\(\int \frac {(d+e x)^{3/2} (f+g x)}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx\) [2263]

Optimal result

Integrand size = 46, antiderivative size = 193 \[ \int \frac {(d+e x)^{3/2} (f+g x)}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=-\frac {4 (2 c d-b e) (5 c e f+3 c d g-4 b e g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{15 c^3 e^2 \sqrt {d+e x}}-\frac {2 (5 c e f+3 c d g-4 b e g) \sqrt {d+e x} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{15 c^2 e^2}-\frac {2 g (d+e x)^{3/2} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{5 c e^2} \]

[Out]

-2/5*g*(e*x+d)^(3/2)*(d*(-b*e+c*d)-b*e^2*x-c*e^2*x^2)^(1/2)/c/e^2-4/15*(-b*e+2*c*d)*(-4*b*e*g+3*c*d*g+5*c*e*f)
*(d*(-b*e+c*d)-b*e^2*x-c*e^2*x^2)^(1/2)/c^3/e^2/(e*x+d)^(1/2)-2/15*(-4*b*e*g+3*c*d*g+5*c*e*f)*(e*x+d)^(1/2)*(d
*(-b*e+c*d)-b*e^2*x-c*e^2*x^2)^(1/2)/c^2/e^2

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {808, 670, 662} \[ \int \frac {(d+e x)^{3/2} (f+g x)}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=-\frac {4 (2 c d-b e) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2} (-4 b e g+3 c d g+5 c e f)}{15 c^3 e^2 \sqrt {d+e x}}-\frac {2 \sqrt {d+e x} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2} (-4 b e g+3 c d g+5 c e f)}{15 c^2 e^2}-\frac {2 g (d+e x)^{3/2} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{5 c e^2} \]

[In]

Int[((d + e*x)^(3/2)*(f + g*x))/Sqrt[c*d^2 - b*d*e - b*e^2*x - c*e^2*x^2],x]

[Out]

(-4*(2*c*d - b*e)*(5*c*e*f + 3*c*d*g - 4*b*e*g)*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2])/(15*c^3*e^2*Sqrt[d
+ e*x]) - (2*(5*c*e*f + 3*c*d*g - 4*b*e*g)*Sqrt[d + e*x]*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2])/(15*c^2*e^
2) - (2*g*(d + e*x)^(3/2)*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2])/(5*c*e^2)

Rule 662

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*
((a + b*x + c*x^2)^(p + 1)/(c*(p + 1))), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c
*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p, 0]

Rule 670

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*
((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Dist[Simplify[m + p]*((2*c*d - b*e)/(c*(m + 2*p + 1))), In
t[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && E
qQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && IGtQ[Simplify[m + p], 0]

Rule 808

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), x] + Dist[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)
*(2*c*f - b*g))/(c*e*(m + 2*p + 2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g
, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[m + 2*p + 2, 0] && (NeQ[m, 2] || Eq
Q[d, 0])

Rubi steps \begin{align*} \text {integral}& = -\frac {2 g (d+e x)^{3/2} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{5 c e^2}-\frac {\left (2 \left (\frac {1}{2} e \left (-2 c e^2 f+b e^2 g\right )+\frac {3}{2} \left (-c e^3 f+\left (-c d e^2+b e^3\right ) g\right )\right )\right ) \int \frac {(d+e x)^{3/2}}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx}{5 c e^3} \\ & = -\frac {2 (5 c e f+3 c d g-4 b e g) \sqrt {d+e x} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{15 c^2 e^2}-\frac {2 g (d+e x)^{3/2} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{5 c e^2}+\frac {(2 (2 c d-b e) (5 c e f+3 c d g-4 b e g)) \int \frac {\sqrt {d+e x}}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx}{15 c^2 e} \\ & = -\frac {4 (2 c d-b e) (5 c e f+3 c d g-4 b e g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{15 c^3 e^2 \sqrt {d+e x}}-\frac {2 (5 c e f+3 c d g-4 b e g) \sqrt {d+e x} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{15 c^2 e^2}-\frac {2 g (d+e x)^{3/2} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{5 c e^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.61 \[ \int \frac {(d+e x)^{3/2} (f+g x)}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\frac {2 \sqrt {d+e x} (-c d+b e+c e x) \left (8 b^2 e^2 g-2 b c e (5 e f+13 d g+2 e g x)+c^2 \left (18 d^2 g+e^2 x (5 f+3 g x)+d e (25 f+9 g x)\right )\right )}{15 c^3 e^2 \sqrt {(d+e x) (-b e+c (d-e x))}} \]

[In]

Integrate[((d + e*x)^(3/2)*(f + g*x))/Sqrt[c*d^2 - b*d*e - b*e^2*x - c*e^2*x^2],x]

[Out]

