3.2105 \(\int \frac {(a+b x) (a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^{5/2}} \, dx\)

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

[Out]

-2/3*(-a*e+b*d)^4*((b*x+a)^2)^(1/2)/e^5/(b*x+a)/(e*x+d)^(3/2)-8/3*b^3*(-a*e+b*d)*(e*x+d)^(3/2)*((b*x+a)^2)^(1/
2)/e^5/(b*x+a)+2/5*b^4*(e*x+d)^(5/2)*((b*x+a)^2)^(1/2)/e^5/(b*x+a)+8*b*(-a*e+b*d)^3*((b*x+a)^2)^(1/2)/e^5/(b*x
+a)/(e*x+d)^(1/2)+12*b^2*(-a*e+b*d)^2*(e*x+d)^(1/2)*((b*x+a)^2)^(1/2)/e^5/(b*x+a)

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

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*d - a*e)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^5*(a + b*x)*(d + e*x)^(3/2)) + (8*b*(b*d - a*e)^3*Sqrt[a
^2 + 2*a*b*x + b^2*x^2])/(e^5*(a + b*x)*Sqrt[d + e*x]) + (12*b^2*(b*d - a*e)^2*Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*
x + b^2*x^2])/(e^5*(a + b*x)) - (8*b^3*(b*d - a*e)*(d + e*x)^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^5*(a +
b*x)) + (2*b^4*(d + e*x)^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*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)^{5/2}} \, 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)^{5/2}} \, 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)^{5/2}} \, 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)^{5/2}}-\frac {4 b (b d-a e)^3}{e^4 (d+e x)^{3/2}}+\frac {6 b^2 (b d-a e)^2}{e^4 \sqrt {d+e x}}-\frac {4 b^3 (b d-a e) \sqrt {d+e x}}{e^4}+\frac {b^4 (d+e x)^{3/2}}{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}}{3 e^5 (a+b x) (d+e x)^{3/2}}+\frac {8 b (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x) \sqrt {d+e x}}+\frac {12 b^2 (b d-a e)^2 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x)}-\frac {8 b^3 (b d-a e) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^5 (a+b x)}+\frac {2 b^4 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^5 (a+b x)}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 171, normalized size = 0.66 \[ \frac {2 \sqrt {(a+b x)^2} \left (-5 a^4 e^4-20 a^3 b e^3 (2 d+3 e x)+30 a^2 b^2 e^2 \left (8 d^2+12 d e x+3 e^2 x^2\right )+20 a b^3 e \left (-16 d^3-24 d^2 e x-6 d e^2 x^2+e^3 x^3\right )+b^4 \left (128 d^4+192 d^3 e x+48 d^2 e^2 x^2-8 d e^3 x^3+3 e^4 x^4\right )\right )}{15 e^5 (a+b x) (d+e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.78, size = 203, normalized size = 0.78 \[ \frac {2 \, {\left (3 \, b^{4} e^{4} x^{4} + 128 \, b^{4} d^{4} - 320 \, a b^{3} d^{3} e + 240 \, a^{2} b^{2} d^{2} e^{2} - 40 \, a^{3} b d e^{3} - 5 \, a^{4} e^{4} - 4 \, {\left (2 \, b^{4} d e^{3} - 5 \, a b^{3} e^{4}\right )} x^{3} + 6 \, {\left (8 \, b^{4} d^{2} e^{2} - 20 \, a b^{3} d e^{3} + 15 \, a^{2} b^{2} e^{4}\right )} x^{2} + 12 \, {\left (16 \, b^{4} d^{3} e - 40 \, a b^{3} d^{2} e^{2} + 30 \, a^{2} b^{2} d e^{3} - 5 \, a^{3} b e^{4}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} e^{5}\right )}} \]

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)^(5/2),x, algorithm="fricas")

[Out]

2/15*(3*b^4*e^4*x^4 + 128*b^4*d^4 - 320*a*b^3*d^3*e + 240*a^2*b^2*d^2*e^2 - 40*a^3*b*d*e^3 - 5*a^4*e^4 - 4*(2*
b^4*d*e^3 - 5*a*b^3*e^4)*x^3 + 6*(8*b^4*d^2*e^2 - 20*a*b^3*d*e^3 + 15*a^2*b^2*e^4)*x^2 + 12*(16*b^4*d^3*e - 40
*a*b^3*d^2*e^2 + 30*a^2*b^2*d*e^3 - 5*a^3*b*e^4)*x)*sqrt(e*x + d)/(e^7*x^2 + 2*d*e^6*x + d^2*e^5)

