\(\int (A+B x) \sqrt {d+e x} (b x+c x^2) \, dx\) [1216]

Optimal result

Integrand size = 24, antiderivative size = 126 \[ \int (A+B x) \sqrt {d+e x} \left (b x+c x^2\right ) \, dx=-\frac {2 d (B d-A e) (c d-b e) (d+e x)^{3/2}}{3 e^4}+\frac {2 (B d (3 c d-2 b e)-A e (2 c d-b e)) (d+e x)^{5/2}}{5 e^4}-\frac {2 (3 B c d-b B e-A c e) (d+e x)^{7/2}}{7 e^4}+\frac {2 B c (d+e x)^{9/2}}{9 e^4} \]

[Out]

-2/3*d*(-A*e+B*d)*(-b*e+c*d)*(e*x+d)^(3/2)/e^4+2/5*(B*d*(-2*b*e+3*c*d)-A*e*(-b*e+2*c*d))*(e*x+d)^(5/2)/e^4-2/7
*(-A*c*e-B*b*e+3*B*c*d)*(e*x+d)^(7/2)/e^4+2/9*B*c*(e*x+d)^(9/2)/e^4

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {785} \[ \int (A+B x) \sqrt {d+e x} \left (b x+c x^2\right ) \, dx=-\frac {2 (d+e x)^{7/2} (-A c e-b B e+3 B c d)}{7 e^4}+\frac {2 (d+e x)^{5/2} (B d (3 c d-2 b e)-A e (2 c d-b e))}{5 e^4}-\frac {2 d (d+e x)^{3/2} (B d-A e) (c d-b e)}{3 e^4}+\frac {2 B c (d+e x)^{9/2}}{9 e^4} \]

[In]

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

[Out]

(-2*d*(B*d - A*e)*(c*d - b*e)*(d + e*x)^(3/2))/(3*e^4) + (2*(B*d*(3*c*d - 2*b*e) - A*e*(2*c*d - b*e))*(d + e*x
)^(5/2))/(5*e^4) - (2*(3*B*c*d - b*B*e - A*c*e)*(d + e*x)^(7/2))/(7*e^4) + (2*B*c*(d + e*x)^(9/2))/(9*e^4)

Rule 785

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(d + e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && N
eQ[b^2 - 4*a*c, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {d (B d-A e) (c d-b e) \sqrt {d+e x}}{e^3}+\frac {(B d (3 c d-2 b e)-A e (2 c d-b e)) (d+e x)^{3/2}}{e^3}+\frac {(-3 B c d+b B e+A c e) (d+e x)^{5/2}}{e^3}+\frac {B c (d+e x)^{7/2}}{e^3}\right ) \, dx \\ & = -\frac {2 d (B d-A e) (c d-b e) (d+e x)^{3/2}}{3 e^4}+\frac {2 (B d (3 c d-2 b e)-A e (2 c d-b e)) (d+e x)^{5/2}}{5 e^4}-\frac {2 (3 B c d-b B e-A c e) (d+e x)^{7/2}}{7 e^4}+\frac {2 B c (d+e x)^{9/2}}{9 e^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.90 \[ \int (A+B x) \sqrt {d+e x} \left (b x+c x^2\right ) \, dx=\frac {2 (d+e x)^{3/2} \left (3 A e \left (7 b e (-2 d+3 e x)+c \left (8 d^2-12 d e x+15 e^2 x^2\right )\right )+B \left (3 b e \left (8 d^2-12 d e x+15 e^2 x^2\right )+c \left (-16 d^3+24 d^2 e x-30 d e^2 x^2+35 e^3 x^3\right )\right )\right )}{315 e^4} \]

[In]

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

[Out]

(2*(d + e*x)^(3/2)*(3*A*e*(7*b*e*(-2*d + 3*e*x) + c*(8*d^2 - 12*d*e*x + 15*e^2*x^2)) + B*(3*b*e*(8*d^2 - 12*d*
e*x + 15*e^2*x^2) + c*(-16*d^3 + 24*d^2*e*x - 30*d*e^2*x^2 + 35*e^3*x^3))))/(315*e^4)

