\(\int (A+B x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2} \, dx\) [1838]

Optimal result

Integrand size = 35, antiderivative size = 164 \[ \int (A+B x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 (b d-a e) (B d-A e) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x)}-\frac {2 (2 b B d-A b e-a B e) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^3 (a+b x)}+\frac {2 b B (d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}{7 e^3 (a+b x)} \]

[Out]

2/3*(-a*e+b*d)*(-A*e+B*d)*(e*x+d)^(3/2)*((b*x+a)^2)^(1/2)/e^3/(b*x+a)-2/5*(-A*b*e-B*a*e+2*B*b*d)*(e*x+d)^(5/2)
*((b*x+a)^2)^(1/2)/e^3/(b*x+a)+2/7*b*B*(e*x+d)^(7/2)*((b*x+a)^2)^(1/2)/e^3/(b*x+a)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 164, 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} \[ \int (A+B x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (-a B e-A b e+2 b B d)}{5 e^3 (a+b x)}+\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e) (B d-A e)}{3 e^3 (a+b x)}+\frac {2 b B \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{7/2}}{7 e^3 (a+b x)} \]

[In]

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

[Out]

(2*(b*d - a*e)*(B*d - A*e)*(d + e*x)^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^3*(a + b*x)) - (2*(2*b*B*d - A*
b*e - a*B*e)*(d + e*x)^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^3*(a + b*x)) + (2*b*B*(d + e*x)^(7/2)*Sqrt[a^
2 + 2*a*b*x + b^2*x^2])/(7*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 \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) \sqrt {d+e x}}{e^2}+\frac {b (-2 b B d+A b e+a B e) (d+e x)^{3/2}}{e^2}+\frac {b^2 B (d+e x)^{5/2}}{e^2}\right ) \, dx}{a b+b^2 x} \\ & = \frac {2 (b d-a e) (B d-A 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 (2 b B d-A b e-a B e) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^3 (a+b x)}+\frac {2 b B (d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}{7 e^3 (a+b x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.54 \[ \int (A+B x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \sqrt {(a+b x)^2} (d+e x)^{3/2} \left (7 A b e (-2 d+3 e x)+7 a e (-2 B d+5 A e+3 B e x)+b B \left (8 d^2-12 d e x+15 e^2 x^2\right )\right )}{105 e^3 (a+b x)} \]

[In]

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

[Out]

(2*Sqrt[(a + b*x)^2]*(d + e*x)^(3/2)*(7*A*b*e*(-2*d + 3*e*x) + 7*a*e*(-2*B*d + 5*A*e + 3*B*e*x) + b*B*(8*d^2 -
12*d*e*x + 15*e^2*x^2)))/(105*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 = 0.48

method	result	size
default	\(\frac {2 \,\operatorname {csgn}\left (b x +a \right ) \left (e x +d \right )^{\frac {3}{2}} \left (15 B b \,e^{2} x^{2}+21 A b \,e^{2} x +21 B a \,e^{2} x -12 B b d e x +35 A a \,e^{2}-14 A b d e -14 B a d e +8 B b \,d^{2}\right )}{105 e^{3}}\)	\(79\)
gosper	\(\frac {2 \left (e x +d \right )^{\frac {3}{2}} \left (15 B b \,e^{2} x^{2}+21 A b \,e^{2} x +21 B a \,e^{2} x -12 B b d e x +35 A a \,e^{2}-14 A b d e -14 B a d e +8 B b \,d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{105 e^{3} \left (b x +a \right )}\)	\(89\)
risch	\(\frac {2 \sqrt {\left (b x +a \right )^{2}}\, \left (15 x^{3} B b \,e^{3}+21 A b \,e^{3} x^{2}+21 B \,x^{2} a \,e^{3}+3 B b d \,e^{2} x^{2}+35 A a \,e^{3} x +7 A b d \,e^{2} x +7 B a d \,e^{2} x -4 B b \,d^{2} e x +35 A a d \,e^{2}-14 A b \,d^{2} e -14 B a \,d^{2} e +8 B b \,d^{3}\right ) \sqrt {e x +d}}{105 \left (b x +a \right ) e^{3}}\)	\(137\)

[In]

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

[Out]

