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

Optimal. Leaf size=249 \[ -\frac {(b d-a e)^2 (-4 a B e+3 A b e+b B d)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(A b-a B) (b d-a e)^3}{2 b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 e (a+b x) (b d-a e) \log (a+b x) (-2 a B e+A b e+b B d)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^2 x (a+b x) (-3 a B e+A b e+3 b B d)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B e^3 x^2 (a+b x)}{2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.22, antiderivative size = 249, 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 {e^2 x (a+b x) (-3 a B e+A b e+3 b B d)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(b d-a e)^2 (-4 a B e+3 A b e+b B d)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(A b-a B) (b d-a e)^3}{2 b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 e (a+b x) (b d-a e) \log (a+b x) (-2 a B e+A b e+b B d)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B e^3 x^2 (a+b x)}{2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.16, size = 256, normalized size = 1.03 \begin {gather*} \frac {-A b \left (5 a^3 e^3+a^2 b e^2 (4 e x-9 d)+a b^2 e \left (3 d^2-12 d e x-4 e^2 x^2\right )+b^3 \left (d^3+6 d^2 e x-2 e^3 x^3\right )\right )+B \left (7 a^4 e^3+a^3 b e^2 (2 e x-15 d)+a^2 b^2 e \left (9 d^2-12 d e x-11 e^2 x^2\right )-a b^3 \left (d^3-12 d^2 e x-12 d e^2 x^2+4 e^3 x^3\right )+b^4 x \left (-2 d^3+6 d e^2 x^2+e^3 x^3\right )\right )+6 e (a+b x)^2 (b d-a e) \log (a+b x) (-2 a B e+A b e+b B d)}{2 b^5 (a+b x) \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(A*b*(5*a^3*e^3 + a^2*b*e^2*(-9*d + 4*e*x) + a*b^2*e*(3*d^2 - 12*d*e*x - 4*e^2*x^2) + b^3*(d^3 + 6*d^2*e*x -
2*e^3*x^3))) + B*(7*a^4*e^3 + a^3*b*e^2*(-15*d + 2*e*x) + a^2*b^2*e*(9*d^2 - 12*d*e*x - 11*e^2*x^2) + b^4*x*(
-2*d^3 + 6*d*e^2*x^2 + e^3*x^3) - a*b^3*(d^3 - 12*d^2*e*x - 12*d*e^2*x^2 + 4*e^3*x^3)) + 6*e*(b*d - a*e)*(b*B*
d + A*b*e - 2*a*B*e)*(a + b*x)^2*Log[a + b*x])/(2*b^5*(a + b*x)*Sqrt[(a + b*x)^2])

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IntegrateAlgebraic [B]  time = 9.48, size = 7046, normalized size = 28.30 \begin {gather*} \text {Result too large to show} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Result too large to show

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fricas [B]  time = 0.43, size = 442, normalized size = 1.78 \begin {gather*} \frac {B b^{4} e^{3} x^{4} - {\left (B a b^{3} + A b^{4}\right )} d^{3} + 3 \, {\left (3 \, B a^{2} b^{2} - A a b^{3}\right )} d^{2} e - 3 \, {\left (5 \, B a^{3} b - 3 \, A a^{2} b^{2}\right )} d e^{2} + {\left (7 \, B a^{4} - 5 \, A a^{3} b\right )} e^{3} + 2 \, {\left (3 \, B b^{4} d e^{2} - {\left (2 \, B a b^{3} - A b^{4}\right )} e^{3}\right )} x^{3} + {\left (12 \, B a b^{3} d e^{2} - {\left (11 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )} e^{3}\right )} x^{2} - 2 \, {\left (B b^{4} d^{3} - 3 \, {\left (2 \, B a b^{3} - A b^{4}\right )} d^{2} e + 6 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} d e^{2} - {\left (B a^{3} b - 2 \, A a^{2} b^{2}\right )} e^{3}\right )} x + 6 \, {\left (B a^{2} b^{2} d^{2} e - {\left (3 \, B a^{3} b - A a^{2} b^{2}\right )} d e^{2} + {\left (2 \, B a^{4} - A a^{3} b\right )} e^{3} + {\left (B b^{4} d^{2} e - {\left (3 \, B a b^{3} - A b^{4}\right )} d e^{2} + {\left (2 \, B a^{2} b^{2} - A a b^{3}\right )} e^{3}\right )} x^{2} + 2 \, {\left (B a b^{3} d^{2} e - {\left (3 \, B a^{2} b^{2} - A a b^{3}\right )} d e^{2} + {\left (2 \, B a^{3} b - A a^{2} b^{2}\right )} e^{3}\right )} x\right )} \log \left (b x + a\right )}{2 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(B*b^4*e^3*x^4 - (B*a*b^3 + A*b^4)*d^3 + 3*(3*B*a^2*b^2 - A*a*b^3)*d^2*e - 3*(5*B*a^3*b - 3*A*a^2*b^2)*d*e
