\(\int \frac {(A+B x) (b x+c x^2)}{(d+e x)^{3/2}} \, dx\) [1218]

Optimal result

Integrand size = 24, antiderivative size = 122 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^{3/2}} \, dx=\frac {2 d (B d-A e) (c d-b e)}{e^4 \sqrt {d+e x}}+\frac {2 (B d (3 c d-2 b e)-A e (2 c d-b e)) \sqrt {d+e x}}{e^4}-\frac {2 (3 B c d-b B e-A c e) (d+e x)^{3/2}}{3 e^4}+\frac {2 B c (d+e x)^{5/2}}{5 e^4} \]

[Out]

-2/3*(-A*c*e-B*b*e+3*B*c*d)*(e*x+d)^(3/2)/e^4+2/5*B*c*(e*x+d)^(5/2)/e^4+2*d*(-A*e+B*d)*(-b*e+c*d)/e^4/(e*x+d)^
(1/2)+2*(B*d*(-2*b*e+3*c*d)-A*e*(-b*e+2*c*d))*(e*x+d)^(1/2)/e^4

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 122, 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} \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^{3/2}} \, dx=-\frac {2 (d+e x)^{3/2} (-A c e-b B e+3 B c d)}{3 e^4}+\frac {2 \sqrt {d+e x} (B d (3 c d-2 b e)-A e (2 c d-b e))}{e^4}+\frac {2 d (B d-A e) (c d-b e)}{e^4 \sqrt {d+e x}}+\frac {2 B c (d+e x)^{5/2}}{5 e^4} \]

[In]

Int[((A + B*x)*(b*x + c*x^2))/(d + e*x)^(3/2),x]

[Out]

(2*d*(B*d - A*e)*(c*d - b*e))/(e^4*Sqrt[d + e*x]) + (2*(B*d*(3*c*d - 2*b*e) - A*e*(2*c*d - b*e))*Sqrt[d + e*x]
)/e^4 - (2*(3*B*c*d - b*B*e - A*c*e)*(d + e*x)^(3/2))/(3*e^4) + (2*B*c*(d + e*x)^(5/2))/(5*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)}{e^3 (d+e x)^{3/2}}+\frac {B d (3 c d-2 b e)-A e (2 c d-b e)}{e^3 \sqrt {d+e x}}+\frac {(-3 B c d+b B e+A c e) \sqrt {d+e x}}{e^3}+\frac {B c (d+e x)^{3/2}}{e^3}\right ) \, dx \\ & = \frac {2 d (B d-A e) (c d-b e)}{e^4 \sqrt {d+e x}}+\frac {2 (B d (3 c d-2 b e)-A e (2 c d-b e)) \sqrt {d+e x}}{e^4}-\frac {2 (3 B c d-b B e-A c e) (d+e x)^{3/2}}{3 e^4}+\frac {2 B c (d+e x)^{5/2}}{5 e^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.90 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^{3/2}} \, dx=\frac {2 \left (5 A e \left (3 b e (2 d+e x)+c \left (-8 d^2-4 d e x+e^2 x^2\right )\right )+B \left (5 b e \left (-8 d^2-4 d e x+e^2 x^2\right )+3 c \left (16 d^3+8 d^2 e x-2 d e^2 x^2+e^3 x^3\right )\right )\right )}{15 e^4 \sqrt {d+e x}} \]

[In]

Integrate[((A + B*x)*(b*x + c*x^2))/(d + e*x)^(3/2),x]

[Out]

