3.19.81 \(\int \frac {(a+b x) (a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^{7/2}} \, dx\)

Optimal. Leaf size=260 \[ -\frac {12 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^5 (a+b x) \sqrt {d+e x}}+\frac {8 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}{3 e^5 (a+b x) (d+e x)^{3/2}}-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}{5 e^5 (a+b x) (d+e x)^{5/2}}+\frac {2 b^4 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2}}{3 e^5 (a+b x)}-\frac {8 b^3 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)}{e^5 (a+b x)} \]

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Rubi [A]  time = 0.10, antiderivative size = 260, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {770, 21, 43} \begin {gather*} \frac {2 b^4 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2}}{3 e^5 (a+b x)}-\frac {8 b^3 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)}{e^5 (a+b x)}-\frac {12 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^5 (a+b x) \sqrt {d+e x}}+\frac {8 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}{3 e^5 (a+b x) (d+e x)^{3/2}}-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}{5 e^5 (a+b x) (d+e x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^(7/2),x]

[Out]

(-2*(b*d - a*e)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^5*(a + b*x)*(d + e*x)^(5/2)) + (8*b*(b*d - a*e)^3*Sqrt[a
^2 + 2*a*b*x + b^2*x^2])/(3*e^5*(a + b*x)*(d + e*x)^(3/2)) - (12*b^2*(b*d - a*e)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^
2])/(e^5*(a + b*x)*Sqrt[d + e*x]) - (8*b^3*(b*d - a*e)*Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^5*(a +
b*x)) + (2*b^4*(d + e*x)^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^5*(a + b*x))

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, 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 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{7/2}} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x) \left (a b+b^2 x\right )^3}{(d+e x)^{7/2}} \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac {\left (b \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \frac {(a+b x)^4}{(d+e x)^{7/2}} \, dx}{a b+b^2 x}\\ &=\frac {\left (b \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \left (\frac {(-b d+a e)^4}{e^4 (d+e x)^{7/2}}-\frac {4 b (b d-a e)^3}{e^4 (d+e x)^{5/2}}+\frac {6 b^2 (b d-a e)^2}{e^4 (d+e x)^{3/2}}-\frac {4 b^3 (b d-a e)}{e^4 \sqrt {d+e x}}+\frac {b^4 \sqrt {d+e x}}{e^4}\right ) \, dx}{a b+b^2 x}\\ &=-\frac {2 (b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^5 (a+b x) (d+e x)^{5/2}}+\frac {8 b (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^5 (a+b x) (d+e x)^{3/2}}-\frac {12 b^2 (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x) \sqrt {d+e x}}-\frac {8 b^3 (b d-a e) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x)}+\frac {2 b^4 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^5 (a+b x)}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 172, normalized size = 0.66 \begin {gather*} -\frac {2 \sqrt {(a+b x)^2} \left (3 a^4 e^4+4 a^3 b e^3 (2 d+5 e x)+6 a^2 b^2 e^2 \left (8 d^2+20 d e x+15 e^2 x^2\right )-12 a b^3 e \left (16 d^3+40 d^2 e x+30 d e^2 x^2+5 e^3 x^3\right )+b^4 \left (128 d^4+320 d^3 e x+240 d^2 e^2 x^2+40 d e^3 x^3-5 e^4 x^4\right )\right )}{15 e^5 (a+b x) (d+e x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^(7/2),x]

[Out]

(-2*Sqrt[(a + b*x)^2]*(3*a^4*e^4 + 4*a^3*b*e^3*(2*d + 5*e*x) + 6*a^2*b^2*e^2*(8*d^2 + 20*d*e*x + 15*e^2*x^2) -
12*a*b^3*e*(16*d^3 + 40*d^2*e*x + 30*d*e^2*x^2 + 5*e^3*x^3) + b^4*(128*d^4 + 320*d^3*e*x + 240*d^2*e^2*x^2 +
40*d*e^3*x^3 - 5*e^4*x^4)))/(15*e^5*(a + b*x)*(d + e*x)^(5/2))