^2 + (7*B*a^4 - 5*A*a^3*b)*e^3 + 2*(3*B*b^4*d*e^2 - (2*B*a*b^3 - A*b^4)*e^3)*x^3 + (12*B*a*b^3*d*e^2 - (11*B*a
^2*b^2 - 4*A*a*b^3)*e^3)*x^2 - 2*(B*b^4*d^3 - 3*(2*B*a*b^3 - A*b^4)*d^2*e + 6*(B*a^2*b^2 - A*a*b^3)*d*e^2 - (B
*a^3*b - 2*A*a^2*b^2)*e^3)*x + 6*(B*a^2*b^2*d^2*e - (3*B*a^3*b - A*a^2*b^2)*d*e^2 + (2*B*a^4 - A*a^3*b)*e^3 +
(B*b^4*d^2*e - (3*B*a*b^3 - A*b^4)*d*e^2 + (2*B*a^2*b^2 - A*a*b^3)*e^3)*x^2 + 2*(B*a*b^3*d^2*e - (3*B*a^2*b^2
- A*a*b^3)*d*e^2 + (2*B*a^3*b - A*a^2*b^2)*e^3)*x)*log(b*x + a))/(b^7*x^2 + 2*a*b^6*x + a^2*b^5)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{0} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [B]  time = 0.14, size = 556, normalized size = 2.23 \begin {gather*} -\frac {\left (-B \,b^{4} e^{3} x^{4}+6 A a \,b^{3} e^{3} x^{2} \ln \left (b x +a \right )-6 A \,b^{4} d \,e^{2} x^{2} \ln \left (b x +a \right )-2 A \,b^{4} e^{3} x^{3}-12 B \,a^{2} b^{2} e^{3} x^{2} \ln \left (b x +a \right )+18 B a \,b^{3} d \,e^{2} x^{2} \ln \left (b x +a \right )+4 B a \,b^{3} e^{3} x^{3}-6 B \,b^{4} d^{2} e \,x^{2} \ln \left (b x +a \right )-6 B \,b^{4} d \,e^{2} x^{3}+12 A \,a^{2} b^{2} e^{3} x \ln \left (b x +a \right )-12 A a \,b^{3} d \,e^{2} x \ln \left (b x +a \right )-4 A a \,b^{3} e^{3} x^{2}-24 B \,a^{3} b \,e^{3} x \ln \left (b x +a \right )+36 B \,a^{2} b^{2} d \,e^{2} x \ln \left (b x +a \right )+11 B \,a^{2} b^{2} e^{3} x^{2}-12 B a \,b^{3} d^{2} e x \ln \left (b x +a \right )-12 B a \,b^{3} d \,e^{2} x^{2}+6 A \,a^{3} b \,e^{3} \ln \left (b x +a \right )-6 A \,a^{2} b^{2} d \,e^{2} \ln \left (b x +a \right )+4 A \,a^{2} b^{2} e^{3} x -12 A a \,b^{3} d \,e^{2} x +6 A \,b^{4} d^{2} e x -12 B \,a^{4} e^{3} \ln \left (b x +a \right )+18 B \,a^{3} b d \,e^{2} \ln \left (b x +a \right )-2 B \,a^{3} b \,e^{3} x -6 B \,a^{2} b^{2} d^{2} e \ln \left (b x +a \right )+12 B \,a^{2} b^{2} d \,e^{2} x -12 B a \,b^{3} d^{2} e x +2 B \,b^{4} d^{3} x +5 A \,a^{3} b \,e^{3}-9 A \,a^{2} b^{2} d \,e^{2}+3 A a \,b^{3} d^{2} e +A \,b^{4} d^{3}-7 B \,a^{4} e^{3}+15 B \,a^{3} b d \,e^{2}-9 B \,a^{2} b^{2} d^{2} e +B a \,b^{3} d^{3}\right ) \left (b x +a \right )}{2 \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(12*A*ln(b*x+a)*x*a^2*b^2*e^3-24*B*ln(b*x+a)*x*a^3*b*e^3+6*A*ln(b*x+a)*x^2*a*b^3*e^3-6*A*ln(b*x+a)*x^2*b^
4*d*e^2-12*B*ln(b*x+a)*x^2*a^2*b^2*e^3-6*B*ln(b*x+a)*x^2*b^4*d^2*e+18*B*ln(b*x+a)*x^2*a*b^3*d*e^2-12*A*ln(b*x+
a)*x*a*b^3*d*e^2+36*B*ln(b*x+a)*x*a^2*b^2*d*e^2-12*B*ln(b*x+a)*x*a*b^3*d^2*e-12*B*a*b^3*d^2*e*x+5*A*a^3*b*e^3+
12*B*a^2*b^2*d*e^2*x+18*B*a^3*b*d*e^2*ln(b*x+a)-6*B*a^2*b^2*d^2*e*ln(b*x+a)-12*A*a*b^3*d*e^2*x-6*A*a^2*b^2*d*e
^2*ln(b*x+a)-12*B*a*b^3*d*e^2*x^2+A*b^4*d^3-7*B*a^4*e^3+15*B*a^3*b*d*e^2+B*a*b^3*d^3-2*A*b^4*e^3*x^3-12*B*a^4*
e^3*ln(b*x+a)+2*B*b^4*d^3*x-B*b^4*e^3*x^4+6*A*a^3*b*e^3*ln(b*x+a)+4*A*a^2*b^2*e^3*x-9*b^2*B*a^2*d^2*e+3*A*a*b^
