\(\int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx\) [1839]

Optimal result

Integrand size = 35, antiderivative size = 162 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 (b d-a e) (B d-A e) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x)}-\frac {2 (2 b B d-A b e-a B e) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x)}+\frac {2 b B (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^3 (a+b x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {784, 78} \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (-a B e-A b e+2 b B d)}{3 e^3 (a+b x)}+\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e) (B d-A e)}{e^3 (a+b x)}+\frac {2 b B \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2}}{5 e^3 (a+b x)} \]

[In]

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

[Out]

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

Rule 78

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.53 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 \sqrt {(a+b x)^2} \sqrt {d+e x} \left (5 A b e (-2 d+e x)+5 a e (-2 B d+3 A e+B e x)+b B \left (8 d^2-4 d e x+3 e^2 x^2\right )\right )}{15 e^3 (a+b x)} \]

[In]

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

[Out]

(2*Sqrt[(a + b*x)^2]*Sqrt[d + e*x]*(5*A*b*e*(-2*d + e*x) + 5*a*e*(-2*B*d + 3*A*e + B*e*x) + b*B*(8*d^2 - 4*d*e
*x + 3*e^2*x^2)))/(15*e^3*(a + b*x))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 2.

Time = 0.22 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.49

method	result	size
default	\(\frac {2 \,\operatorname {csgn}\left (b x +a \right ) \sqrt {e x +d}\, \left (3 B b \,e^{2} x^{2}+5 A b \,e^{2} x +5 B a \,e^{2} x -4 B b d e x +15 A a \,e^{2}-10 A b d e -10 B a d e +8 B b \,d^{2}\right )}{15 e^{3}}\)	\(79\)
gosper	\(\frac {2 \sqrt {e x +d}\, \left (3 B b \,e^{2} x^{2}+5 A b \,e^{2} x +5 B a \,e^{2} x -4 B b d e x +15 A a \,e^{2}-10 A b d e -10 B a d e +8 B b \,d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{15 e^{3} \left (b x +a \right )}\)	\(89\)
risch	\(\frac {2 \sqrt {e x +d}\, \left (3 B b \,e^{2} x^{2}+5 A b \,e^{2} x +5 B a \,e^{2} x -4 B b d e x +15 A a \,e^{2}-10 A b d e -10 B a d e +8 B b \,d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{15 e^{3} \left (b x +a \right )}\)	\(89\)

[In]

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

[Out]

2/15*csgn(b*x+a)*(e*x+d)^(1/2)*(3*B*b*e^2*x^2+5*A*b*e^2*x+5*B*a*e^2*x-4*B*b*d*e*x+15*A*a*e^2-10*A*b*d*e-10*B*a
*d*e+8*B*b*d^2)/e^3

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.43 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (3 \, B b e^{2} x^{2} + 8 \, B b d^{2} + 15 \, A a e^{2} - 10 \, {\left (B a + A b\right )} d e - {\left (4 \, B b d e - 5 \, {\left (B a + A b\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{15 \, e^{3}} \]

[In]

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

[Out]

2/15*(3*B*b*e^2*x^2 + 8*B*b*d^2 + 15*A*a*e^2 - 10*(B*a + A*b)*d*e - (4*B*b*d*e - 5*(B*a + A*b)*e^2)*x)*sqrt(e*
x + d)/e^3

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.73 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (b e^{2} x^{2} - 2 \, b d^{2} + 3 \, a d e - {\left (b d e - 3 \, a e^{2}\right )} x\right )} A}{3 \, \sqrt {e x + d} e^{2}} + \frac {2 \, {\left (3 \, b e^{3} x^{3} + 8 \, b d^{3} - 10 \, a d^{2} e - {\left (b d e^{2} - 5 \, a e^{3}\right )} x^{2} + {\left (4 \, b d^{2} e - 5 \, a d e^{2}\right )} x\right )} B}{15 \, \sqrt {e x + d} e^{3}} \]

[In]

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

[Out]

2/3*(b*e^2*x^2 - 2*b*d^2 + 3*a*d*e - (b*d*e - 3*a*e^2)*x)*A/(sqrt(e*x + d)*e^2) + 2/15*(3*b*e^3*x^3 + 8*b*d^3
- 10*a*d^2*e - (b*d*e^2 - 5*a*e^3)*x^2 + (4*b*d^2*e - 5*a*d*e^2)*x)*B/(sqrt(e*x + d)*e^3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.80 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (15 \, \sqrt {e x + d} A a \mathrm {sgn}\left (b x + a\right ) + \frac {5 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} B a \mathrm {sgn}\left (b x + a\right )}{e} + \frac {5 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} A b \mathrm {sgn}\left (b x + a\right )}{e} + \frac {{\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} B b \mathrm {sgn}\left (b x + a\right )}{e^{2}}\right )}}{15 \, e} \]

[In]

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

[Out]

2/15*(15*sqrt(e*x + d)*A*a*sgn(b*x + a) + 5*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*B*a*sgn(b*x + a)/e + 5*((e*x
+ d)^(3/2) - 3*sqrt(e*x + d)*d)*A*b*sgn(b*x + a)/e + (3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x
+ d)*d^2)*B*b*sgn(b*x + a)/e^2)/e

Mupad [B] (verification not implemented)

Time = 11.07 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.96 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (\frac {2\,B\,x^3}{5}+\frac {16\,B\,b\,d^3+30\,A\,a\,d\,e^2-20\,A\,b\,d^2\,e-20\,B\,a\,d^2\,e}{15\,b\,e^3}+\frac {x\,\left (30\,A\,a\,e^3-10\,A\,b\,d\,e^2-10\,B\,a\,d\,e^2+8\,B\,b\,d^2\,e\right )}{15\,b\,e^3}+\frac {x^2\,\left (10\,A\,b\,e^3+10\,B\,a\,e^3-2\,B\,b\,d\,e^2\right )}{15\,b\,e^3}\right )}{x\,\sqrt {d+e\,x}+\frac {a\,\sqrt {d+e\,x}}{b}} \]

[In]

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

[Out]

(((a + b*x)^2)^(1/2)*((2*B*x^3)/5 + (16*B*b*d^3 + 30*A*a*d*e^2 - 20*A*b*d^2*e - 20*B*a*d^2*e)/(15*b*e^3) + (x*
(30*A*a*e^3 - 10*A*b*d*e^2 - 10*B*a*d*e^2 + 8*B*b*d^2*e))/(15*b*e^3) + (x^2*(10*A*b*e^3 + 10*B*a*e^3 - 2*B*b*d
*e^2))/(15*b*e^3)))/(x*(d + e*x)^(1/2) + (a*(d + e*x)^(1/2))/b)