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.72

method	result	size
pseudoelliptic	\(-\frac {4 \left (e x +d \right )^{\frac {3}{2}} \left (-\frac {3 \left (\frac {5 B c \,x^{2}}{9}+\frac {5 \left (A c +B b \right ) x}{7}+A b \right ) x \,e^{3}}{2}+\left (\frac {5 B c \,x^{2}}{7}+\frac {6 \left (A c +B b \right ) x}{7}+A b \right ) d \,e^{2}-\frac {4 d^{2} \left (B c x +A c +B b \right ) e}{7}+\frac {8 B c \,d^{3}}{21}\right )}{15 e^{4}}\)	\(91\)
default	\(\frac {\frac {2 B c \left (e x +d \right )^{\frac {9}{2}}}{9}+\frac {2 \left (\left (A e -2 B d \right ) c +B \left (b e -c d \right )\right ) \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (\left (-A e +B d \right ) d c +\left (A e -2 B d \right ) \left (b e -c d \right )\right ) \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 \left (-A e +B d \right ) d \left (b e -c d \right ) \left (e x +d \right )^{\frac {3}{2}}}{3}}{e^{4}}\)	\(112\)
derivativedivides	\(\frac {\frac {2 B c \left (e x +d \right )^{\frac {9}{2}}}{9}+\frac {2 \left (\left (A e -2 B d \right ) c +B \left (b e -c d \right )\right ) \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (-\left (A e -B d \right ) d c +\left (A e -2 B d \right ) \left (b e -c d \right )\right ) \left (e x +d \right )^{\frac {5}{2}}}{5}-\frac {2 \left (A e -B d \right ) d \left (b e -c d \right ) \left (e x +d \right )^{\frac {3}{2}}}{3}}{e^{4}}\)	\(113\)
gosper	\(-\frac {2 \left (e x +d \right )^{\frac {3}{2}} \left (-35 B c \,x^{3} e^{3}-45 A c \,e^{3} x^{2}-45 B \,x^{2} b \,e^{3}+30 B \,x^{2} c d \,e^{2}-63 A b \,e^{3} x +36 A c d \,e^{2} x +36 B x b d \,e^{2}-24 B c \,d^{2} e x +42 A b d \,e^{2}-24 A c \,d^{2} e -24 B b \,d^{2} e +16 B c \,d^{3}\right )}{315 e^{4}}\)	\(121\)
trager	\(-\frac {2 \left (-35 B \,e^{4} c \,x^{4}-45 A c \,e^{4} x^{3}-45 B \,e^{4} b \,x^{3}-5 B c d \,e^{3} x^{3}-63 A b \,e^{4} x^{2}-9 A c d \,e^{3} x^{2}-9 B b d \,e^{3} x^{2}+6 B c \,d^{2} e^{2} x^{2}-21 A b d \,e^{3} x +12 A c \,d^{2} e^{2} x +12 B b \,d^{2} e^{2} x -8 B c \,d^{3} e x +42 A b \,d^{2} e^{2}-24 A c \,d^{3} e -24 B b \,d^{3} e +16 B c \,d^{4}\right ) \sqrt {e x +d}}{315 e^{4}}\)	\(173\)
risch	\(-\frac {2 \left (-35 B \,e^{4} c \,x^{4}-45 A c \,e^{4} x^{3}-45 B \,e^{4} b \,x^{3}-5 B c d \,e^{3} x^{3}-63 A b \,e^{4} x^{2}-9 A c d \,e^{3} x^{2}-9 B b d \,e^{3} x^{2}+6 B c \,d^{2} e^{2} x^{2}-21 A b d \,e^{3} x +12 A c \,d^{2} e^{2} x +12 B b \,d^{2} e^{2} x -8 B c \,d^{3} e x +42 A b \,d^{2} e^{2}-24 A c \,d^{3} e -24 B b \,d^{3} e +16 B c \,d^{4}\right ) \sqrt {e x +d}}{315 e^{4}}\)	\(173\)

