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

Optimal. Leaf size=132 \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) (B d-A e) \log (d+e x)}{e^3 (a+b x)}-\frac {b x \sqrt {a^2+2 a b x+b^2 x^2} (B d-A e)}{e^2 (a+b x)}+\frac {B (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{2 b e} \]

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Rubi [A]  time = 0.09, antiderivative size = 132, 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 {b x \sqrt {a^2+2 a b x+b^2 x^2} (B d-A e)}{e^2 (a+b x)}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) (B d-A e) \log (d+e x)}{e^3 (a+b x)}+\frac {B (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{2 b e} \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),x]

[Out]

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

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

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Mathematica [A]  time = 0.04, size = 74, normalized size = 0.56 \begin {gather*} \frac {\sqrt {(a+b x)^2} (e x (2 a B e+b (2 A e-2 B d+B e x))+2 (b d-a e) (B d-A e) \log (d+e x))}{2 e^3 (a+b x)} \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),x]

[Out]

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

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IntegrateAlgebraic [F]  time = 1.54, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{d+e x} \, dx \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),x]

[Out]

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

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fricas [A]  time = 0.42, size = 68, normalized size = 0.52 \begin {gather*} \frac {B b e^{2} x^{2} - 2 \, {\left (B b d e - {\left (B a + A b\right )} e^{2}\right )} x + 2 \, {\left (B b d^{2} + A a e^{2} - {\left (B a + A b\right )} d e\right )} \log \left (e x + d\right )}{2 \, e^{3}} \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),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 119, normalized size = 0.90 \begin {gather*} {\left (B b d^{2} \mathrm {sgn}\left (b x + a\right ) - B a d e \mathrm {sgn}\left (b x + a\right ) - A b d e \mathrm {sgn}\left (b x + a\right ) + A a e^{2} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-3\right )} \log \left ({\left | x e + d \right |}\right ) + \frac {1}{2} \, {\left (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 )\right )} e^{\left (-2\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),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.07, size = 146, normalized size = 1.11 \begin {gather*} \frac {\left (B \,b^{2} e^{2} x^{2}+2 A a b \,e^{2} \ln \left (b e x +b d \right )-2 A \,b^{2} d e \ln \left (b e x +b d \right )+2 A \,b^{2} e^{2} x -2 B a b d e \ln \left (b e x +b d \right )+2 B a b \,e^{2} x +2 B \,b^{2} d^{2} \ln \left (b e x +b d \right )-2 B \,b^{2} d e x +2 A a b \,e^{2}+B \,a^{2} e^{2}-2 B a b d e \right ) \mathrm {csgn}\left (b x +a \right )}{2 b \,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),x)

[Out]

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

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

[Out]

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

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sympy [A]  time = 0.28, size = 53, normalized size = 0.40 \begin {gather*} \frac {B b x^{2}}{2 e} + x \left (\frac {A b}{e} + \frac {B a}{e} - \frac {B b d}{e^{2}}\right ) - \frac {\left (- A e + B d\right ) \left (a e - b d\right ) \log {\left (d + e x \right )}}{e^{3}} \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),x)

[Out]

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

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