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giac [A]  time = 0.26, size = 319, normalized size = 1.23 \[ \frac {2}{15} \, {\left (3 \, {\left (x e + d\right )}^{\frac {5}{2}} b^{4} e^{20} \mathrm {sgn}\left (b x + a\right ) - 20 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{4} d e^{20} \mathrm {sgn}\left (b x + a\right ) + 90 \, \sqrt {x e + d} b^{4} d^{2} e^{20} \mathrm {sgn}\left (b x + a\right ) + 20 \, {\left (x e + d\right )}^{\frac {3}{2}} a b^{3} e^{21} \mathrm {sgn}\left (b x + a\right ) - 180 \, \sqrt {x e + d} a b^{3} d e^{21} \mathrm {sgn}\left (b x + a\right ) + 90 \, \sqrt {x e + d} a^{2} b^{2} e^{22} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-25\right )} + \frac {2 \, {\left (12 \, {\left (x e + d\right )} b^{4} d^{3} \mathrm {sgn}\left (b x + a\right ) - b^{4} d^{4} \mathrm {sgn}\left (b x + a\right ) - 36 \, {\left (x e + d\right )} a b^{3} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 4 \, a b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 36 \, {\left (x e + d\right )} a^{2} b^{2} d e^{2} \mathrm {sgn}\left (b x + a\right ) - 6 \, a^{2} b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 12 \, {\left (x e + d\right )} a^{3} b e^{3} \mathrm {sgn}\left (b x + a\right ) + 4 \, a^{3} b d e^{3} \mathrm {sgn}\left (b x + a\right ) - a^{4} e^{4} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-5\right )}}{3 \, {\left (x e + d\right )}^{\frac {3}{2}}} \]

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)^(5/2),x, algorithm="giac")

[Out]

2/15*(3*(x*e + d)^(5/2)*b^4*e^20*sgn(b*x + a) - 20*(x*e + d)^(3/2)*b^4*d*e^20*sgn(b*x + a) + 90*sqrt(x*e + d)*
b^4*d^2*e^20*sgn(b*x + a) + 20*(x*e + d)^(3/2)*a*b^3*e^21*sgn(b*x + a) - 180*sqrt(x*e + d)*a*b^3*d*e^21*sgn(b*
x + a) + 90*sqrt(x*e + d)*a^2*b^2*e^22*sgn(b*x + a))*e^(-25) + 2/3*(12*(x*e + d)*b^4*d^3*sgn(b*x + a) - b^4*d^
4*sgn(b*x + a) - 36*(x*e + d)*a*b^3*d^2*e*sgn(b*x + a) + 4*a*b^3*d^3*e*sgn(b*x + a) + 36*(x*e + d)*a^2*b^2*d*e
^2*sgn(b*x + a) - 6*a^2*b^2*d^2*e^2*sgn(b*x + a) - 12*(x*e + d)*a^3*b*e^3*sgn(b*x + a) + 4*a^3*b*d*e^3*sgn(b*x
+ a) - a^4*e^4*sgn(b*x + a))*e^(-5)/(x*e + d)^(3/2)

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

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)^(5/2),x)

[Out]

-2/15/(e*x+d)^(3/2)*(-3*b^4*e^4*x^4-20*a*b^3*e^4*x^3+8*b^4*d*e^3*x^3-90*a^2*b^2*e^4*x^2+120*a*b^3*d*e^3*x^2-48
*b^4*d^2*e^2*x^2+60*a^3*b*e^4*x-360*a^2*b^2*d*e^3*x+480*a*b^3*d^2*e^2*x-192*b^4*d^3*e*x+5*a^4*e^4+40*a^3*b*d*e
^3-240*a^2*b^2*d^2*e^2+320*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 = 0.73, size = 304, normalized size = 1.17 \[ \frac {2 \, {\left (b^{3} e^{3} x^{3} - 16 \, b^{3} d^{3} + 24 \, a b^{2} d^{2} e - 6 \, a^{2} b d e^{2} - a^{3} e^{3} - 3 \, {\left (2 \, b^{3} d e^{2} - 3 \, a b^{2} e^{3}\right )} x^{2} - 3 \, {\left (8 \, b^{3} d^{2} e - 12 \, a b^{2} d e^{2} + 3 \, a^{2} b e^{3}\right )} x\right )} a}{3 \, {\left (e^{5} x + d e^{4}\right )} \sqrt {e x + d}} + \frac {2 \, {\left (3 \, b^{3} e^{4} x^{4} + 128 \, b^{3} d^{4} - 240 \, a b^{2} d^{3} e + 120 \, a^{2} b d^{2} e^{2} - 10 \, a^{3} d e^{3} - {\left (8 \, b^{3} d e^{3} - 15 \, a b^{2} e^{4}\right )} x^{3} + 3 \, {\left (16 \, b^{3} d^{2} e^{2} - 30 \, a b^{2} d e^{3} + 15 \, a^{2} b e^{4}\right )} x^{2} + 3 \, {\left (64 \, b^{3} d^{3} e - 120 \, a b^{2} d^{2} e^{2} + 60 \, a^{2} b d e^{3} - 5 \, a^{3} e^{4}\right )} x\right )} b}{15 \, {\left (e^{6} x + d e^{5}\right )} \sqrt {e x + d}} \]

