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

Optimal. Leaf size=134 \[ \frac {e (a+b x) (A+B x)^2}{2 b B \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (A b-a B) (b d-a e) \log (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B x (a+b x) (b d-a e)}{b^2 \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.08, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {770, 77} \begin {gather*} \frac {(a+b x) (A b-a B) (b d-a e) \log (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e (a+b x) (A+B x)^2}{2 b B \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B x (a+b x) (b d-a e)}{b^2 \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))/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

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

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Mathematica [A]  time = 0.04, size = 72, normalized size = 0.54 \begin {gather*} \frac {(a+b x) (b x (b (2 A e+2 B d+B e x)-2 a B e)+2 (A b-a B) (b d-a e) \log (a+b x))}{2 b^3 \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [B]  time = 0.72, size = 344, normalized size = 2.57 \begin {gather*} \frac {\sqrt {a^2+2 a b x+b^2 x^2} (-3 a B e+2 A b e+2 b B d+b B e x)}{4 b^3}+\frac {\log \left (\sqrt {a^2+2 a b x+b^2 x^2}-a-\sqrt {b^2} x\right ) \left (-a^2 \sqrt {b^2} B e-a^2 b B e+a A b^2 e+a A \sqrt {b^2} b e+a b^2 B d+a \sqrt {b^2} b B d-A b^3 d-A \left (b^2\right )^{3/2} d\right )}{2 b^3 \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^2 \sqrt {b^2} B e-a^2 b B e+a A b^2 e-a A \sqrt {b^2} b e+a b^2 B d-a \sqrt {b^2} b B d-A b^3 d+A \left (b^2\right )^{3/2} d\right )}{2 b^3 \sqrt {b^2}}+\frac {2 a B e x-2 A b e x-2 b B d x-b B e x^2}{4 b \sqrt {b^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 75, normalized size = 0.56 \begin {gather*} \frac {B b^{2} e x^{2} + 2 \, {\left (B b^{2} d - {\left (B a b - A b^{2}\right )} e\right )} x - 2 \, {\left ({\left (B a b - A b^{2}\right )} d - {\left (B a^{2} - A a b\right )} e\right )} \log \left (b x + a\right )}{2 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 122, normalized size = 0.91 \begin {gather*} \frac {B b x^{2} e \mathrm {sgn}\left (b x + a\right ) + 2 \, B b d x \mathrm {sgn}\left (b x + a\right ) - 2 \, B a x e \mathrm {sgn}\left (b x + a\right ) + 2 \, A b x e \mathrm {sgn}\left (b x + a\right )}{2 \, b^{2}} - \frac {{\left (B a b d \mathrm {sgn}\left (b x + a\right ) - A b^{2} d \mathrm {sgn}\left (b x + a\right ) - B a^{2} e \mathrm {sgn}\left (b x + a\right ) + A a b e \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | b x + a \right |}\right )}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 104, normalized size = 0.78 \begin {gather*} -\frac {\left (b x +a \right ) \left (-B \,b^{2} e \,x^{2}+2 A a b e \ln \left (b x +a \right )-2 A \,b^{2} d \ln \left (b x +a \right )-2 A \,b^{2} e x -2 B \,a^{2} e \ln \left (b x +a \right )+2 B a b d \ln \left (b x +a \right )+2 B a b e x -2 B \,b^{2} d x \right )}{2 \sqrt {\left (b x +a \right )^{2}}\, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.49, size = 101, normalized size = 0.75 \begin {gather*} \frac {B e x^{2}}{2 \, b} - \frac {B a e x}{b^{2}} + \frac {A d \log \left (x + \frac {a}{b}\right )}{b} + \frac {B a^{2} e \log \left (x + \frac {a}{b}\right )}{b^{3}} - \frac {{\left (B d + A 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 + A e\right )}}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*B*e*x^2/b - B*a*e*x/b^2 + A*d*log(x + a/b)/b + B*a^2*e*log(x + a/b)/b^3 - (B*d + A*e)*a*log(x + a/b)/b^2 +
sqrt(b^2*x^2 + 2*a*b*x + a^2)*(B*d + A*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 )}{\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))/((a + b*x)^2)^(1/2),x)

[Out]

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

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sympy [A]  time = 0.29, size = 53, normalized size = 0.40 \begin {gather*} \frac {B e x^{2}}{2 b} + x \left (\frac {A e}{b} - \frac {B a e}{b^{2}} + \frac {B d}{b}\right ) + \frac {\left (- A b + B a\right ) \left (a e - b d\right ) \log {\left (a + b x \right )}}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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