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IntegrateAlgebraic [A]  time = 23.23, size = 241, normalized size = 0.93 \begin {gather*} \frac {2 \sqrt {\frac {(a e+b e x)^2}{e^2}} \left (-3 a^4 e^4-20 a^3 b e^3 (d+e x)+12 a^3 b d e^3-18 a^2 b^2 d^2 e^2-90 a^2 b^2 e^2 (d+e x)^2+60 a^2 b^2 d e^2 (d+e x)+12 a b^3 d^3 e-60 a b^3 d^2 e (d+e x)+60 a b^3 e (d+e x)^3+180 a b^3 d e (d+e x)^2-3 b^4 d^4+20 b^4 d^3 (d+e x)-90 b^4 d^2 (d+e x)^2+5 b^4 (d+e x)^4-60 b^4 d (d+e x)^3\right )}{15 e^4 (d+e x)^{5/2} (a e+b e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^(7/2),x]

[Out]

(2*Sqrt[(a*e + b*e*x)^2/e^2]*(-3*b^4*d^4 + 12*a*b^3*d^3*e - 18*a^2*b^2*d^2*e^2 + 12*a^3*b*d*e^3 - 3*a^4*e^4 +
20*b^4*d^3*(d + e*x) - 60*a*b^3*d^2*e*(d + e*x) + 60*a^2*b^2*d*e^2*(d + e*x) - 20*a^3*b*e^3*(d + e*x) - 90*b^4
*d^2*(d + e*x)^2 + 180*a*b^3*d*e*(d + e*x)^2 - 90*a^2*b^2*e^2*(d + e*x)^2 - 60*b^4*d*(d + e*x)^3 + 60*a*b^3*e*
(d + e*x)^3 + 5*b^4*(d + e*x)^4))/(15*e^4*(d + e*x)^(5/2)*(a*e + b*e*x))

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fricas [A]  time = 0.42, size = 213, normalized size = 0.82 \begin {gather*} \frac {2 \, {\left (5 \, b^{4} e^{4} x^{4} - 128 \, b^{4} d^{4} + 192 \, a b^{3} d^{3} e - 48 \, a^{2} b^{2} d^{2} e^{2} - 8 \, a^{3} b d e^{3} - 3 \, a^{4} e^{4} - 20 \, {\left (2 \, b^{4} d e^{3} - 3 \, a b^{3} e^{4}\right )} x^{3} - 30 \, {\left (8 \, b^{4} d^{2} e^{2} - 12 \, a b^{3} d e^{3} + 3 \, a^{2} b^{2} e^{4}\right )} x^{2} - 20 \, {\left (16 \, b^{4} d^{3} e - 24 \, a b^{3} d^{2} e^{2} + 6 \, a^{2} b^{2} d e^{3} + a^{3} b e^{4}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (e^{8} x^{3} + 3 \, d e^{7} x^{2} + 3 \, d^{2} e^{6} x + d^{3} e^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(7/2),x, algorithm="fricas")

[Out]

2/15*(5*b^4*e^4*x^4 - 128*b^4*d^4 + 192*a*b^3*d^3*e - 48*a^2*b^2*d^2*e^2 - 8*a^3*b*d*e^3 - 3*a^4*e^4 - 20*(2*b
^4*d*e^3 - 3*a*b^3*e^4)*x^3 - 30*(8*b^4*d^2*e^2 - 12*a*b^3*d*e^3 + 3*a^2*b^2*e^4)*x^2 - 20*(16*b^4*d^3*e - 24*
a*b^3*d^2*e^2 + 6*a^2*b^2*d*e^3 + a^3*b*e^4)*x)*sqrt(e*x + d)/(e^8*x^3 + 3*d*e^7*x^2 + 3*d^2*e^6*x + d^3*e^5)

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giac [A]  time = 0.22, size = 316, normalized size = 1.22 \begin {gather*} \frac {2}{3} \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} b^{4} e^{10} \mathrm {sgn}\left (b x + a\right ) - 12 \, \sqrt {x e + d} b^{4} d e^{10} \mathrm {sgn}\left (b x + a\right ) + 12 \, \sqrt {x e + d} a b^{3} e^{11} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-15\right )} - \frac {2 \, {\left (90 \, {\left (x e + d\right )}^{2} b^{4} d^{2} \mathrm {sgn}\left (b x + a\right ) - 20 \, {\left (x e + d\right )} b^{4} d^{3} \mathrm {sgn}\left (b x + a\right ) + 3 \, b^{4} d^{4} \mathrm {sgn}\left (b x + a\right ) - 180 \, {\left (x e + d\right )}^{2} a b^{3} d e \mathrm {sgn}\left (b x + a\right ) + 60 \, {\left (x e + d\right )} a b^{3} d^{2} e \mathrm {sgn}\left (b x + a\right ) - 12 \, a b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 90 \, {\left (x e + d\right )}^{2} a^{2} b^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 60 \, {\left (x e + d\right )} a^{2} b^{2} d e^{2} \mathrm {sgn}\left (b x + a\right ) + 18 \, a^{2} b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 20 \, {\left (x e + d\right )} a^{3} b e^{3} \mathrm {sgn}\left (b x + a\right ) - 12 \, a^{3} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{4} e^{4} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-5\right )}}{15 \, {\left (x e + d\right )}^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(7/2),x, algorithm="giac")