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

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.87

method	result	size
pseudoelliptic	\(\frac {\frac {2 \left (3 \left (e^{3} x^{3}-2 d \,e^{2} x^{2}+8 d^{2} e x +16 d^{3}\right ) c -40 \left (-\frac {1}{8} e^{2} x^{2}+\frac {1}{2} d e x +d^{2}\right ) e b \right ) B}{15}+4 A \left (\frac {\left (\frac {1}{2} e^{2} x^{2}-2 d e x -4 d^{2}\right ) c}{3}+b e \left (\frac {e x}{2}+d \right )\right ) e}{\sqrt {e x +d}\, e^{4}}\)	\(106\)
risch	\(\frac {2 \left (3 B c \,x^{2} e^{2}+5 A c x \,e^{2}+5 B b x \,e^{2}-9 B c d e x +15 A b \,e^{2}-25 A c d e -25 B b d e +33 B c \,d^{2}\right ) \sqrt {e x +d}}{15 e^{4}}+\frac {2 d \left (A b \,e^{2}-A c d e -B b d e +B c \,d^{2}\right )}{e^{4} \sqrt {e x +d}}\)	\(112\)
gosper	\(\frac {\frac {2}{5} B c \,x^{3} e^{3}+\frac {2}{3} A c \,e^{3} x^{2}+\frac {2}{3} B \,x^{2} b \,e^{3}-\frac {4}{5} B \,x^{2} c d \,e^{2}+2 A b \,e^{3} x -\frac {8}{3} A c d \,e^{2} x -\frac {8}{3} B x b d \,e^{2}+\frac {16}{5} B c \,d^{2} e x +4 A b d \,e^{2}-\frac {16}{3} A c \,d^{2} e -\frac {16}{3} B b \,d^{2} e +\frac {32}{5} B c \,d^{3}}{\sqrt {e x +d}\, e^{4}}\)	\(121\)
trager	\(\frac {\frac {2}{5} B c \,x^{3} e^{3}+\frac {2}{3} A c \,e^{3} x^{2}+\frac {2}{3} B \,x^{2} b \,e^{3}-\frac {4}{5} B \,x^{2} c d \,e^{2}+2 A b \,e^{3} x -\frac {8}{3} A c d \,e^{2} x -\frac {8}{3} B x b d \,e^{2}+\frac {16}{5} B c \,d^{2} e x +4 A b d \,e^{2}-\frac {16}{3} A c \,d^{2} e -\frac {16}{3} B b \,d^{2} e +\frac {32}{5} B c \,d^{3}}{\sqrt {e x +d}\, e^{4}}\)	\(121\)
derivativedivides	\(\frac {\frac {2 B c \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 A c e \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {2 B b e \left (e x +d \right )^{\frac {3}{2}}}{3}-2 B c d \left (e x +d \right )^{\frac {3}{2}}+2 A b \,e^{2} \sqrt {e x +d}-4 A c d e \sqrt {e x +d}-4 B b d e \sqrt {e x +d}+6 B c \,d^{2} \sqrt {e x +d}+\frac {2 d \left (A b \,e^{2}-A c d e -B b d e +B c \,d^{2}\right )}{\sqrt {e x +d}}}{e^{4}}\)	\(141\)
default	\(\frac {\frac {2 B c \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 A c e \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {2 B b e \left (e x +d \right )^{\frac {3}{2}}}{3}-2 B c d \left (e x +d \right )^{\frac {3}{2}}+2 A b \,e^{2} \sqrt {e x +d}-4 A c d e \sqrt {e x +d}-4 B b d e \sqrt {e x +d}+6 B c \,d^{2} \sqrt {e x +d}+\frac {2 d \left (A b \,e^{2}-A c d e -B b d e +B c \,d^{2}\right )}{\sqrt {e x +d}}}{e^{4}}\)	\(141\)

[In]

int((B*x+A)*(c*x^2+b*x)/(e*x+d)^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/15*((3*(e^3*x^3-2*d*e^2*x^2+8*d^2*e*x+16*d^3)*c-40*(-1/8*e^2*x^2+1/2*d*e*x+d^2)*e*b)*B+30*A*(1/3*(1/2*e^2*x^
2-2*d*e*x-4*d^2)*c+b*e*(1/2*e*x+d))*e)/(e*x+d)^(1/2)/e^4

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.97 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (3 \, B c e^{3} x^{3} + 48 \, B c d^{3} + 30 \, A b d e^{2} - 40 \, {\left (B b + A c\right )} d^{2} e - {\left (6 \, B c d e^{2} - 5 \, {\left (B b + A c\right )} e^{3}\right )} x^{2} + {\left (24 \, B c d^{2} e + 15 \, A b e^{3} - 20 \, {\left (B b + A c\right )} d e^{2}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (e^{5} x + d e^{4}\right )}} \]

[In]

integrate((B*x+A)*(c*x^2+b*x)/(e*x+d)^(3/2),x, algorithm="fricas")

[Out]

2/15*(3*B*c*e^3*x^3 + 48*B*c*d^3 + 30*A*b*d*e^2 - 40*(B*b + A*c)*d^2*e - (6*B*c*d*e^2 - 5*(B*b + A*c)*e^3)*x^2
+ (24*B*c*d^2*e + 15*A*b*e^3 - 20*(B*b + A*c)*d*e^2)*x)*sqrt(e*x + d)/(e^5*x + d*e^4)

Sympy [A] (verification not implemented)

