3.16.11 \(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^6} \, dx\)

Optimal. Leaf size=106 \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^3 (A b-a B)}{4 (d+e x)^4 (b d-a e)^2}+\frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2} (B d-A e)}{5 (d+e x)^5 (b d-a e)^2} \]

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Rubi [A]  time = 0.06, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {769, 646, 37} \begin {gather*} \frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^3 (A b-a B)}{4 (d+e x)^4 (b d-a e)^2}+\frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2} (B d-A e)}{5 (d+e x)^5 (b d-a e)^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)^6,x]

[Out]

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

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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]

Rule 769

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(-2*c*(e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(2*c*d - b*e)^2), x] + Dist[(2*c*f -
b*g)/(2*c*d - b*e), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x]
&& EqQ[b^2 - 4*a*c, 0] && EqQ[m + 2*p + 3, 0] && NeQ[2*c*f - b*g, 0] && NeQ[2*c*d - b*e, 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)^6} \, dx &=\frac {(B d-A e) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 (b d-a e)^2 (d+e x)^5}+\frac {(A b-a B) \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^5} \, dx}{b d-a e}\\ &=\frac {(B d-A e) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 (b d-a e)^2 (d+e x)^5}+\frac {\left ((A b-a B) \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \frac {\left (a b+b^2 x\right )^3}{(d+e x)^5} \, dx}{b^2 (b d-a e) \left (a b+b^2 x\right )}\\ &=\frac {(A b-a B) (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (b d-a e)^2 (d+e x)^4}+\frac {(B d-A e) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 (b d-a e)^2 (d+e x)^5}\\ \end {align*}

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Mathematica [B]  time = 0.10, size = 229, normalized size = 2.16 \begin {gather*} -\frac {\sqrt {(a+b x)^2} \left (a^3 e^3 (4 A e+B (d+5 e x))+a^2 b e^2 \left (3 A e (d+5 e x)+2 B \left (d^2+5 d e x+10 e^2 x^2\right )\right )+a b^2 e \left (2 A e \left (d^2+5 d e x+10 e^2 x^2\right )+3 B \left (d^3+5 d^2 e x+10 d e^2 x^2+10 e^3 x^3\right )\right )+b^3 \left (A e \left (d^3+5 d^2 e x+10 d e^2 x^2+10 e^3 x^3\right )+4 B \left (d^4+5 d^3 e x+10 d^2 e^2 x^2+10 d e^3 x^3+5 e^4 x^4\right )\right )\right )}{20 e^5 (a+b x) (d+e x)^5} \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)^6,x]

[Out]

-1/20*(Sqrt[(a + b*x)^2]*(a^3*e^3*(4*A*e + B*(d + 5*e*x)) + a^2*b*e^2*(3*A*e*(d + 5*e*x) + 2*B*(d^2 + 5*d*e*x
+ 10*e^2*x^2)) + a*b^2*e*(2*A*e*(d^2 + 5*d*e*x + 10*e^2*x^2) + 3*B*(d^3 + 5*d^2*e*x + 10*d*e^2*x^2 + 10*e^3*x^
3)) + b^3*(A*e*(d^3 + 5*d^2*e*x + 10*d*e^2*x^2 + 10*e^3*x^3) + 4*B*(d^4 + 5*d^3*e*x + 10*d^2*e^2*x^2 + 10*d*e^
3*x^3 + 5*e^4*x^4))))/(e^5*(a + b*x)*(d + e*x)^5)

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IntegrateAlgebraic [F]  time = 180.02, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^6,x]

[Out]

$Aborted

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fricas [B]  time = 0.41, size = 304, normalized size = 2.87 \begin {gather*} -\frac {20 \, B b^{3} e^{4} x^{4} + 4 \, B b^{3} d^{4} + 4 \, A a^{3} e^{4} + {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 2 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} + {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} + 10 \, {\left (4 \, B b^{3} d e^{3} + {\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} + 10 \, {\left (4 \, B b^{3} d^{2} e^{2} + {\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} + 2 \, {\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} + 5 \, {\left (4 \, B b^{3} d^{3} e + {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + 2 \, {\left (B a^{2} b + A a b^{2}\right )} d e^{3} + {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{4}\right )} x}{20 \, {\left (e^{10} x^{5} + 5 \, d e^{9} x^{4} + 10 \, d^{2} e^{8} x^{3} + 10 \, d^{3} e^{7} x^{2} + 5 \, d^{4} e^{6} x + d^{5} 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)^6,x, algorithm="fricas")

