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

Optimal. Leaf size=191 \[ \frac {(a+b x) (d+e x)^2 (A b-a B)}{2 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (A b-a B) (b d-a e)^2 \log (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e x (a+b x) (A b-a B) (b d-a e)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B (a+b x) (d+e x)^3}{3 b e \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.11, antiderivative size = 191, 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 {(a+b x) (d+e x)^2 (A b-a B)}{2 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e x (a+b x) (A b-a B) (b d-a e)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (A b-a B) (b d-a e)^2 \log (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B (a+b x) (d+e x)^3}{3 b e \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)^2)/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

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

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Mathematica [A]  time = 0.08, size = 118, normalized size = 0.62 \begin {gather*} \frac {(a+b x) \left (b x \left (6 a^2 B e^2-3 a b e (2 A e+4 B d+B e x)+b^2 \left (3 A e (4 d+e x)+2 B \left (3 d^2+3 d e x+e^2 x^2\right )\right )\right )+6 (A b-a B) (b d-a e)^2 \log (a+b x)\right )}{6 b^4 \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [B]  time = 1.11, size = 610, normalized size = 3.19 \begin {gather*} \frac {-6 a^2 B e^2 x+6 a A b e^2 x+12 a b B d e x+3 a b B e^2 x^2-12 A b^2 d e x-3 A b^2 e^2 x^2-6 b^2 B d^2 x-6 b^2 B d e x^2-2 b^2 B e^2 x^3}{12 \left (b^2\right )^{3/2}}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (11 a^2 B e^2-9 a A b e^2-18 a b B d e-5 a b B e^2 x+12 A b^2 d e+3 A b^2 e^2 x+6 b^2 B d^2+6 b^2 B d e x+2 b^2 B e^2 x^2\right )}{12 b^4}+\frac {\log \left (\sqrt {a^2+2 a b x+b^2 x^2}-a-\sqrt {b^2} x\right ) \left (a^3 \sqrt {b^2} B e^2+a^3 b B e^2-a^2 A b^2 e^2-a^2 A \sqrt {b^2} b e^2-2 a^2 b^2 B d e-2 a^2 \sqrt {b^2} b B d e+2 a A b^3 d e+2 a A \left (b^2\right )^{3/2} d e+a b^3 B d^2+a \left (b^2\right )^{3/2} B d^2-A b^4 d^2-A \sqrt {b^2} b^3 d^2\right )}{2 b^4 \sqrt {b^2}}+\frac {\log \left (\sqrt {a^2+2 a b x+b^2 x^2}+a-\sqrt {b^2} x\right ) \left (-a^3 \sqrt {b^2} B e^2+a^3 b B e^2-a^2 A b^2 e^2+a^2 A \sqrt {b^2} b e^2-2 a^2 b^2 B d e+2 a^2 \sqrt {b^2} b B d e+2 a A b^3 d e-2 a A \left (b^2\right )^{3/2} d e+a b^3 B d^2-a \left (b^2\right )^{3/2} B d^2-A b^4 d^2+A \sqrt {b^2} b^3 d^2\right )}{2 b^4 \sqrt {b^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(6*b^2*B*d^2 + 12*A*b^2*d*e - 18*a*b*B*d*e - 9*a*A*b*e^2 + 11*a^2*B*e^2 + 6*b^2
*B*d*e*x + 3*A*b^2*e^2*x - 5*a*b*B*e^2*x + 2*b^2*B*e^2*x^2))/(12*b^4) + (-6*b^2*B*d^2*x - 12*A*b^2*d*e*x + 12*
a*b*B*d*e*x + 6*a*A*b*e^2*x - 6*a^2*B*e^2*x - 6*b^2*B*d*e*x^2 - 3*A*b^2*e^2*x^2 + 3*a*b*B*e^2*x^2 - 2*b^2*B*e^
2*x^3)/(12*(b^2)^(3/2)) + ((-(A*b^4*d^2) - A*b^3*Sqrt[b^2]*d^2 + a*b^3*B*d^2 + a*(b^2)^(3/2)*B*d^2 + 2*a*A*b^3
*d*e + 2*a*A*(b^2)^(3/2)*d*e - 2*a^2*b^2*B*d*e - 2*a^2*b*Sqrt[b^2]*B*d*e - a^2*A*b^2*e^2 - a^2*A*b*Sqrt[b^2]*e
^2 + a^3*b*B*e^2 + a^3*Sqrt[b^2]*B*e^2)*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(2*b^4*Sqrt[b^2
]) + ((-(A*b^4*d^2) + A*b^3*Sqrt[b^2]*d^2 + a*b^3*B*d^2 - a*(b^2)^(3/2)*B*d^2 + 2*a*A*b^3*d*e - 2*a*A*(b^2)^(3
/2)*d*e - 2*a^2*b^2*B*d*e + 2*a^2*b*Sqrt[b^2]*B*d*e - a^2*A*b^2*e^2 + a^2*A*b*Sqrt[b^2]*e^2 + a^3*b*B*e^2 - a^
3*Sqrt[b^2]*B*e^2)*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(2*b^4*Sqrt[b^2])