[Out]

2/3*((x*e + d)^(3/2)*b^4*e^10*sgn(b*x + a) - 12*sqrt(x*e + d)*b^4*d*e^10*sgn(b*x + a) + 12*sqrt(x*e + d)*a*b^3
*e^11*sgn(b*x + a))*e^(-15) - 2/15*(90*(x*e + d)^2*b^4*d^2*sgn(b*x + a) - 20*(x*e + d)*b^4*d^3*sgn(b*x + a) +
3*b^4*d^4*sgn(b*x + a) - 180*(x*e + d)^2*a*b^3*d*e*sgn(b*x + a) + 60*(x*e + d)*a*b^3*d^2*e*sgn(b*x + a) - 12*a
*b^3*d^3*e*sgn(b*x + a) + 90*(x*e + d)^2*a^2*b^2*e^2*sgn(b*x + a) - 60*(x*e + d)*a^2*b^2*d*e^2*sgn(b*x + a) +
18*a^2*b^2*d^2*e^2*sgn(b*x + a) + 20*(x*e + d)*a^3*b*e^3*sgn(b*x + a) - 12*a^3*b*d*e^3*sgn(b*x + a) + 3*a^4*e^
4*sgn(b*x + a))*e^(-5)/(x*e + d)^(5/2)

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maple [A]  time = 0.05, size = 202, normalized size = 0.78 \begin {gather*} -\frac {2 \left (-5 b^{4} e^{4} x^{4}-60 a \,b^{3} e^{4} x^{3}+40 b^{4} d \,e^{3} x^{3}+90 a^{2} b^{2} e^{4} x^{2}-360 a \,b^{3} d \,e^{3} x^{2}+240 b^{4} d^{2} e^{2} x^{2}+20 a^{3} b \,e^{4} x +120 a^{2} b^{2} d \,e^{3} x -480 a \,b^{3} d^{2} e^{2} x +320 b^{4} d^{3} e x +3 a^{4} e^{4}+8 a^{3} b d \,e^{3}+48 a^{2} b^{2} d^{2} e^{2}-192 a \,b^{3} d^{3} e +128 b^{4} d^{4}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{15 \left (e x +d \right )^{\frac {5}{2}} \left (b x +a \right )^{3} e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(7/2),x)

[Out]

-2/15/(e*x+d)^(5/2)*(-5*b^4*e^4*x^4-60*a*b^3*e^4*x^3+40*b^4*d*e^3*x^3+90*a^2*b^2*e^4*x^2-360*a*b^3*d*e^3*x^2+2
40*b^4*d^2*e^2*x^2+20*a^3*b*e^4*x+120*a^2*b^2*d*e^3*x-480*a*b^3*d^2*e^2*x+320*b^4*d^3*e*x+3*a^4*e^4+8*a^3*b*d*
e^3+48*a^2*b^2*d^2*e^2-192*a*b^3*d^3*e+128*b^4*d^4)*((b*x+a)^2)^(3/2)/e^5/(b*x+a)^3

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maxima [A]  time = 0.69, size = 326, normalized size = 1.25 \begin {gather*} \frac {2 \, {\left (5 \, b^{3} e^{3} x^{3} + 16 \, b^{3} d^{3} - 8 \, a b^{2} d^{2} e - 2 \, a^{2} b d e^{2} - a^{3} e^{3} + 15 \, {\left (2 \, b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 5 \, {\left (8 \, b^{3} d^{2} e - 4 \, a b^{2} d e^{2} - a^{2} b e^{3}\right )} x\right )} a}{5 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )} \sqrt {e x + d}} + \frac {2 \, {\left (5 \, b^{3} e^{4} x^{4} - 128 \, b^{3} d^{4} + 144 \, a b^{2} d^{3} e - 24 \, a^{2} b d^{2} e^{2} - 2 \, a^{3} d e^{3} - 5 \, {\left (8 \, b^{3} d e^{3} - 9 \, a b^{2} e^{4}\right )} x^{3} - 15 \, {\left (16 \, b^{3} d^{2} e^{2} - 18 \, a b^{2} d e^{3} + 3 \, a^{2} b e^{4}\right )} x^{2} - 5 \, {\left (64 \, b^{3} d^{3} e - 72 \, a b^{2} d^{2} e^{2} + 12 \, a^{2} b d e^{3} + a^{3} e^{4}\right )} x\right )} b}{15 \, {\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} e^{5}\right )} \sqrt {e x + d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(7/2),x, algorithm="maxima")

