\(\int \frac {a+b x+c x^2}{(d+e x)^{7/2}} \, dx\) [2274]

Optimal result

Integrand size = 20, antiderivative size = 73 \[ \int \frac {a+b x+c x^2}{(d+e x)^{7/2}} \, dx=-\frac {2 \left (c d^2-b d e+a e^2\right )}{5 e^3 (d+e x)^{5/2}}+\frac {2 (2 c d-b e)}{3 e^3 (d+e x)^{3/2}}-\frac {2 c}{e^3 \sqrt {d+e x}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {712} \[ \int \frac {a+b x+c x^2}{(d+e x)^{7/2}} \, dx=-\frac {2 \left (a e^2-b d e+c d^2\right )}{5 e^3 (d+e x)^{5/2}}+\frac {2 (2 c d-b e)}{3 e^3 (d+e x)^{3/2}}-\frac {2 c}{e^3 \sqrt {d+e x}} \]

[In]

Int[(a + b*x + c*x^2)/(d + e*x)^(7/2),x]

[Out]

(-2*(c*d^2 - b*d*e + a*e^2))/(5*e^3*(d + e*x)^(5/2)) + (2*(2*c*d - b*e))/(3*e^3*(d + e*x)^(3/2)) - (2*c)/(e^3*
Sqrt[d + e*x])

Rule 712

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

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.74 \[ \int \frac {a+b x+c x^2}{(d+e x)^{7/2}} \, dx=-\frac {2 \left (e (2 b d+3 a e+5 b e x)+c \left (8 d^2+20 d e x+15 e^2 x^2\right )\right )}{15 e^3 (d+e x)^{5/2}} \]

[In]

Integrate[(a + b*x + c*x^2)/(d + e*x)^(7/2),x]

[Out]

(-2*(e*(2*b*d + 3*a*e + 5*b*e*x) + c*(8*d^2 + 20*d*e*x + 15*e^2*x^2)))/(15*e^3*(d + e*x)^(5/2))

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.66

method	result	size
pseudoelliptic	\(\frac {\frac {2 \left (-15 c \,x^{2}-5 b x -3 a \right ) e^{2}}{15}-\frac {4 d \left (10 c x +b \right ) e}{15}-\frac {16 c \,d^{2}}{15}}{\left (e x +d \right )^{\frac {5}{2}} e^{3}}\)	\(48\)
gosper	\(-\frac {2 \left (15 c \,x^{2} e^{2}+5 b \,e^{2} x +20 c d e x +3 a \,e^{2}+2 b d e +8 c \,d^{2}\right )}{15 \left (e x +d \right )^{\frac {5}{2}} e^{3}}\)	\(53\)
trager	\(-\frac {2 \left (15 c \,x^{2} e^{2}+5 b \,e^{2} x +20 c d e x +3 a \,e^{2}+2 b d e +8 c \,d^{2}\right )}{15 \left (e x +d \right )^{\frac {5}{2}} e^{3}}\)	\(53\)
derivativedivides	\(\frac {-\frac {2 \left (a \,e^{2}-b d e +c \,d^{2}\right )}{5 \left (e x +d \right )^{\frac {5}{2}}}-\frac {2 c}{\sqrt {e x +d}}-\frac {2 \left (b e -2 c d \right )}{3 \left (e x +d \right )^{\frac {3}{2}}}}{e^{3}}\)	\(59\)
default	\(\frac {-\frac {2 \left (a \,e^{2}-b d e +c \,d^{2}\right )}{5 \left (e x +d \right )^{\frac {5}{2}}}-\frac {2 c}{\sqrt {e x +d}}-\frac {2 \left (b e -2 c d \right )}{3 \left (e x +d \right )^{\frac {3}{2}}}}{e^{3}}\)	\(59\)

[In]

int((c*x^2+b*x+a)/(e*x+d)^(7/2),x,method=_RETURNVERBOSE)

[Out]

2/15*((-15*c*x^2-5*b*x-3*a)*e^2-2*d*(10*c*x+b)*e-8*c*d^2)/(e*x+d)^(5/2)/e^3

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.16 \[ \int \frac {a+b x+c x^2}{(d+e x)^{7/2}} \, dx=-\frac {2 \, {\left (15 \, c e^{2} x^{2} + 8 \, c d^{2} + 2 \, b d e + 3 \, a e^{2} + 5 \, {\left (4 \, c d e + b e^{2}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}} \]

[In]

integrate((c*x^2+b*x+a)/(e*x+d)^(7/2),x, algorithm="fricas")

[Out]

-2/15*(15*c*e^2*x^2 + 8*c*d^2 + 2*b*d*e + 3*a*e^2 + 5*(4*c*d*e + b*e^2)*x)*sqrt(e*x + d)/(e^6*x^3 + 3*d*e^5*x^
2 + 3*d^2*e^4*x + d^3*e^3)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 376 vs. \(2 (76) = 152\).