2/105*csgn(b*x+a)*(e*x+d)^(3/2)*(15*B*b*e^2*x^2+21*A*b*e^2*x+21*B*a*e^2*x-12*B*b*d*e*x+35*A*a*e^2-14*A*b*d*e-1
4*B*a*d*e+8*B*b*d^2)/e^3

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.66 \[ \int (A+B x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \, {\left (15 \, B b e^{3} x^{3} + 8 \, B b d^{3} + 35 \, A a d e^{2} - 14 \, {\left (B a + A b\right )} d^{2} e + 3 \, {\left (B b d e^{2} + 7 \, {\left (B a + A b\right )} e^{3}\right )} x^{2} - {\left (4 \, B b d^{2} e - 35 \, A a e^{3} - 7 \, {\left (B a + A b\right )} d e^{2}\right )} x\right )} \sqrt {e x + d}}{105 \, e^{3}} \]

[In]

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

[Out]

2/105*(15*B*b*e^3*x^3 + 8*B*b*d^3 + 35*A*a*d*e^2 - 14*(B*a + A*b)*d^2*e + 3*(B*b*d*e^2 + 7*(B*a + A*b)*e^3)*x^
2 - (4*B*b*d^2*e - 35*A*a*e^3 - 7*(B*a + A*b)*d*e^2)*x)*sqrt(e*x + d)/e^3

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.73 \[ \int (A+B x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \, {\left (3 \, b e^{2} x^{2} - 2 \, b d^{2} + 5 \, a d e + {\left (b d e + 5 \, a e^{2}\right )} x\right )} \sqrt {e x + d} A}{15 \, e^{2}} + \frac {2 \, {\left (15 \, b e^{3} x^{3} + 8 \, b d^{3} - 14 \, a d^{2} e + 3 \, {\left (b d e^{2} + 7 \, a e^{3}\right )} x^{2} - {\left (4 \, b d^{2} e - 7 \, a d e^{2}\right )} x\right )} \sqrt {e x + d} B}{105 \, e^{3}} \]

[In]

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

[Out]

2/15*(3*b*e^2*x^2 - 2*b*d^2 + 5*a*d*e + (b*d*e + 5*a*e^2)*x)*sqrt(e*x + d)*A/e^2 + 2/105*(15*b*e^3*x^3 + 8*b*d
^3 - 14*a*d^2*e + 3*(b*d*e^2 + 7*a*e^3)*x^2 - (4*b*d^2*e - 7*a*d*e^2)*x)*sqrt(e*x + d)*B/e^3

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 309 vs. \(2 (119) = 238\).

Time = 0.28 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.88 \[ \int (A+B x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \, {\left (105 \, \sqrt {e x + d} A a d \mathrm {sgn}\left (b x + a\right ) + 35 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} A a \mathrm {sgn}\left (b x + a\right ) + \frac {35 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} B a d \mathrm {sgn}\left (b x + a\right )}{e} + \frac {35 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} A b d \mathrm {sgn}\left (b x + a\right )}{e} + \frac {7 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} B b d \mathrm {sgn}\left (b x + a\right )}{e^{2}} + \frac {7 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} B a \mathrm {sgn}\left (b x + a\right )}{e} + \frac {7 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} A b \mathrm {sgn}\left (b x + a\right )}{e} + \frac {3 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} B b \mathrm {sgn}\left (b x + a\right )}{e^{2}}\right )}}{105 \, e} \]

[In]

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

[Out]

2/105*(105*sqrt(e*x + d)*A*a*d*sgn(b*x + a) + 35*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*A*a*sgn(b*x + a) + 35*(
(e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*B*a*d*sgn(b*x + a)/e + 35*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*A*b*d*sgn
(b*x + a)/e + 7*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*B*b*d*sgn(b*x + a)/e^2 + 7*(
3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*B*a*sgn(b*x + a)/e + 7*(3*(e*x + d)^(5/2) - 1
0*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*A*b*sgn(b*x + a)/e + 3*(5*(e*x + d)^(7/2) - 21*(e*x + d)^(5/2)*d +
35*(e*x + d)^(3/2)*d^2 - 35*sqrt(e*x + d)*d^3)*B*b*sgn(b*x + a)/e^2)/e

Mupad [F(-1)]

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

[In]

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

[Out]

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