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)^(5/2),x, algorithm="maxima")

[Out]

2/3*(b^3*e^3*x^3 - 16*b^3*d^3 + 24*a*b^2*d^2*e - 6*a^2*b*d*e^2 - a^3*e^3 - 3*(2*b^3*d*e^2 - 3*a*b^2*e^3)*x^2 -
3*(8*b^3*d^2*e - 12*a*b^2*d*e^2 + 3*a^2*b*e^3)*x)*a/((e^5*x + d*e^4)*sqrt(e*x + d)) + 2/15*(3*b^3*e^4*x^4 + 1
28*b^3*d^4 - 240*a*b^2*d^3*e + 120*a^2*b*d^2*e^2 - 10*a^3*d*e^3 - (8*b^3*d*e^3 - 15*a*b^2*e^4)*x^3 + 3*(16*b^3
*d^2*e^2 - 30*a*b^2*d*e^3 + 15*a^2*b*e^4)*x^2 + 3*(64*b^3*d^3*e - 120*a*b^2*d^2*e^2 + 60*a^2*b*d*e^3 - 5*a^3*e
^4)*x)*b/((e^6*x + d*e^5)*sqrt(e*x + d))

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mupad [B]  time = 2.89, size = 254, normalized size = 0.98 \[ \frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (\frac {2\,b^3\,x^4}{5\,e^2}-\frac {10\,a^4\,e^4+80\,a^3\,b\,d\,e^3-480\,a^2\,b^2\,d^2\,e^2+640\,a\,b^3\,d^3\,e-256\,b^4\,d^4}{15\,b\,e^6}-\frac {x\,\left (120\,a^3\,b\,e^4-720\,a^2\,b^2\,d\,e^3+960\,a\,b^3\,d^2\,e^2-384\,b^4\,d^3\,e\right )}{15\,b\,e^6}+\frac {8\,b^2\,x^3\,\left (5\,a\,e-2\,b\,d\right )}{15\,e^3}+\frac {4\,b\,x^2\,\left (15\,a^2\,e^2-20\,a\,b\,d\,e+8\,b^2\,d^2\right )}{5\,e^4}\right )}{x^2\,\sqrt {d+e\,x}+\frac {a\,d\,\sqrt {d+e\,x}}{b\,e}+\frac {x\,\left (15\,a\,e^6+15\,b\,d\,e^5\right )\,\sqrt {d+e\,x}}{15\,b\,e^6}} \]

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)^(5/2),x)

[Out]

((a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*((2*b^3*x^4)/(5*e^2) - (10*a^4*e^4 - 256*b^4*d^4 - 480*a^2*b^2*d^2*e^2 + 640*
a*b^3*d^3*e + 80*a^3*b*d*e^3)/(15*b*e^6) - (x*(120*a^3*b*e^4 - 384*b^4*d^3*e + 960*a*b^3*d^2*e^2 - 720*a^2*b^2
*d*e^3))/(15*b*e^6) + (8*b^2*x^3*(5*a*e - 2*b*d))/(15*e^3) + (4*b*x^2*(15*a^2*e^2 + 8*b^2*d^2 - 20*a*b*d*e))/(
5*e^4)))/(x^2*(d + e*x)^(1/2) + (a*d*(d + e*x)^(1/2))/(b*e) + (x*(15*a*e^6 + 15*b*d*e^5)*(d + e*x)^(1/2))/(15*
b*e^6))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{\frac {5}{2}}}\, dx \]

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)**(5/2),x)

[Out]

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

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