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fricas [A]  time = 0.43, size = 158, normalized size = 0.83 \begin {gather*} \frac {2 \, B b^{3} e^{2} x^{3} + 3 \, {\left (2 \, B b^{3} d e - {\left (B a b^{2} - A b^{3}\right )} e^{2}\right )} x^{2} + 6 \, {\left (B b^{3} d^{2} - 2 \, {\left (B a b^{2} - A b^{3}\right )} d e + {\left (B a^{2} b - A a b^{2}\right )} e^{2}\right )} x - 6 \, {\left ({\left (B a b^{2} - A b^{3}\right )} d^{2} - 2 \, {\left (B a^{2} b - A a b^{2}\right )} d e + {\left (B a^{3} - A a^{2} b\right )} e^{2}\right )} \log \left (b x + a\right )}{6 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.22, size = 254, normalized size = 1.33 \begin {gather*} \frac {2 \, B b^{2} x^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) + 6 \, B b^{2} d x^{2} e \mathrm {sgn}\left (b x + a\right ) + 6 \, B b^{2} d^{2} x \mathrm {sgn}\left (b x + a\right ) - 3 \, B a b x^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, A b^{2} x^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 12 \, B a b d x e \mathrm {sgn}\left (b x + a\right ) + 12 \, A b^{2} d x e \mathrm {sgn}\left (b x + a\right ) + 6 \, B a^{2} x e^{2} \mathrm {sgn}\left (b x + a\right ) - 6 \, A a b x e^{2} \mathrm {sgn}\left (b x + a\right )}{6 \, b^{3}} - \frac {{\left (B a b^{2} d^{2} \mathrm {sgn}\left (b x + a\right ) - A b^{3} d^{2} \mathrm {sgn}\left (b x + a\right ) - 2 \, B a^{2} b d e \mathrm {sgn}\left (b x + a\right ) + 2 \, A a b^{2} d e \mathrm {sgn}\left (b x + a\right ) + B a^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) - A a^{2} b e^{2} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | b x + a \right |}\right )}{b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 212, normalized size = 1.11 \begin {gather*} \frac {\left (b x +a \right ) \left (2 B \,b^{3} e^{2} x^{3}+3 A \,b^{3} e^{2} x^{2}-3 B a \,b^{2} e^{2} x^{2}+6 B \,b^{3} d e \,x^{2}+6 A \,a^{2} b \,e^{2} \ln \left (b x +a \right )-12 A a \,b^{2} d e \ln \left (b x +a \right )-6 A a \,b^{2} e^{2} x +6 A \,b^{3} d^{2} \ln \left (b x +a \right )+12 A \,b^{3} d e x -6 B \,a^{3} e^{2} \ln \left (b x +a \right )+12 B \,a^{2} b d e \ln \left (b x +a \right )+6 B \,a^{2} b \,e^{2} x -6 B a \,b^{2} d^{2} \ln \left (b x +a \right )-12 B a \,b^{2} d e x +6 B \,b^{3} d^{2} x \right )}{6 \sqrt {\left (b x +a \right )^{2}}\, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.62, size = 244, normalized size = 1.28 \begin {gather*} -\frac {5 \, B a e^{2} x^{2}}{6 \, b^{2}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B e^{2} x^{2}}{3 \, b^{2}} + \frac {5 \, B a^{2} e^{2} x}{3 \, b^{3}} + \frac {A d^{2} \log \left (x + \frac {a}{b}\right )}{b} - \frac {B a^{3} e^{2} \log \left (x + \frac {a}{b}\right )}{b^{4}} + \frac {{\left (2 \, B d e + A e^{2}\right )} x^{2}}{2 \, b} - \frac {2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{2} e^{2}}{3 \, b^{4}} - \frac {{\left (2 \, B d e + A e^{2}\right )} a x}{b^{2}} + \frac {{\left (2 \, B d e + A e^{2}\right )} a^{2} \log \left (x + \frac {a}{b}\right )}{b^{3}} - \frac {{\left (B d^{2} + 2 \, A d e\right )} a \log \left (x + \frac {a}{b}\right )}{b^{2}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} {\left (B d^{2} + 2 \, A d e\right )}}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (A+B\,x\right )\,{\left (d+e\,x\right )}^2}{\sqrt {{\left (a+b\,x\right )}^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.49, size = 117, normalized size = 0.61 \begin {gather*} \frac {B e^{2} x^{3}}{3 b} + x^{2} \left (\frac {A e^{2}}{2 b} - \frac {B a e^{2}}{2 b^{2}} + \frac {B d e}{b}\right ) + x \left (- \frac {A a e^{2}}{b^{2}} + \frac {2 A d e}{b} + \frac {B a^{2} e^{2}}{b^{3}} - \frac {2 B a d e}{b^{2}} + \frac {B d^{2}}{b}\right ) - \frac {\left (- A b + B a\right ) \left (a e - b d\right )^{2} \log {\left (a + b x \right )}}{b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*e**2*x**3/(3*b) + x**2*(A*e**2/(2*b) - B*a*e**2/(2*b**2) + B*d*e/b) + x*(-A*a*e**2/b**2 + 2*A*d*e/b + B*a**2
*e**2/b**3 - 2*B*a*d*e/b**2 + B*d**2/b) - (-A*b + B*a)*(a*e - b*d)**2*log(a + b*x)/b**4

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