[Out]

2/5*(5*b^3*e^3*x^3 + 16*b^3*d^3 - 8*a*b^2*d^2*e - 2*a^2*b*d*e^2 - a^3*e^3 + 15*(2*b^3*d*e^2 - a*b^2*e^3)*x^2 +
5*(8*b^3*d^2*e - 4*a*b^2*d*e^2 - a^2*b*e^3)*x)*a/((e^6*x^2 + 2*d*e^5*x + d^2*e^4)*sqrt(e*x + d)) + 2/15*(5*b^
3*e^4*x^4 - 128*b^3*d^4 + 144*a*b^2*d^3*e - 24*a^2*b*d^2*e^2 - 2*a^3*d*e^3 - 5*(8*b^3*d*e^3 - 9*a*b^2*e^4)*x^3
- 15*(16*b^3*d^2*e^2 - 18*a*b^2*d*e^3 + 3*a^2*b*e^4)*x^2 - 5*(64*b^3*d^3*e - 72*a*b^2*d^2*e^2 + 12*a^2*b*d*e^
3 + a^3*e^4)*x)*b/((e^7*x^2 + 2*d*e^6*x + d^2*e^5)*sqrt(e*x + d))

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mupad [B]  time = 2.98, size = 283, normalized size = 1.09 \begin {gather*} -\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (\frac {6\,a^4\,e^4+16\,a^3\,b\,d\,e^3+96\,a^2\,b^2\,d^2\,e^2-384\,a\,b^3\,d^3\,e+256\,b^4\,d^4}{15\,b\,e^7}-\frac {2\,b^3\,x^4}{3\,e^3}+\frac {x\,\left (40\,a^3\,b\,e^4+240\,a^2\,b^2\,d\,e^3-960\,a\,b^3\,d^2\,e^2+640\,b^4\,d^3\,e\right )}{15\,b\,e^7}-\frac {8\,b^2\,x^3\,\left (3\,a\,e-2\,b\,d\right )}{3\,e^4}+\frac {4\,b\,x^2\,\left (3\,a^2\,e^2-12\,a\,b\,d\,e+8\,b^2\,d^2\right )}{e^5}\right )}{x^3\,\sqrt {d+e\,x}+\frac {a\,d^2\,\sqrt {d+e\,x}}{b\,e^2}+\frac {x^2\,\left (15\,a\,e^7+30\,b\,d\,e^6\right )\,\sqrt {d+e\,x}}{15\,b\,e^7}+\frac {d\,x\,\left (2\,a\,e+b\,d\right )\,\sqrt {d+e\,x}}{b\,e^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(d + e*x)^(7/2),x)

[Out]

-((a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*((6*a^4*e^4 + 256*b^4*d^4 + 96*a^2*b^2*d^2*e^2 - 384*a*b^3*d^3*e + 16*a^3*b*
d*e^3)/(15*b*e^7) - (2*b^3*x^4)/(3*e^3) + (x*(40*a^3*b*e^4 + 640*b^4*d^3*e - 960*a*b^3*d^2*e^2 + 240*a^2*b^2*d
*e^3))/(15*b*e^7) - (8*b^2*x^3*(3*a*e - 2*b*d))/(3*e^4) + (4*b*x^2*(3*a^2*e^2 + 8*b^2*d^2 - 12*a*b*d*e))/e^5))
/(x^3*(d + e*x)^(1/2) + (a*d^2*(d + e*x)^(1/2))/(b*e^2) + (x^2*(15*a*e^7 + 30*b*d*e^6)*(d + e*x)^(1/2))/(15*b*
e^7) + (d*x*(2*a*e + b*d)*(d + e*x)^(1/2))/(b*e^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{\frac {7}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(b**2*x**2+2*a*b*x+a**2)**(3/2)/(e*x+d)**(7/2),x)

[Out]

Integral((a + b*x)*((a + b*x)**2)**(3/2)/(d + e*x)**(7/2), x)

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