)^3} \, 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)^3,x]

[Out]

Defer[IntegrateAlgebraic][(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/(d + e*x)^3, x]

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fricas [A]  time = 0.40, size = 416, normalized size = 1.41 \begin {gather*} \frac {2 \, b^{5} e^{5} x^{5} - 27 \, b^{5} d^{5} + 105 \, a b^{4} d^{4} e - 150 \, a^{2} b^{3} d^{3} e^{2} + 90 \, a^{3} b^{2} d^{2} e^{3} - 15 \, a^{4} b d e^{4} - 3 \, a^{5} e^{5} - 5 \, {\left (b^{5} d e^{4} - 3 \, a b^{4} e^{5}\right )} x^{4} + 20 \, {\left (b^{5} d^{2} e^{3} - 3 \, a b^{4} d e^{4} + 3 \, a^{2} b^{3} e^{5}\right )} x^{3} + 3 \, {\left (21 \, b^{5} d^{3} e^{2} - 55 \, a b^{4} d^{2} e^{3} + 40 \, a^{2} b^{3} d e^{4}\right )} x^{2} + 6 \, {\left (b^{5} d^{4} e + 5 \, a b^{4} d^{3} e^{2} - 20 \, 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} - 3 \, a b^{4} d^{4} e + 3 \, a^{2} b^{3} d^{3} e^{2} - a^{3} b^{2} d^{2} e^{3} + {\left (b^{5} d^{3} e^{2} - 3 \, a b^{4} d^{2} e^{3} + 3 \, a^{2} b^{3} d e^{4} - a^{3} b^{2} e^{5}\right )} x^{2} + 2 \, {\left (b^{5} d^{4} e - 3 \, a b^{4} d^{3} e^{2} + 3 \, a^{2} b^{3} d^{2} e^{3} - a^{3} b^{2} d e^{4}\right )} x\right )} \log \left (e x + d\right )}{6 \, {\left (e^{8} x^{2} + 2 \, d e^{7} x + d^{2} 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)^3,x, algorithm="fricas")

[Out]

1/6*(2*b^5*e^5*x^5 - 27*b^5*d^5 + 105*a*b^4*d^4*e - 150*a^2*b^3*d^3*e^2 + 90*a^3*b^2*d^2*e^3 - 15*a^4*b*d*e^4
- 3*a^5*e^5 - 5*(b^5*d*e^4 - 3*a*b^4*e^5)*x^4 + 20*(b^5*d^2*e^3 - 3*a*b^4*d*e^4 + 3*a^2*b^3*e^5)*x^3 + 3*(21*b
^5*d^3*e^2 - 55*a*b^4*d^2*e^3 + 40*a^2*b^3*d*e^4)*x^2 + 6*(b^5*d^4*e + 5*a*b^4*d^3*e^2 - 20*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 - 3*a*b^4*d^4*e + 3*a^2*b^3*d^3*e^2 - a^3*b^2*d^2*e^3 + (b^5*d
^3*e^2 - 3*a*b^4*d^2*e^3 + 3*a^2*b^3*d*e^4 - a^3*b^2*e^5)*x^2 + 2*(b^5*d^4*e - 3*a*b^4*d^3*e^2 + 3*a^2*b^3*d^2
*e^3 - a^3*b^2*d*e^4)*x)*log(e*x + d))/(e^8*x^2 + 2*d*e^7*x + d^2*e^6)

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giac [A]  time = 0.18, size = 376, normalized size = 1.27 \begin {gather*} -10 \, {\left (b^{5} d^{3} \mathrm {sgn}\left (b x + a\right ) - 3 \, a b^{4} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b^{3} d e^{2} \mathrm {sgn}\left (b x + a\right ) - a^{3} b^{2} e^{3} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-6\right )} \log \left ({\left | x e + d \right |}\right ) + \frac {1}{6} \, {\left (2 \, b^{5} x^{3} e^{6} \mathrm {sgn}\left (b x + a\right ) - 9 \, b^{5} d x^{2} e^{5} \mathrm {sgn}\left (b x + a\right ) + 36 \, b^{5} d^{2} x e^{4} \mathrm {sgn}\left (b x + a\right ) + 15 \, a b^{4} x^{2} e^{6} \mathrm {sgn}\left (b x + a\right ) - 90 \, a b^{4} d x e^{5} \mathrm {sgn}\left (b x + a\right ) + 60 \, a^{2} b^{3} x e^{6} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-9\right )} - \frac {{\left (9 \, b^{5} d^{5} \mathrm {sgn}\left (b x + a\right ) - 35 \, a b^{4} d^{4} e \mathrm {sgn}\left (b x + a\right ) + 50 \, a^{2} b^{3} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) - 30 \, 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 ) + 10 \, {\left (b^{5} d^{4} e \mathrm {sgn}\left (b x + a\right ) - 4 \, a b^{4} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) + 6 \, 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 )}}{2 \, {\left (x e + d\right )}^{2}} \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)^3,x, algorithm="giac")

[Out]

