3.15.99 \(\int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^7} \, dx\)

Optimal. Leaf size=158 \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2} (-a B e-A b e+2 b B d)}{5 e^3 (a+b x) (d+e x)^5}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) (B d-A e)}{6 e^3 (a+b x) (d+e x)^6}-\frac {b B \sqrt {a^2+2 a b x+b^2 x^2}}{4 e^3 (a+b x) (d+e x)^4} \]

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Rubi [A]  time = 0.10, 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)}{5 e^3 (a+b x) (d+e x)^5}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) (B d-A e)}{6 e^3 (a+b x) (d+e x)^6}-\frac {b B \sqrt {a^2+2 a b x+b^2 x^2}}{4 e^3 (a+b x) (d+e x)^4} \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)^7,x]

[Out]

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

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)^7} \, 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)^7} \, 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)^7}+\frac {b (-2 b B d+A b e+a B e)}{e^2 (d+e x)^6}+\frac {b^2 B}{e^2 (d+e x)^5}\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}}{6 e^3 (a+b x) (d+e x)^6}+\frac {(2 b B d-A b e-a B e) \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^3 (a+b x) (d+e x)^5}-\frac {b B \sqrt {a^2+2 a b x+b^2 x^2}}{4 e^3 (a+b x) (d+e x)^4}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 82, normalized size = 0.52 \begin {gather*} -\frac {\sqrt {(a+b x)^2} \left (2 a e (5 A e+B (d+6 e x))+b \left (2 A e (d+6 e x)+B \left (d^2+6 d e x+15 e^2 x^2\right )\right )\right )}{60 e^3 (a+b x) (d+e x)^6} \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)^7,x]

[Out]

-1/60*(Sqrt[(a + b*x)^2]*(2*a*e*(5*A*e + B*(d + 6*e*x)) + b*(2*A*e*(d + 6*e*x) + B*(d^2 + 6*d*e*x + 15*e^2*x^2
))))/(e^3*(a + b*x)*(d + e*x)^6)

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IntegrateAlgebraic [F]  time = 180.55, 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)^7,x]

[Out]

$Aborted

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fricas [A]  time = 0.42, size = 126, normalized size = 0.80 \begin {gather*} -\frac {15 \, B b e^{2} x^{2} + B b d^{2} + 10 \, A a e^{2} + 2 \, {\left (B a + A b\right )} d e + 6 \, {\left (B b d e + 2 \, {\left (B a + A b\right )} e^{2}\right )} x}{60 \, {\left (e^{9} x^{6} + 6 \, d e^{8} x^{5} + 15 \, d^{2} e^{7} x^{4} + 20 \, d^{3} e^{6} x^{3} + 15 \, d^{4} e^{5} x^{2} + 6 \, d^{5} e^{4} x + d^{6} 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)^7,x, algorithm="fricas")

[Out]

-1/60*(15*B*b*e^2*x^2 + B*b*d^2 + 10*A*a*e^2 + 2*(B*a + A*b)*d*e + 6*(B*b*d*e + 2*(B*a + A*b)*e^2)*x)/(e^9*x^6
+ 6*d*e^8*x^5 + 15*d^2*e^7*x^4 + 20*d^3*e^6*x^3 + 15*d^4*e^5*x^2 + 6*d^5*e^4*x + d^6*e^3)

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giac [A]  time = 0.16, size = 118, normalized size = 0.75 \begin {gather*} -\frac {{\left (15 \, B b x^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 6 \, B b d x e \mathrm {sgn}\left (b x + a\right ) + B b d^{2} \mathrm {sgn}\left (b x + a\right ) + 12 \, B a x e^{2} \mathrm {sgn}\left (b x + a\right ) + 12 \, A b x e^{2} \mathrm {sgn}\left (b x + a\right ) + 2 \, B a d e \mathrm {sgn}\left (b x + a\right ) + 2 \, A b d e \mathrm {sgn}\left (b x + a\right ) + 10 \, A a e^{2} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-3\right )}}{60 \, {\left (x e + d\right )}^{6}} \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)^7,x, algorithm="giac")

[Out]

-1/60*(15*B*b*x^2*e^2*sgn(b*x + a) + 6*B*b*d*x*e*sgn(b*x + a) + B*b*d^2*sgn(b*x + a) + 12*B*a*x*e^2*sgn(b*x +
a) + 12*A*b*x*e^2*sgn(b*x + a) + 2*B*a*d*e*sgn(b*x + a) + 2*A*b*d*e*sgn(b*x + a) + 10*A*a*e^2*sgn(b*x + a))*e^
(-3)/(x*e + d)^6

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maple [A]  time = 0.05, size = 88, normalized size = 0.56 \begin {gather*} -\frac {\left (15 B b \,e^{2} x^{2}+12 A b \,e^{2} x +12 B a \,e^{2} x +6 B b d e x +10 A a \,e^{2}+2 A b d e +2 B a d e +B b \,d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{60 \left (e x +d \right )^{6} \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)^7,x)

[Out]

-1/60/e^3*(15*B*b*e^2*x^2+12*A*b*e^2*x+12*B*a*e^2*x+6*B*b*d*e*x+10*A*a*e^2+2*A*b*d*e+2*B*a*d*e+B*b*d^2)*((b*x+
a)^2)^(1/2)/(e*x+d)^6/(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)^7,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 = 2.16, size = 87, normalized size = 0.55 \begin {gather*} -\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (10\,A\,a\,e^2+B\,b\,d^2+12\,A\,b\,e^2\,x+12\,B\,a\,e^2\,x+15\,B\,b\,e^2\,x^2+2\,A\,b\,d\,e+2\,B\,a\,d\,e+6\,B\,b\,d\,e\,x\right )}{60\,e^3\,\left (a+b\,x\right )\,{\left (d+e\,x\right )}^6} \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)^7,x)

[Out]

-(((a + b*x)^2)^(1/2)*(10*A*a*e^2 + B*b*d^2 + 12*A*b*e^2*x + 12*B*a*e^2*x + 15*B*b*e^2*x^2 + 2*A*b*d*e + 2*B*a
*d*e + 6*B*b*d*e*x))/(60*e^3*(a + b*x)*(d + e*x)^6)

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sympy [A]  time = 6.57, size = 144, normalized size = 0.91 \begin {gather*} \frac {- 10 A a e^{2} - 2 A b d e - 2 B a d e - B b d^{2} - 15 B b e^{2} x^{2} + x \left (- 12 A b e^{2} - 12 B a e^{2} - 6 B b d e\right )}{60 d^{6} e^{3} + 360 d^{5} e^{4} x + 900 d^{4} e^{5} x^{2} + 1200 d^{3} e^{6} x^{3} + 900 d^{2} e^{7} x^{4} + 360 d e^{8} x^{5} + 60 e^{9} x^{6}} \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)**7,x)

[Out]

(-10*A*a*e**2 - 2*A*b*d*e - 2*B*a*d*e - B*b*d**2 - 15*B*b*e**2*x**2 + x*(-12*A*b*e**2 - 12*B*a*e**2 - 6*B*b*d*
e))/(60*d**6*e**3 + 360*d**5*e**4*x + 900*d**4*e**5*x**2 + 1200*d**3*e**6*x**3 + 900*d**2*e**7*x**4 + 360*d*e*
*8*x**5 + 60*e**9*x**6)

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