[Out]

-1/20*(20*B*b^3*e^4*x^4 + 4*B*b^3*d^4 + 4*A*a^3*e^4 + (3*B*a*b^2 + A*b^3)*d^3*e + 2*(B*a^2*b + A*a*b^2)*d^2*e^
2 + (B*a^3 + 3*A*a^2*b)*d*e^3 + 10*(4*B*b^3*d*e^3 + (3*B*a*b^2 + A*b^3)*e^4)*x^3 + 10*(4*B*b^3*d^2*e^2 + (3*B*
a*b^2 + A*b^3)*d*e^3 + 2*(B*a^2*b + A*a*b^2)*e^4)*x^2 + 5*(4*B*b^3*d^3*e + (3*B*a*b^2 + A*b^3)*d^2*e^2 + 2*(B*
a^2*b + A*a*b^2)*d*e^3 + (B*a^3 + 3*A*a^2*b)*e^4)*x)/(e^10*x^5 + 5*d*e^9*x^4 + 10*d^2*e^8*x^3 + 10*d^3*e^7*x^2
+ 5*d^4*e^6*x + d^5*e^5)

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giac [B]  time = 0.19, size = 425, normalized size = 4.01 \begin {gather*} -\frac {{\left (20 \, B b^{3} x^{4} e^{4} \mathrm {sgn}\left (b x + a\right ) + 40 \, B b^{3} d x^{3} e^{3} \mathrm {sgn}\left (b x + a\right ) + 40 \, B b^{3} d^{2} x^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 20 \, B b^{3} d^{3} x e \mathrm {sgn}\left (b x + a\right ) + 4 \, B b^{3} d^{4} \mathrm {sgn}\left (b x + a\right ) + 30 \, B a b^{2} x^{3} e^{4} \mathrm {sgn}\left (b x + a\right ) + 10 \, A b^{3} x^{3} e^{4} \mathrm {sgn}\left (b x + a\right ) + 30 \, B a b^{2} d x^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 10 \, A b^{3} d x^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 15 \, B a b^{2} d^{2} x e^{2} \mathrm {sgn}\left (b x + a\right ) + 5 \, A b^{3} d^{2} x e^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, B a b^{2} d^{3} e \mathrm {sgn}\left (b x + a\right ) + A b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 20 \, B a^{2} b x^{2} e^{4} \mathrm {sgn}\left (b x + a\right ) + 20 \, A a b^{2} x^{2} e^{4} \mathrm {sgn}\left (b x + a\right ) + 10 \, B a^{2} b d x e^{3} \mathrm {sgn}\left (b x + a\right ) + 10 \, A a b^{2} d x e^{3} \mathrm {sgn}\left (b x + a\right ) + 2 \, B a^{2} b d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 2 \, A a b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 5 \, B a^{3} x e^{4} \mathrm {sgn}\left (b x + a\right ) + 15 \, A a^{2} b x e^{4} \mathrm {sgn}\left (b x + a\right ) + B a^{3} d e^{3} \mathrm {sgn}\left (b x + a\right ) + 3 \, A a^{2} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + 4 \, A a^{3} e^{4} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-5\right )}}{20 \, {\left (x e + d\right )}^{5}} \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)^6,x, algorithm="giac")

[Out]