[In]

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

[Out]

-4/15*(e*x+d)^(3/2)*(-3/2*(5/9*B*c*x^2+5/7*(A*c+B*b)*x+A*b)*x*e^3+(5/7*B*c*x^2+6/7*(A*c+B*b)*x+A*b)*d*e^2-4/7*
d^2*(B*c*x+A*c+B*b)*e+8/21*B*c*d^3)/e^4

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.17 \[ \int (A+B x) \sqrt {d+e x} \left (b x+c x^2\right ) \, dx=\frac {2 \, {\left (35 \, B c e^{4} x^{4} - 16 \, B c d^{4} - 42 \, A b d^{2} e^{2} + 24 \, {\left (B b + A c\right )} d^{3} e + 5 \, {\left (B c d e^{3} + 9 \, {\left (B b + A c\right )} e^{4}\right )} x^{3} - 3 \, {\left (2 \, B c d^{2} e^{2} - 21 \, A b e^{4} - 3 \, {\left (B b + A c\right )} d e^{3}\right )} x^{2} + {\left (8 \, B c d^{3} e + 21 \, A b d e^{3} - 12 \, {\left (B b + A c\right )} d^{2} e^{2}\right )} x\right )} \sqrt {e x + d}}{315 \, e^{4}} \]

[In]

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

[Out]

2/315*(35*B*c*e^4*x^4 - 16*B*c*d^4 - 42*A*b*d^2*e^2 + 24*(B*b + A*c)*d^3*e + 5*(B*c*d*e^3 + 9*(B*b + A*c)*e^4)
*x^3 - 3*(2*B*c*d^2*e^2 - 21*A*b*e^4 - 3*(B*b + A*c)*d*e^3)*x^2 + (8*B*c*d^3*e + 21*A*b*d*e^3 - 12*(B*b + A*c)
*d^2*e^2)*x)*sqrt(e*x + d)/e^4

Sympy [A] (verification not implemented)

Time = 0.91 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.43 \[ \int (A+B x) \sqrt {d+e x} \left (b x+c x^2\right ) \, dx=\begin {cases} \frac {2 \left (\frac {B c \left (d + e x\right )^{\frac {9}{2}}}{9 e^{3}} + \frac {\left (d + e x\right )^{\frac {7}{2}} \left (A c e + B b e - 3 B c d\right )}{7 e^{3}} + \frac {\left (d + e x\right )^{\frac {5}{2}} \left (A b e^{2} - 2 A c d e - 2 B b d e + 3 B c d^{2}\right )}{5 e^{3}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \left (- A b d e^{2} + A c d^{2} e + B b d^{2} e - B c d^{3}\right )}{3 e^{3}}\right )}{e} & \text {for}\: e \neq 0 \\\sqrt {d} \left (\frac {A b x^{2}}{2} + \frac {B c x^{4}}{4} + \frac {x^{3} \left (A c + B b\right )}{3}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*(B*c*(d + e*x)**(9/2)/(9*e**3) + (d + e*x)**(7/2)*(A*c*e + B*b*e - 3*B*c*d)/(7*e**3) + (d + e*x)*
*(5/2)*(A*b*e**2 - 2*A*c*d*e - 2*B*b*d*e + 3*B*c*d**2)/(5*e**3) + (d + e*x)**(3/2)*(-A*b*d*e**2 + A*c*d**2*e +
B*b*d**2*e - B*c*d**3)/(3*e**3))/e, Ne(e, 0)), (sqrt(d)*(A*b*x**2/2 + B*c*x**4/4 + x**3*(A*c + B*b)/3), True)
)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.89 \[ \int (A+B x) \sqrt {d+e x} \left (b x+c x^2\right ) \, dx=\frac {2 \, {\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} B c - 45 \, {\left (3 \, B c d - {\left (B b + A c\right )} e\right )} {\left (e x + d\right )}^{\frac {7}{2}} + 63 \, {\left (3 \, B c d^{2} + A b e^{2} - 2 \, {\left (B b + A c\right )} d e\right )} {\left (e x + d\right )}^{\frac {5}{2}} - 105 \, {\left (B c d^{3} + A b d e^{2} - {\left (B b + A c\right )} d^{2} e\right )} {\left (e x + d\right )}^{\frac {3}{2}}\right )}}{315 \, e^{4}} \]