-10*(b^5*d^3*sgn(b*x + a) - 3*a*b^4*d^2*e*sgn(b*x + a) + 3*a^2*b^3*d*e^2*sgn(b*x + a) - a^3*b^2*e^3*sgn(b*x +
a))*e^(-6)*log(abs(x*e + d)) + 1/6*(2*b^5*x^3*e^6*sgn(b*x + a) - 9*b^5*d*x^2*e^5*sgn(b*x + a) + 36*b^5*d^2*x*e
^4*sgn(b*x + a) + 15*a*b^4*x^2*e^6*sgn(b*x + a) - 90*a*b^4*d*x*e^5*sgn(b*x + a) + 60*a^2*b^3*x*e^6*sgn(b*x + a
))*e^(-9) - 1/2*(9*b^5*d^5*sgn(b*x + a) - 35*a*b^4*d^4*e*sgn(b*x + a) + 50*a^2*b^3*d^3*e^2*sgn(b*x + a) - 30*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) + 10*(b^5*d^4*e*sgn(b*x + a) -
4*a*b^4*d^3*e^2*sgn(b*x + a) + 6*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)^2

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maple [B]  time = 0.07, size = 502, normalized size = 1.70 \begin {gather*} \frac {\left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} \left (2 b^{5} e^{5} x^{5}+15 a \,b^{4} e^{5} x^{4}-5 b^{5} d \,e^{4} x^{4}+60 a^{3} b^{2} e^{5} x^{2} \ln \left (e x +d \right )-180 a^{2} b^{3} d \,e^{4} x^{2} \ln \left (e x +d \right )+60 a^{2} b^{3} e^{5} x^{3}+180 a \,b^{4} d^{2} e^{3} x^{2} \ln \left (e x +d \right )-60 a \,b^{4} d \,e^{4} x^{3}-60 b^{5} d^{3} e^{2} x^{2} \ln \left (e x +d \right )+20 b^{5} d^{2} e^{3} x^{3}+120 a^{3} b^{2} d \,e^{4} x \ln \left (e x +d \right )-360 a^{2} b^{3} d^{2} e^{3} x \ln \left (e x +d \right )+120 a^{2} b^{3} d \,e^{4} x^{2}+360 a \,b^{4} d^{3} e^{2} x \ln \left (e x +d \right )-165 a \,b^{4} d^{2} e^{3} x^{2}-120 b^{5} d^{4} e x \ln \left (e x +d \right )+63 b^{5} d^{3} e^{2} x^{2}-30 a^{4} b \,e^{5} x +60 a^{3} b^{2} d^{2} e^{3} \ln \left (e x +d \right )+120 a^{3} b^{2} d \,e^{4} x -180 a^{2} b^{3} d^{3} e^{2} \ln \left (e x +d \right )-120 a^{2} b^{3} d^{2} e^{3} x +180 a \,b^{4} d^{4} e \ln \left (e x +d \right )+30 a \,b^{4} d^{3} e^{2} x -60 b^{5} d^{5} \ln \left (e x +d \right )+6 b^{5} d^{4} e x -3 a^{5} e^{5}-15 a^{4} b d \,e^{4}+90 a^{3} b^{2} d^{2} e^{3}-150 a^{2} b^{3} d^{3} e^{2}+105 a \,b^{4} d^{4} e -27 b^{5} d^{5}\right )}{6 \left (b x +a \right )^{5} \left (e x +d \right )^{2} 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)^3,x)

[Out]

1/6*((b*x+a)^2)^(5/2)*(-15*a^4*b*d*e^4+90*a^3*b^2*d^2*e^3-150*a^2*b^3*d^3*e^2+105*a*b^4*d^4*e-180*ln(e*x+d)*x^
2*a^2*b^3*d*e^4+180*ln(e*x+d)*x^2*a*b^4*d^2*e^3+360*a*b^4*d^3*e^2*x*ln(e*x+d)+120*a^3*b^2*d*e^4*x*ln(e*x+d)-36
0*a^2*b^3*d^2*e^3*x*ln(e*x+d)-3*a^5*e^5-180*a^2*b^3*d^3*e^2*ln(e*x+d)-27*b^5*d^5+63*b^5*d^3*e^2*x^2-30*a^4*b*e
^5*x+6*b^5*d^4*e*x+15*a*b^4*e^5*x^4-5*b^5*d*e^4*x^4+60*a^2*b^3*e^5*x^3+20*b^5*d^2*e^3*x^3+60*ln(e*x+d)*x^2*a^3
*b^2*e^5-60*ln(e*x+d)*x^2*b^5*d^3*e^2+180*a*b^4*d^4*e*ln(e*x+d)+120*a^3*b^2*d*e^4*x-120*a^2*b^3*d^2*e^3*x+30*a
*b^4*d^3*e^2*x+2*b^5*e^5*x^5-60*b^5*d^5*ln(e*x+d)+120*a^2*b^3*d*e^4*x^2-165*a*b^4*d^2*e^3*x^2+60*a^3*b^2*d^2*e
^3*ln(e*x+d)-60*a*b^4*d*e^4*x^3-120*b^5*d^4*e*x*ln(e*x+d))/(b*x+a)^5/e^6/(e*x+d)^2

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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)^3,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 )}^3} \,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)^3,x)

[Out]

int((a^2 + b^2*x^2 + 2*a*b*x)^(5/2)/(d + e*x)^3, 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 )^{3}}\, 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)**3,x)

[Out]

Integral(((a + b*x)**2)**(5/2)/(d + e*x)**3, x)

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