3*d^2*e-2*B*a^3*b*e^3*x+11*B*a^2*b^2*e^3*x^2+4*B*a*b^3*e^3*x^3-6*B*b^4*d*e^2*x^3-4*A*a*b^3*e^3*x^2-9*A*b^2*a^2
*d*e^2+6*A*b^4*d^2*e*x)*(b*x+a)/b^5/((b*x+a)^2)^(3/2)

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maxima [B]  time = 0.51, size = 487, normalized size = 1.96 \begin {gather*} \frac {B e^{3} x^{3}}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} - \frac {5 \, B a e^{3} x^{2}}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{3}} + \frac {6 \, B a^{2} e^{3} \log \left (x + \frac {a}{b}\right )}{b^{5}} - \frac {5 \, B a^{3} e^{3}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{5}} + \frac {{\left (3 \, B d e^{2} + A e^{3}\right )} x^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac {12 \, B a^{3} e^{3} x}{b^{6} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {A d^{3}}{2 \, b^{3} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {23 \, B a^{4} e^{3}}{2 \, b^{7} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {3 \, {\left (3 \, B d e^{2} + A e^{3}\right )} a \log \left (x + \frac {a}{b}\right )}{b^{4}} + \frac {3 \, {\left (B d^{2} e + A d e^{2}\right )} \log \left (x + \frac {a}{b}\right )}{b^{3}} + \frac {2 \, {\left (3 \, B d e^{2} + A e^{3}\right )} a^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4}} - \frac {B d^{3} + 3 \, A d^{2} e}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} - \frac {6 \, {\left (3 \, B d e^{2} + A e^{3}\right )} a^{2} x}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {6 \, {\left (B d^{2} e + A d e^{2}\right )} a x}{b^{4} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {11 \, {\left (3 \, B d e^{2} + A e^{3}\right )} a^{3}}{2 \, b^{6} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {9 \, {\left (B d^{2} e + A d e^{2}\right )} a^{2}}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {{\left (B d^{3} + 3 \, A d^{2} e\right )} a}{2 \, b^{4} {\left (x + \frac {a}{b}\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*B*e^3*x^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) - 5/2*B*a*e^3*x^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^3) + 6*B*
a^2*e^3*log(x + a/b)/b^5 - 5*B*a^3*e^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^5) + (3*B*d*e^2 + A*e^3)*x^2/(sqrt(b^2
*x^2 + 2*a*b*x + a^2)*b^2) + 12*B*a^3*e^3*x/(b^6*(x + a/b)^2) - 1/2*A*d^3/(b^3*(x + a/b)^2) + 23/2*B*a^4*e^3/(
b^7*(x + a/b)^2) - 3*(3*B*d*e^2 + A*e^3)*a*log(x + a/b)/b^4 + 3*(B*d^2*e + A*d*e^2)*log(x + a/b)/b^3 + 2*(3*B*
d*e^2 + A*e^3)*a^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^4) - (B*d^3 + 3*A*d^2*e)/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^
2) - 6*(3*B*d*e^2 + A*e^3)*a^2*x/(b^5*(x + a/b)^2) + 6*(B*d^2*e + A*d*e^2)*a*x/(b^4*(x + a/b)^2) - 11/2*(3*B*d
*e^2 + A*e^3)*a^3/(b^6*(x + a/b)^2) + 9/2*(B*d^2*e + A*d*e^2)*a^2/(b^5*(x + a/b)^2) + 1/2*(B*d^3 + 3*A*d^2*e)*
a/(b^4*(x + a/b)^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (A+B\,x\right )\,{\left (d+e\,x\right )}^3}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(((A + B*x)*(d + e*x)^3)/(a^2 + b^2*x^2 + 2*a*b*x)^(3/2), 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 (d + e x\right )^{3}}{\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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