[In]

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

[Out]

2/315*(35*(e*x + d)^(9/2)*B*c - 45*(3*B*c*d - (B*b + A*c)*e)*(e*x + d)^(7/2) + 63*(3*B*c*d^2 + A*b*e^2 - 2*(B*
b + A*c)*d*e)*(e*x + d)^(5/2) - 105*(B*c*d^3 + A*b*d*e^2 - (B*b + A*c)*d^2*e)*(e*x + d)^(3/2))/e^4

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 367 vs. \(2 (110) = 220\).

Time = 0.27 (sec) , antiderivative size = 367, normalized size of antiderivative = 2.91 \[ \int (A+B x) \sqrt {d+e x} \left (b x+c x^2\right ) \, dx=\frac {2 \, {\left (\frac {105 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} A b d}{e} + \frac {21 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} B b d}{e^{2}} + \frac {21 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} A c d}{e^{2}} + \frac {21 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} A b}{e} + \frac {9 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} B c d}{e^{3}} + \frac {9 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} B b}{e^{2}} + \frac {9 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} A c}{e^{2}} + \frac {{\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} - 180 \, {\left (e x + d\right )}^{\frac {7}{2}} d + 378 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2} - 420 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{3} + 315 \, \sqrt {e x + d} d^{4}\right )} B c}{e^{3}}\right )}}{315 \, e} \]

[In]

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

[Out]

2/315*(105*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*A*b*d/e + 21*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*s
qrt(e*x + d)*d^2)*B*b*d/e^2 + 21*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*A*c*d/e^2 +
21*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*A*b/e + 9*(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*c*d/e^3 + 9*(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/e^2 + 9*(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)*A*c/e^2 + (35*(e*x + d)^(9/2) - 180*(e*x + d)^(7/2)*d + 378*(
e*x + d)^(5/2)*d^2 - 420*(e*x + d)^(3/2)*d^3 + 315*sqrt(e*x + d)*d^4)*B*c/e^3)/e

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.88 \[ \int (A+B x) \sqrt {d+e x} \left (b x+c x^2\right ) \, dx=\frac {{\left (d+e\,x\right )}^{5/2}\,\left (2\,A\,b\,e^2+6\,B\,c\,d^2-4\,A\,c\,d\,e-4\,B\,b\,d\,e\right )}{5\,e^4}+\frac {{\left (d+e\,x\right )}^{7/2}\,\left (2\,A\,c\,e+2\,B\,b\,e-6\,B\,c\,d\right )}{7\,e^4}+\frac {2\,B\,c\,{\left (d+e\,x\right )}^{9/2}}{9\,e^4}-\frac {2\,d\,\left (A\,e-B\,d\right )\,\left (b\,e-c\,d\right )\,{\left (d+e\,x\right )}^{3/2}}{3\,e^4} \]

[In]

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

[Out]

((d + e*x)^(5/2)*(2*A*b*e^2 + 6*B*c*d^2 - 4*A*c*d*e - 4*B*b*d*e))/(5*e^4) + ((d + e*x)^(7/2)*(2*A*c*e + 2*B*b*
e - 6*B*c*d))/(7*e^4) + (2*B*c*(d + e*x)^(9/2))/(9*e^4) - (2*d*(A*e - B*d)*(b*e - c*d)*(d + e*x)^(3/2))/(3*e^4
)