Time = 2.45 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.31 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^{3/2}} \, dx=\begin {cases} \frac {2 \left (\frac {B c \left (d + e x\right )^{\frac {5}{2}}}{5 e^{3}} - \frac {d \left (- A e + B d\right ) \left (b e - c d\right )}{e^{3} \sqrt {d + e x}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \left (A c e + B b e - 3 B c d\right )}{3 e^{3}} + \frac {\sqrt {d + e x} \left (A b e^{2} - 2 A c d e - 2 B b d e + 3 B c d^{2}\right )}{e^{3}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {\frac {A b x^{2}}{2} + \frac {B c x^{4}}{4} + \frac {x^{3} \left (A c + B b\right )}{3}}{d^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

[In]

integrate((B*x+A)*(c*x**2+b*x)/(e*x+d)**(3/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.98 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (\frac {3 \, {\left (e x + d\right )}^{\frac {5}{2}} B c - 5 \, {\left (3 \, B c d - {\left (B b + A c\right )} e\right )} {\left (e x + d\right )}^{\frac {3}{2}} + 15 \, {\left (3 \, B c d^{2} + A b e^{2} - 2 \, {\left (B b + A c\right )} d e\right )} \sqrt {e x + d}}{e^{3}} + \frac {15 \, {\left (B c d^{3} + A b d e^{2} - {\left (B b + A c\right )} d^{2} e\right )}}{\sqrt {e x + d} e^{3}}\right )}}{15 \, e} \]

[In]

integrate((B*x+A)*(c*x^2+b*x)/(e*x+d)^(3/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.37 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (B c d^{3} - B b d^{2} e - A c d^{2} e + A b d e^{2}\right )}}{\sqrt {e x + d} e^{4}} + \frac {2 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} B c e^{16} - 15 \, {\left (e x + d\right )}^{\frac {3}{2}} B c d e^{16} + 45 \, \sqrt {e x + d} B c d^{2} e^{16} + 5 \, {\left (e x + d\right )}^{\frac {3}{2}} B b e^{17} + 5 \, {\left (e x + d\right )}^{\frac {3}{2}} A c e^{17} - 30 \, \sqrt {e x + d} B b d e^{17} - 30 \, \sqrt {e x + d} A c d e^{17} + 15 \, \sqrt {e x + d} A b e^{18}\right )}}{15 \, e^{20}} \]

[In]

integrate((B*x+A)*(c*x^2+b*x)/(e*x+d)^(3/2),x, algorithm="giac")

[Out]

2*(B*c*d^3 - B*b*d^2*e - A*c*d^2*e + A*b*d*e^2)/(sqrt(e*x + d)*e^4) + 2/15*(3*(e*x + d)^(5/2)*B*c*e^16 - 15*(e
*x + d)^(3/2)*B*c*d*e^16 + 45*sqrt(e*x + d)*B*c*d^2*e^16 + 5*(e*x + d)^(3/2)*B*b*e^17 + 5*(e*x + d)^(3/2)*A*c*
e^17 - 30*sqrt(e*x + d)*B*b*d*e^17 - 30*sqrt(e*x + d)*A*c*d*e^17 + 15*sqrt(e*x + d)*A*b*e^18)/e^20

Mupad [B] (verification not implemented)

Time = 10.40 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.02 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^{3/2}} \, dx=\frac {\sqrt {d+e\,x}\,\left (2\,A\,b\,e^2+6\,B\,c\,d^2-4\,A\,c\,d\,e-4\,B\,b\,d\,e\right )}{e^4}+\frac {{\left (d+e\,x\right )}^{3/2}\,\left (2\,A\,c\,e+2\,B\,b\,e-6\,B\,c\,d\right )}{3\,e^4}+\frac {2\,B\,c\,d^3+2\,A\,b\,d\,e^2-2\,A\,c\,d^2\,e-2\,B\,b\,d^2\,e}{e^4\,\sqrt {d+e\,x}}+\frac {2\,B\,c\,{\left (d+e\,x\right )}^{5/2}}{5\,e^4} \]

[In]

int(((b*x + c*x^2)*(A + B*x))/(d + e*x)^(3/2),x)

[Out]

((d + e*x)^(1/2)*(2*A*b*e^2 + 6*B*c*d^2 - 4*A*c*d*e - 4*B*b*d*e))/e^4 + ((d + e*x)^(3/2)*(2*A*c*e + 2*B*b*e -
6*B*c*d))/(3*e^4) + (2*B*c*d^3 + 2*A*b*d*e^2 - 2*A*c*d^2*e - 2*B*b*d^2*e)/(e^4*(d + e*x)^(1/2)) + (2*B*c*(d +
e*x)^(5/2))/(5*e^4)