\(\int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^2} \, dx\) [1961]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [F(-2)]
   Giac [A] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 33, antiderivative size = 132 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^2} \, dx=\frac {b^2 x \sqrt {a^2+2 a b x+b^2 x^2}}{e^2 (a+b x)}-\frac {(b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x) (d+e x)}-\frac {2 b (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^3 (a+b x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {784, 21, 45} \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^2} \, dx=-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^3 (a+b x) (d+e x)}-\frac {2 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) \log (d+e x)}{e^3 (a+b x)}+\frac {b^2 x \sqrt {a^2+2 a b x+b^2 x^2}}{e^2 (a+b x)} \]

[In]

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

[Out]

(b^2*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^2*(a + b*x)) - ((b*d - a*e)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^3*(a
+ b*x)*(d + e*x)) - (2*b*(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 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
 n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
 d*x, a + b*x])

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 784

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*} \text {integral}& = \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x) \left (a b+b^2 x\right )}{(d+e x)^2} \, dx}{a b+b^2 x} \\ & = \frac {\left (b \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \frac {(a+b x)^2}{(d+e x)^2} \, dx}{a b+b^2 x} \\ & = \frac {\left (b \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \left (\frac {b^2}{e^2}+\frac {(-b d+a e)^2}{e^2 (d+e x)^2}-\frac {2 b (b d-a e)}{e^2 (d+e x)}\right ) \, dx}{a b+b^2 x} \\ & = \frac {b^2 x \sqrt {a^2+2 a b x+b^2 x^2}}{e^2 (a+b x)}-\frac {(b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x) (d+e x)}-\frac {2 b (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*}

Mathematica [A] (verified)

Time = 1.04 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.66 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^2} \, dx=\frac {\sqrt {(a+b x)^2} \left (2 a b d e-a^2 e^2+b^2 \left (-d^2+d e x+e^2 x^2\right )-2 b (b d-a e) (d+e x) \log (d+e x)\right )}{e^3 (a+b x) (d+e x)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.84

method result size
risch \(\frac {b^{2} x \sqrt {\left (b x +a \right )^{2}}}{e^{2} \left (b x +a \right )}-\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right )}{\left (b x +a \right ) e^{3} \left (e x +d \right )}+\frac {2 \sqrt {\left (b x +a \right )^{2}}\, \left (a e -b d \right ) b \ln \left (e x +d \right )}{\left (b x +a \right ) e^{3}}\) \(111\)
default \(\frac {\operatorname {csgn}\left (b x +a \right ) \left (2 \ln \left (-b e x -b d \right ) a b \,e^{2} x -2 \ln \left (-b e x -b d \right ) b^{2} d e x +b^{2} e^{2} x^{2}+2 \ln \left (-b e x -b d \right ) a b d e -2 \ln \left (-b e x -b d \right ) b^{2} d^{2}+a b \,e^{2} x +b^{2} d e x -e^{2} a^{2}+3 a b d e -b^{2} d^{2}\right )}{e^{3} \left (e x +d \right )}\) \(139\)

[In]

int((b*x+a)*((b*x+a)^2)^(1/2)/(e*x+d)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.70 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^2} \, dx=\frac {b^{2} e^{2} x^{2} + b^{2} d e x - b^{2} d^{2} + 2 \, a b d e - a^{2} e^{2} - 2 \, {\left (b^{2} d^{2} - a b d e + {\left (b^{2} d e - a b e^{2}\right )} x\right )} \log \left (e x + d\right )}{e^{4} x + d e^{3}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^2} \, dx=\int \frac {\left (a + b x\right ) \sqrt {\left (a + b x\right )^{2}}}{\left (d + e x\right )^{2}}\, dx \]

[In]

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

[Out]

Integral((a + b*x)*sqrt((a + b*x)**2)/(d + e*x)**2, x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((b*x+a)*((b*x+a)^2)^(1/2)/(e*x+d)^2,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 detail

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.77 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^2} \, dx=\frac {b^{2} x \mathrm {sgn}\left (b x + a\right )}{e^{2}} - \frac {2 \, {\left (b^{2} d \mathrm {sgn}\left (b x + a\right ) - a b e \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | e x + d \right |}\right )}{e^{3}} - \frac {b^{2} d^{2} \mathrm {sgn}\left (b x + a\right ) - 2 \, a b d e \mathrm {sgn}\left (b x + a\right ) + a^{2} e^{2} \mathrm {sgn}\left (b x + a\right )}{{\left (e x + d\right )} e^{3}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^2} \, dx=\int \frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (a+b\,x\right )}{{\left (d+e\,x\right )}^2} \,d x \]

[In]

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

[Out]

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