-1/20*(20*B*b^3*x^4*e^4*sgn(b*x + a) + 40*B*b^3*d*x^3*e^3*sgn(b*x + a) + 40*B*b^3*d^2*x^2*e^2*sgn(b*x + a) + 2
0*B*b^3*d^3*x*e*sgn(b*x + a) + 4*B*b^3*d^4*sgn(b*x + a) + 30*B*a*b^2*x^3*e^4*sgn(b*x + a) + 10*A*b^3*x^3*e^4*s
gn(b*x + a) + 30*B*a*b^2*d*x^2*e^3*sgn(b*x + a) + 10*A*b^3*d*x^2*e^3*sgn(b*x + a) + 15*B*a*b^2*d^2*x*e^2*sgn(b
*x + a) + 5*A*b^3*d^2*x*e^2*sgn(b*x + a) + 3*B*a*b^2*d^3*e*sgn(b*x + a) + A*b^3*d^3*e*sgn(b*x + a) + 20*B*a^2*
b*x^2*e^4*sgn(b*x + a) + 20*A*a*b^2*x^2*e^4*sgn(b*x + a) + 10*B*a^2*b*d*x*e^3*sgn(b*x + a) + 10*A*a*b^2*d*x*e^
3*sgn(b*x + a) + 2*B*a^2*b*d^2*e^2*sgn(b*x + a) + 2*A*a*b^2*d^2*e^2*sgn(b*x + a) + 5*B*a^3*x*e^4*sgn(b*x + a)
+ 15*A*a^2*b*x*e^4*sgn(b*x + a) + B*a^3*d*e^3*sgn(b*x + a) + 3*A*a^2*b*d*e^3*sgn(b*x + a) + 4*A*a^3*e^4*sgn(b*
x + a))*e^(-5)/(x*e + d)^5

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maple [B]  time = 0.06, size = 315, normalized size = 2.97 \begin {gather*} -\frac {\left (20 B \,b^{3} e^{4} x^{4}+10 A \,b^{3} e^{4} x^{3}+30 B a \,b^{2} e^{4} x^{3}+40 B \,b^{3} d \,e^{3} x^{3}+20 A a \,b^{2} e^{4} x^{2}+10 A \,b^{3} d \,e^{3} x^{2}+20 B \,a^{2} b \,e^{4} x^{2}+30 B a \,b^{2} d \,e^{3} x^{2}+40 B \,b^{3} d^{2} e^{2} x^{2}+15 A \,a^{2} b \,e^{4} x +10 A a \,b^{2} d \,e^{3} x +5 A \,b^{3} d^{2} e^{2} x +5 B \,a^{3} e^{4} x +10 B \,a^{2} b d \,e^{3} x +15 B a \,b^{2} d^{2} e^{2} x +20 B \,b^{3} d^{3} e x +4 A \,a^{3} e^{4}+3 A \,a^{2} b d \,e^{3}+2 A a \,b^{2} d^{2} e^{2}+A \,b^{3} d^{3} e +B \,a^{3} d \,e^{3}+2 B \,a^{2} b \,d^{2} e^{2}+3 B a \,b^{2} d^{3} e +4 B \,b^{3} d^{4}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{20 \left (e x +d \right )^{5} \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)^6,x)

[Out]

