\(\int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{5/2}} \, dx\) [1841]

Optimal result

Integrand size = 35, antiderivative size = 160 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{5/2}} \, dx=-\frac {2 (b d-a e) (B d-A e) \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x) (d+e x)^{3/2}}+\frac {2 (2 b B d-A b e-a B e) \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x) \sqrt {d+e x}}+\frac {2 b B \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x)} \]

[Out]

-2/3*(-a*e+b*d)*(-A*e+B*d)*((b*x+a)^2)^(1/2)/e^3/(b*x+a)/(e*x+d)^(3/2)+2*(-A*b*e-B*a*e+2*B*b*d)*((b*x+a)^2)^(1
/2)/e^3/(b*x+a)/(e*x+d)^(1/2)+2*b*B*(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 = 160, 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} \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{5/2}} \, dx=\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (-a B e-A b e+2 b B d)}{e^3 (a+b x) \sqrt {d+e x}}-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) (B d-A e)}{3 e^3 (a+b x) (d+e x)^{3/2}}+\frac {2 b B \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x}}{e^3 (a+b x)} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.54 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{5/2}} \, dx=-\frac {2 \sqrt {(a+b x)^2} \left (A b e (2 d+3 e x)+a e (2 B d+A e+3 B e x)-b B \left (8 d^2+12 d e x+3 e^2 x^2\right )\right )}{3 e^3 (a+b x) (d+e x)^{3/2}} \]

[In]

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

[Out]

(-2*Sqrt[(a + b*x)^2]*(A*b*e*(2*d + 3*e*x) + a*e*(2*B*d + A*e + 3*B*e*x) - b*B*(8*d^2 + 12*d*e*x + 3*e^2*x^2))
)/(3*e^3*(a + b*x)*(d + e*x)^(3/2))

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 = 78, normalized size of antiderivative = 0.49

method	result	size
default	\(-\frac {2 \,\operatorname {csgn}\left (b x +a \right ) \left (-3 B b \,e^{2} x^{2}+3 A b \,e^{2} x +3 B a \,e^{2} x -12 B b d e x +A a \,e^{2}+2 A b d e +2 B a d e -8 B b \,d^{2}\right )}{3 e^{3} \left (e x +d \right )^{\frac {3}{2}}}\)	\(78\)
gosper	\(-\frac {2 \left (-3 B b \,e^{2} x^{2}+3 A b \,e^{2} x +3 B a \,e^{2} x -12 B b d e x +A a \,e^{2}+2 A b d e +2 B a d e -8 B b \,d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{3 \left (e x +d \right )^{\frac {3}{2}} e^{3} \left (b x +a \right )}\)	\(88\)
risch	\(\frac {2 b B \sqrt {e x +d}\, \sqrt {\left (b x +a \right )^{2}}}{e^{3} \left (b x +a \right )}-\frac {2 \left (3 A b \,e^{2} x +3 B a \,e^{2} x -6 B b d e x +A a \,e^{2}+2 A b d e +2 B a d e -5 B b \,d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{3 e^{3} \left (e x +d \right )^{\frac {3}{2}} \left (b x +a \right )}\)	\(109\)

[In]

int((B*x+A)*((b*x+a)^2)^(1/2)/(e*x+d)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/3*csgn(b*x+a)*(-3*B*b*e^2*x^2+3*A*b*e^2*x+3*B*a*e^2*x-12*B*b*d*e*x+A*a*e^2+2*A*b*d*e+2*B*a*d*e-8*B*b*d^2)/e
^3/(e*x+d)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.57 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (3 \, B b e^{2} x^{2} + 8 \, B b d^{2} - A a e^{2} - 2 \, {\left (B a + A b\right )} d e + 3 \, {\left (4 \, B b d e - {\left (B a + A b\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{3 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} \]

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/(e*x+d)^(5/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{5/2}} \, dx=\int \frac {\left (A + B x\right ) \sqrt {\left (a + b x\right )^{2}}}{\left (d + e x\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate((B*x+A)*((b*x+a)**2)**(1/2)/(e*x+d)**(5/2),x)

[Out]

Integral((A + B*x)*sqrt((a + b*x)**2)/(d + e*x)**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.60 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{5/2}} \, dx=-\frac {2 \, {\left (3 \, b e x + 2 \, b d + a e\right )} A}{3 \, {\left (e^{3} x + d e^{2}\right )} \sqrt {e x + d}} + \frac {2 \, {\left (3 \, b e^{2} x^{2} + 8 \, b d^{2} - 2 \, a d e + 3 \, {\left (4 \, b d e - a e^{2}\right )} x\right )} B}{3 \, {\left (e^{4} x + d e^{3}\right )} \sqrt {e x + d}} \]

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/(e*x+d)^(5/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.81 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{5/2}} \, dx=\frac {2 \, \sqrt {e x + d} B b \mathrm {sgn}\left (b x + a\right )}{e^{3}} + \frac {2 \, {\left (6 \, {\left (e x + d\right )} B b d \mathrm {sgn}\left (b x + a\right ) - B b d^{2} \mathrm {sgn}\left (b x + a\right ) - 3 \, {\left (e x + d\right )} B a e \mathrm {sgn}\left (b x + a\right ) - 3 \, {\left (e x + d\right )} A b e \mathrm {sgn}\left (b x + a\right ) + B a d e \mathrm {sgn}\left (b x + a\right ) + A b d e \mathrm {sgn}\left (b x + a\right ) - A a e^{2} \mathrm {sgn}\left (b x + a\right )\right )}}{3 \, {\left (e x + d\right )}^{\frac {3}{2}} e^{3}} \]

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/(e*x+d)^(5/2),x, algorithm="giac")

[Out]

2*sqrt(e*x + d)*B*b*sgn(b*x + a)/e^3 + 2/3*(6*(e*x + d)*B*b*d*sgn(b*x + a) - B*b*d^2*sgn(b*x + a) - 3*(e*x + d
)*B*a*e*sgn(b*x + a) - 3*(e*x + d)*A*b*e*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*x + d)^(3/2)*e^3)

Mupad [B] (verification not implemented)

Time = 11.22 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.91 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{5/2}} \, dx=-\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (\frac {2\,A\,a\,e^2-16\,B\,b\,d^2+4\,A\,b\,d\,e+4\,B\,a\,d\,e}{3\,b\,e^4}-\frac {2\,B\,x^2}{e^2}+\frac {x\,\left (6\,A\,b\,e^2+6\,B\,a\,e^2-24\,B\,b\,d\,e\right )}{3\,b\,e^4}\right )}{x^2\,\sqrt {d+e\,x}+\frac {a\,d\,\sqrt {d+e\,x}}{b\,e}+\frac {x\,\left (3\,a\,e^4+3\,b\,d\,e^3\right )\,\sqrt {d+e\,x}}{3\,b\,e^4}} \]

[In]

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

[Out]

-(((a + b*x)^2)^(1/2)*((2*A*a*e^2 - 16*B*b*d^2 + 4*A*b*d*e + 4*B*a*d*e)/(3*b*e^4) - (2*B*x^2)/e^2 + (x*(6*A*b*
e^2 + 6*B*a*e^2 - 24*B*b*d*e))/(3*b*e^4)))/(x^2*(d + e*x)^(1/2) + (a*d*(d + e*x)^(1/2))/(b*e) + (x*(3*a*e^4 +
3*b*d*e^3)*(d + e*x)^(1/2))/(3*b*e^4))