(2*Sqrt[d + e*x]*(-(c*d) + b*e + c*e*x)*(8*b^2*e^2*g - 2*b*c*e*(5*e*f + 13*d*g + 2*e*g*x) + c^2*(18*d^2*g + e^
2*x*(5*f + 3*g*x) + d*e*(25*f + 9*g*x))))/(15*c^3*e^2*Sqrt[(d + e*x)*(-(b*e) + c*(d - e*x))])

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.62

method	result	size
default	\(-\frac {2 \sqrt {-\left (e x +d \right ) \left (x c e +b e -c d \right )}\, \left (3 g \,x^{2} c^{2} e^{2}-4 b c \,e^{2} g x +9 c^{2} d e g x +5 c^{2} e^{2} f x +8 b^{2} e^{2} g -26 b c d e g -10 b c \,e^{2} f +18 c^{2} d^{2} g +25 c^{2} d e f \right )}{15 \sqrt {e x +d}\, c^{3} e^{2}}\)	\(119\)
gosper	\(\frac {2 \left (x c e +b e -c d \right ) \left (3 g \,x^{2} c^{2} e^{2}-4 b c \,e^{2} g x +9 c^{2} d e g x +5 c^{2} e^{2} f x +8 b^{2} e^{2} g -26 b c d e g -10 b c \,e^{2} f +18 c^{2} d^{2} g +25 c^{2} d e f \right ) \sqrt {e x +d}}{15 c^{3} e^{2} \sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}}\)	\(139\)

[In]

int((e*x+d)^(3/2)*(g*x+f)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/15/(e*x+d)^(1/2)*(-(e*x+d)*(c*e*x+b*e-c*d))^(1/2)*(3*c^2*e^2*g*x^2-4*b*c*e^2*g*x+9*c^2*d*e*g*x+5*c^2*e^2*f*
x+8*b^2*e^2*g-26*b*c*d*e*g-10*b*c*e^2*f+18*c^2*d^2*g+25*c^2*d*e*f)/c^3/e^2

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.74 \[ \int \frac {(d+e x)^{3/2} (f+g x)}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=-\frac {2 \, {\left (3 \, c^{2} e^{2} g x^{2} + 5 \, {\left (5 \, c^{2} d e - 2 \, b c e^{2}\right )} f + 2 \, {\left (9 \, c^{2} d^{2} - 13 \, b c d e + 4 \, b^{2} e^{2}\right )} g + {\left (5 \, c^{2} e^{2} f + {\left (9 \, c^{2} d e - 4 \, b c e^{2}\right )} g\right )} x\right )} \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} \sqrt {e x + d}}{15 \, {\left (c^{3} e^{3} x + c^{3} d e^{2}\right )}} \]

[In]

integrate((e*x+d)^(3/2)*(g*x+f)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2),x, algorithm="fricas")

[Out]

-2/15*(3*c^2*e^2*g*x^2 + 5*(5*c^2*d*e - 2*b*c*e^2)*f + 2*(9*c^2*d^2 - 13*b*c*d*e + 4*b^2*e^2)*g + (5*c^2*e^2*f
+ (9*c^2*d*e - 4*b*c*e^2)*g)*x)*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*sqrt(e*x + d)/(c^3*e^3*x + c^3*d*e
^2)

Sympy [F]

\[ \int \frac {(d+e x)^{3/2} (f+g x)}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\int \frac {\left (d + e x\right )^{\frac {3}{2}} \left (f + g x\right )}{\sqrt {- \left (d + e x\right ) \left (b e - c d + c e x\right )}}\, dx \]

[In]

integrate((e*x+d)**(3/2)*(g*x+f)/(-c*e**2*x**2-b*e**2*x-b*d*e+c*d**2)**(1/2),x)

[Out]

Integral((d + e*x)**(3/2)*(f + g*x)/sqrt(-(d + e*x)*(b*e - c*d + c*e*x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.04 \[ \int \frac {(d+e x)^{3/2} (f+g x)}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\frac {2 \, {\left (c^{2} e^{2} x^{2} - 5 \, c^{2} d^{2} + 7 \, b c d e - 2 \, b^{2} e^{2} + {\left (4 \, c^{2} d e - b c e^{2}\right )} x\right )} f}{3 \, \sqrt {-c e x + c d - b e} c^{2} e} + \frac {2 \, {\left (3 \, c^{3} e^{3} x^{3} - 18 \, c^{3} d^{3} + 44 \, b c^{2} d^{2} e - 34 \, b^{2} c d e^{2} + 8 \, b^{3} e^{3} + {\left (6 \, c^{3} d e^{2} - b c^{2} e^{3}\right )} x^{2} + {\left (9 \, c^{3} d^{2} e - 13 \, b c^{2} d e^{2} + 4 \, b^{2} c e^{3}\right )} x\right )} g}{15 \, \sqrt {-c e x + c d - b e} c^{3} e^{2}} \]

[In]

integrate((e*x+d)^(3/2)*(g*x+f)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2),x, algorithm="maxima")

[Out]