-1/20*(20*B*b^3*e^4*x^4+10*A*b^3*e^4*x^3+30*B*a*b^2*e^4*x^3+40*B*b^3*d*e^3*x^3+20*A*a*b^2*e^4*x^2+10*A*b^3*d*e
^3*x^2+20*B*a^2*b*e^4*x^2+30*B*a*b^2*d*e^3*x^2+40*B*b^3*d^2*e^2*x^2+15*A*a^2*b*e^4*x+10*A*a*b^2*d*e^3*x+5*A*b^
3*d^2*e^2*x+5*B*a^3*e^4*x+10*B*a^2*b*d*e^3*x+15*B*a*b^2*d^2*e^2*x+20*B*b^3*d^3*e*x+4*A*a^3*e^4+3*A*a^2*b*d*e^3
+2*A*a*b^2*d^2*e^2+A*b^3*d^3*e+B*a^3*d*e^3+2*B*a^2*b*d^2*e^2+3*B*a*b^2*d^3*e+4*B*b^3*d^4)*((b*x+a)^2)^(3/2)/(e
*x+d)^5/e^5/(b*x+a)^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*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^6,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 [B]  time = 2.20, size = 577, normalized size = 5.44 \begin {gather*} -\frac {\left (\frac {A\,b^3\,e-3\,B\,b^3\,d+3\,B\,a\,b^2\,e}{2\,e^5}-\frac {B\,b^3\,d}{2\,e^5}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^2}-\frac {\left (\frac {A\,a^3}{5\,e}-\frac {d\,\left (\frac {B\,a^3+3\,A\,b\,a^2}{5\,e}+\frac {d\,\left (\frac {d\,\left (\frac {A\,b^3+3\,B\,a\,b^2}{5\,e}-\frac {B\,b^3\,d}{5\,e^2}\right )}{e}-\frac {3\,a\,b\,\left (A\,b+B\,a\right )}{5\,e}\right )}{e}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^5}-\frac {\left (\frac {B\,a^3\,e^3-3\,B\,a^2\,b\,d\,e^2+3\,A\,a^2\,b\,e^3+3\,B\,a\,b^2\,d^2\,e-3\,A\,a\,b^2\,d\,e^2-B\,b^3\,d^3+A\,b^3\,d^2\,e}{4\,e^5}-\frac {d\,\left (\frac {3\,B\,a^2\,b\,e^3-3\,B\,a\,b^2\,d\,e^2+3\,A\,a\,b^2\,e^3+B\,b^3\,d^2\,e-A\,b^3\,d\,e^2}{4\,e^5}-\frac {d\,\left (\frac {b^2\,\left (A\,b\,e+3\,B\,a\,e-B\,b\,d\right )}{4\,e^3}-\frac {B\,b^3\,d}{4\,e^3}\right )}{e}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^4}-\frac {\left (\frac {3\,B\,a^2\,b\,e^2-6\,B\,a\,b^2\,d\,e+3\,A\,a\,b^2\,e^2+3\,B\,b^3\,d^2-2\,A\,b^3\,d\,e}{3\,e^5}-\frac {d\,\left (\frac {b^2\,\left (A\,b\,e+3\,B\,a\,e-2\,B\,b\,d\right )}{3\,e^4}-\frac {B\,b^3\,d}{3\,e^4}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^3}-\frac {B\,b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{e^5\,\left (a+b\,x\right )\,\left (d+e\,x\right )} \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)^6,x)

[Out]

- (((A*b^3*e - 3*B*b^3*d + 3*B*a*b^2*e)/(2*e^5) - (B*b^3*d)/(2*e^5))*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/((a + b*
x)*(d + e*x)^2) - (((A*a^3)/(5*e) - (d*((B*a^3 + 3*A*a^2*b)/(5*e) + (d*((d*((A*b^3 + 3*B*a*b^2)/(5*e) - (B*b^3
*d)/(5*e^2)))/e - (3*a*b*(A*b + B*a))/(5*e)))/e))/e)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/((a + b*x)*(d + e*x)^5)
- (((B*a^3*e^3 - B*b^3*d^3 + 3*A*a^2*b*e^3 + A*b^3*d^2*e - 3*A*a*b^2*d*e^2 + 3*B*a*b^2*d^2*e - 3*B*a^2*b*d*e^2
)/(4*e^5) - (d*((3*A*a*b^2*e^3 + 3*B*a^2*b*e^3 - A*b^3*d*e^2 + B*b^3*d^2*e - 3*B*a*b^2*d*e^2)/(4*e^5) - (d*((b
^2*(A*b*e + 3*B*a*e - B*b*d))/(4*e^3) - (B*b^3*d)/(4*e^3)))/e))/e)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/((a + b*x)
*(d + e*x)^4) - (((3*B*b^3*d^2 - 2*A*b^3*d*e + 3*A*a*b^2*e^2 + 3*B*a^2*b*e^2 - 6*B*a*b^2*d*e)/(3*e^5) - (d*((b
^2*(A*b*e + 3*B*a*e - 2*B*b*d))/(3*e^4) - (B*b^3*d)/(3*e^4)))/e)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/((a + b*x)*(
d + e*x)^3) - (B*b^3*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(e^5*(a + b*x)*(d + e*x))

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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 )^{6}}\, 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)**6,x)

[Out]

Integral((A + B*x)*((a + b*x)**2)**(3/2)/(d + e*x)**6, x)

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