Time = 0.47 (sec) , antiderivative size = 376, normalized size of antiderivative = 5.15 \[ \int \frac {a+b x+c x^2}{(d+e x)^{7/2}} \, dx=\begin {cases} - \frac {6 a e^{2}}{15 d^{2} e^{3} \sqrt {d + e x} + 30 d e^{4} x \sqrt {d + e x} + 15 e^{5} x^{2} \sqrt {d + e x}} - \frac {4 b d e}{15 d^{2} e^{3} \sqrt {d + e x} + 30 d e^{4} x \sqrt {d + e x} + 15 e^{5} x^{2} \sqrt {d + e x}} - \frac {10 b e^{2} x}{15 d^{2} e^{3} \sqrt {d + e x} + 30 d e^{4} x \sqrt {d + e x} + 15 e^{5} x^{2} \sqrt {d + e x}} - \frac {16 c d^{2}}{15 d^{2} e^{3} \sqrt {d + e x} + 30 d e^{4} x \sqrt {d + e x} + 15 e^{5} x^{2} \sqrt {d + e x}} - \frac {40 c d e x}{15 d^{2} e^{3} \sqrt {d + e x} + 30 d e^{4} x \sqrt {d + e x} + 15 e^{5} x^{2} \sqrt {d + e x}} - \frac {30 c e^{2} x^{2}}{15 d^{2} e^{3} \sqrt {d + e x} + 30 d e^{4} x \sqrt {d + e x} + 15 e^{5} x^{2} \sqrt {d + e x}} & \text {for}\: e \neq 0 \\\frac {a x + \frac {b x^{2}}{2} + \frac {c x^{3}}{3}}{d^{\frac {7}{2}}} & \text {otherwise} \end {cases} \]

[In]

integrate((c*x**2+b*x+a)/(e*x+d)**(7/2),x)

[Out]

Piecewise((-6*a*e**2/(15*d**2*e**3*sqrt(d + e*x) + 30*d*e**4*x*sqrt(d + e*x) + 15*e**5*x**2*sqrt(d + e*x)) - 4
*b*d*e/(15*d**2*e**3*sqrt(d + e*x) + 30*d*e**4*x*sqrt(d + e*x) + 15*e**5*x**2*sqrt(d + e*x)) - 10*b*e**2*x/(15
*d**2*e**3*sqrt(d + e*x) + 30*d*e**4*x*sqrt(d + e*x) + 15*e**5*x**2*sqrt(d + e*x)) - 16*c*d**2/(15*d**2*e**3*s
qrt(d + e*x) + 30*d*e**4*x*sqrt(d + e*x) + 15*e**5*x**2*sqrt(d + e*x)) - 40*c*d*e*x/(15*d**2*e**3*sqrt(d + e*x
) + 30*d*e**4*x*sqrt(d + e*x) + 15*e**5*x**2*sqrt(d + e*x)) - 30*c*e**2*x**2/(15*d**2*e**3*sqrt(d + e*x) + 30*
d*e**4*x*sqrt(d + e*x) + 15*e**5*x**2*sqrt(d + e*x)), Ne(e, 0)), ((a*x + b*x**2/2 + c*x**3/3)/d**(7/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.77 \[ \int \frac {a+b x+c x^2}{(d+e x)^{7/2}} \, dx=-\frac {2 \, {\left (15 \, {\left (e x + d\right )}^{2} c + 3 \, c d^{2} - 3 \, b d e + 3 \, a e^{2} - 5 \, {\left (2 \, c d - b e\right )} {\left (e x + d\right )}\right )}}{15 \, {\left (e x + d\right )}^{\frac {5}{2}} e^{3}} \]

[In]

integrate((c*x^2+b*x+a)/(e*x+d)^(7/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.79 \[ \int \frac {a+b x+c x^2}{(d+e x)^{7/2}} \, dx=-\frac {2 \, {\left (15 \, {\left (e x + d\right )}^{2} c - 10 \, {\left (e x + d\right )} c d + 3 \, c d^{2} + 5 \, {\left (e x + d\right )} b e - 3 \, b d e + 3 \, a e^{2}\right )}}{15 \, {\left (e x + d\right )}^{\frac {5}{2}} e^{3}} \]

[In]

integrate((c*x^2+b*x+a)/(e*x+d)^(7/2),x, algorithm="giac")

[Out]

-2/15*(15*(e*x + d)^2*c - 10*(e*x + d)*c*d + 3*c*d^2 + 5*(e*x + d)*b*e - 3*b*d*e + 3*a*e^2)/((e*x + d)^(5/2)*e
^3)

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.71 \[ \int \frac {a+b x+c x^2}{(d+e x)^{7/2}} \, dx=-\frac {16\,c\,d^2+40\,c\,d\,e\,x+4\,b\,d\,e+30\,c\,e^2\,x^2+10\,b\,e^2\,x+6\,a\,e^2}{15\,e^3\,{\left (d+e\,x\right )}^{5/2}} \]

[In]

int((a + b*x + c*x^2)/(d + e*x)^(7/2),x)

[Out]

-(6*a*e^2 + 16*c*d^2 + 30*c*e^2*x^2 + 4*b*d*e + 10*b*e^2*x + 40*c*d*e*x)/(15*e^3*(d + e*x)^(5/2))