2/3*(c^2*e^2*x^2 - 5*c^2*d^2 + 7*b*c*d*e - 2*b^2*e^2 + (4*c^2*d*e - b*c*e^2)*x)*f/(sqrt(-c*e*x + c*d - b*e)*c^
2*e) + 2/15*(3*c^3*e^3*x^3 - 18*c^3*d^3 + 44*b*c^2*d^2*e - 34*b^2*c*d*e^2 + 8*b^3*e^3 + (6*c^3*d*e^2 - b*c^2*e
^3)*x^2 + (9*c^3*d^2*e - 13*b*c^2*d*e^2 + 4*b^2*c*e^3)*x)*g/(sqrt(-c*e*x + c*d - b*e)*c^3*e^2)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.55 \[ \int \frac {(d+e x)^{3/2} (f+g x)}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=-\frac {2 \, {\left (\frac {15 \, {\left (2 \, c^{2} d e f - b c e^{2} f + 2 \, c^{2} d^{2} g - 3 \, b c d e g + b^{2} e^{2} g\right )} \sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e}}{c^{3} e} - \frac {2 \, {\left (10 \, \sqrt {2 \, c d - b e} c^{2} d e f - 5 \, \sqrt {2 \, c d - b e} b c e^{2} f + 6 \, \sqrt {2 \, c d - b e} c^{2} d^{2} g - 11 \, \sqrt {2 \, c d - b e} b c d e g + 4 \, \sqrt {2 \, c d - b e} b^{2} e^{2} g\right )}}{c^{3} e} - \frac {5 \, {\left (-{\left (e x + d\right )} c + 2 \, c d - b e\right )}^{\frac {3}{2}} c e f + 15 \, {\left (-{\left (e x + d\right )} c + 2 \, c d - b e\right )}^{\frac {3}{2}} c d g - 10 \, {\left (-{\left (e x + d\right )} c + 2 \, c d - b e\right )}^{\frac {3}{2}} b e g - 3 \, {\left ({\left (e x + d\right )} c - 2 \, c d + b e\right )}^{2} \sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e} g}{c^{3} e}\right )}}{15 \, e} \]

[In]

integrate((e*x+d)^(3/2)*(g*x+f)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2),x, algorithm="giac")

[Out]

-2/15*(15*(2*c^2*d*e*f - b*c*e^2*f + 2*c^2*d^2*g - 3*b*c*d*e*g + b^2*e^2*g)*sqrt(-(e*x + d)*c + 2*c*d - b*e)/(
c^3*e) - 2*(10*sqrt(2*c*d - b*e)*c^2*d*e*f - 5*sqrt(2*c*d - b*e)*b*c*e^2*f + 6*sqrt(2*c*d - b*e)*c^2*d^2*g - 1
1*sqrt(2*c*d - b*e)*b*c*d*e*g + 4*sqrt(2*c*d - b*e)*b^2*e^2*g)/(c^3*e) - (5*(-(e*x + d)*c + 2*c*d - b*e)^(3/2)
*c*e*f + 15*(-(e*x + d)*c + 2*c*d - b*e)^(3/2)*c*d*g - 10*(-(e*x + d)*c + 2*c*d - b*e)^(3/2)*b*e*g - 3*((e*x +
d)*c - 2*c*d + b*e)^2*sqrt(-(e*x + d)*c + 2*c*d - b*e)*g)/(c^3*e))/e

Mupad [B] (verification not implemented)

Time = 11.40 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.77 \[ \int \frac {(d+e x)^{3/2} (f+g x)}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=-\frac {\left (\frac {\sqrt {d+e\,x}\,\left (16\,g\,b^2\,e^2-52\,g\,b\,c\,d\,e-20\,f\,b\,c\,e^2+36\,g\,c^2\,d^2+50\,f\,c^2\,d\,e\right )}{15\,c^3\,e^3}+\frac {2\,g\,x^2\,\sqrt {d+e\,x}}{5\,c\,e}+\frac {2\,x\,\sqrt {d+e\,x}\,\left (9\,c\,d\,g-4\,b\,e\,g+5\,c\,e\,f\right )}{15\,c^2\,e^2}\right )\,\sqrt {c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,x}}{x+\frac {d}{e}} \]

[In]

int(((f + g*x)*(d + e*x)^(3/2))/(c*d^2 - c*e^2*x^2 - b*d*e - b*e^2*x)^(1/2),x)

[Out]

-((((d + e*x)^(1/2)*(16*b^2*e^2*g + 36*c^2*d^2*g - 20*b*c*e^2*f + 50*c^2*d*e*f - 52*b*c*d*e*g))/(15*c^3*e^3) +
(2*g*x^2*(d + e*x)^(1/2))/(5*c*e) + (2*x*(d + e*x)^(1/2)*(9*c*d*g - 4*b*e*g + 5*c*e*f))/(15*c^2*e^2))*(c*d^2
- c*e^2*x^2 - b*d*e - b*e^2*x)^(1/2))/(x + d/e)