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

Optimal. Leaf size=158 \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2} (-a B e-A b e+2 b B d)}{4 e^3 (a+b x) (d+e x)^4}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) (B d-A e)}{5 e^3 (a+b x) (d+e x)^5}-\frac {b B \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x) (d+e x)^3} \]

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Rubi [A]  time = 0.09, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {770, 77} \begin {gather*} \frac {\sqrt {a^2+2 a b x+b^2 x^2} (-a B e-A b e+2 b B d)}{4 e^3 (a+b x) (d+e x)^4}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) (B d-A e)}{5 e^3 (a+b x) (d+e x)^5}-\frac {b B \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x) (d+e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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

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Mathematica [A]  time = 0.04, size = 83, normalized size = 0.53 \begin {gather*} -\frac {\sqrt {(a+b x)^2} \left (3 a e (4 A e+B (d+5 e x))+b \left (3 A e (d+5 e x)+2 B \left (d^2+5 d e x+10 e^2 x^2\right )\right )\right )}{60 e^3 (a+b x) (d+e x)^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/60*(Sqrt[(a + b*x)^2]*(3*a*e*(4*A*e + B*(d + 5*e*x)) + b*(3*A*e*(d + 5*e*x) + 2*B*(d^2 + 5*d*e*x + 10*e^2*x
^2))))/(e^3*(a + b*x)*(d + e*x)^5)

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IntegrateAlgebraic [F]  time = 180.42, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

$Aborted

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fricas [A]  time = 0.42, size = 117, normalized size = 0.74 \begin {gather*} -\frac {20 \, B b e^{2} x^{2} + 2 \, B b d^{2} + 12 \, A a e^{2} + 3 \, {\left (B a + A b\right )} d e + 5 \, {\left (2 \, B b d e + 3 \, {\left (B a + A b\right )} e^{2}\right )} x}{60 \, {\left (e^{8} x^{5} + 5 \, d e^{7} x^{4} + 10 \, d^{2} e^{6} x^{3} + 10 \, d^{3} e^{5} x^{2} + 5 \, d^{4} e^{4} x + d^{5} e^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/(e*x+d)^6,x, algorithm="fricas")

[Out]

-1/60*(20*B*b*e^2*x^2 + 2*B*b*d^2 + 12*A*a*e^2 + 3*(B*a + A*b)*d*e + 5*(2*B*b*d*e + 3*(B*a + A*b)*e^2)*x)/(e^8
*x^5 + 5*d*e^7*x^4 + 10*d^2*e^6*x^3 + 10*d^3*e^5*x^2 + 5*d^4*e^4*x + d^5*e^3)

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giac [A]  time = 0.16, size = 119, normalized size = 0.75 \begin {gather*} -\frac {{\left (20 \, B b x^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 10 \, B b d x e \mathrm {sgn}\left (b x + a\right ) + 2 \, B b d^{2} \mathrm {sgn}\left (b x + a\right ) + 15 \, B a x e^{2} \mathrm {sgn}\left (b x + a\right ) + 15 \, A b x e^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, B a d e \mathrm {sgn}\left (b x + a\right ) + 3 \, A b d e \mathrm {sgn}\left (b x + a\right ) + 12 \, A a e^{2} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-3\right )}}{60 \, {\left (x e + d\right )}^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/(e*x+d)^6,x, algorithm="giac")

[Out]

-1/60*(20*B*b*x^2*e^2*sgn(b*x + a) + 10*B*b*d*x*e*sgn(b*x + a) + 2*B*b*d^2*sgn(b*x + a) + 15*B*a*x*e^2*sgn(b*x
+ a) + 15*A*b*x*e^2*sgn(b*x + a) + 3*B*a*d*e*sgn(b*x + a) + 3*A*b*d*e*sgn(b*x + a) + 12*A*a*e^2*sgn(b*x + a))
*e^(-3)/(x*e + d)^5

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maple [A]  time = 0.05, size = 89, normalized size = 0.56 \begin {gather*} -\frac {\left (20 B b \,e^{2} x^{2}+15 A b \,e^{2} x +15 B a \,e^{2} x +10 B b d e x +12 A a \,e^{2}+3 A b d e +3 B a d e +2 B b \,d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{60 \left (e x +d \right )^{5} \left (b x +a \right ) e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*((b*x+a)^2)^(1/2)/(e*x+d)^6,x)

[Out]

-1/60/e^3*(20*B*b*e^2*x^2+15*A*b*e^2*x+15*B*a*e^2*x+10*B*b*d*e*x+12*A*a*e^2+3*A*b*d*e+3*B*a*d*e+2*B*b*d^2)*((b
*x+a)^2)^(1/2)/(e*x+d)^5/(b*x+a)

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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*x+a)^2)^(1/2)/(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 = 2.17, size = 88, normalized size = 0.56 \begin {gather*} -\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (12\,A\,a\,e^2+2\,B\,b\,d^2+15\,A\,b\,e^2\,x+15\,B\,a\,e^2\,x+20\,B\,b\,e^2\,x^2+3\,A\,b\,d\,e+3\,B\,a\,d\,e+10\,B\,b\,d\,e\,x\right )}{60\,e^3\,\left (a+b\,x\right )\,{\left (d+e\,x\right )}^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((a + b*x)^2)^(1/2)*(A + B*x))/(d + e*x)^6,x)

[Out]

-(((a + b*x)^2)^(1/2)*(12*A*a*e^2 + 2*B*b*d^2 + 15*A*b*e^2*x + 15*B*a*e^2*x + 20*B*b*e^2*x^2 + 3*A*b*d*e + 3*B
*a*d*e + 10*B*b*d*e*x))/(60*e^3*(a + b*x)*(d + e*x)^5)

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sympy [A]  time = 4.36, size = 134, normalized size = 0.85 \begin {gather*} \frac {- 12 A a e^{2} - 3 A b d e - 3 B a d e - 2 B b d^{2} - 20 B b e^{2} x^{2} + x \left (- 15 A b e^{2} - 15 B a e^{2} - 10 B b d e\right )}{60 d^{5} e^{3} + 300 d^{4} e^{4} x + 600 d^{3} e^{5} x^{2} + 600 d^{2} e^{6} x^{3} + 300 d e^{7} x^{4} + 60 e^{8} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)**2)**(1/2)/(e*x+d)**6,x)

[Out]

(-12*A*a*e**2 - 3*A*b*d*e - 3*B*a*d*e - 2*B*b*d**2 - 20*B*b*e**2*x**2 + x*(-15*A*b*e**2 - 15*B*a*e**2 - 10*B*b
*d*e))/(60*d**5*e**3 + 300*d**4*e**4*x + 600*d**3*e**5*x**2 + 600*d**2*e**6*x**3 + 300*d*e**7*x**4 + 60*e**8*x
**5)

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