\(\int \frac {a+c x^2}{(d+e x)^{7/2}} \, dx\) [597]

Optimal result

Integrand size = 17, antiderivative size = 61 \[ \int \frac {a+c x^2}{(d+e x)^{7/2}} \, dx=-\frac {2 \left (c d^2+a e^2\right )}{5 e^3 (d+e x)^{5/2}}+\frac {4 c d}{3 e^3 (d+e x)^{3/2}}-\frac {2 c}{e^3 \sqrt {d+e x}} \]

[Out]

-2/5*(a*e^2+c*d^2)/e^3/(e*x+d)^(5/2)+4/3*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 = 61, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {711} \[ \int \frac {a+c x^2}{(d+e x)^{7/2}} \, dx=-\frac {2 \left (a e^2+c d^2\right )}{5 e^3 (d+e x)^{5/2}}-\frac {2 c}{e^3 \sqrt {d+e x}}+\frac {4 c d}{3 e^3 (d+e x)^{3/2}} \]

[In]

Int[(a + c*x^2)/(d + e*x)^(7/2),x]

[Out]

(-2*(c*d^2 + a*e^2))/(5*e^3*(d + e*x)^(5/2)) + (4*c*d)/(3*e^3*(d + e*x)^(3/2)) - (2*c)/(e^3*Sqrt[d + e*x])

Rule 711

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

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.72 \[ \int \frac {a+c x^2}{(d+e x)^{7/2}} \, dx=-\frac {2 \left (3 a e^2+c \left (8 d^2+20 d e x+15 e^2 x^2\right )\right )}{15 e^3 (d+e x)^{5/2}} \]

[In]

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

[Out]

(-2*(3*a*e^2 + 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 = 1.90 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.62

method	result	size
pseudoelliptic	\(-\frac {2 \left (\left (5 c \,x^{2}+a \right ) e^{2}+\frac {20 x c d e}{3}+\frac {8 c \,d^{2}}{3}\right )}{5 \left (e x +d \right )^{\frac {5}{2}} e^{3}}\)	\(38\)
gosper	\(-\frac {2 \left (15 c \,x^{2} e^{2}+20 x c d e +3 e^{2} a +8 c \,d^{2}\right )}{15 \left (e x +d \right )^{\frac {5}{2}} e^{3}}\)	\(41\)
trager	\(-\frac {2 \left (15 c \,x^{2} e^{2}+20 x c d e +3 e^{2} a +8 c \,d^{2}\right )}{15 \left (e x +d \right )^{\frac {5}{2}} e^{3}}\)	\(41\)
derivativedivides	\(\frac {-\frac {2 \left (e^{2} a +c \,d^{2}\right )}{5 \left (e x +d \right )^{\frac {5}{2}}}-\frac {2 c}{\sqrt {e x +d}}+\frac {4 c d}{3 \left (e x +d \right )^{\frac {3}{2}}}}{e^{3}}\)	\(48\)
default	\(\frac {-\frac {2 \left (e^{2} a +c \,d^{2}\right )}{5 \left (e x +d \right )^{\frac {5}{2}}}-\frac {2 c}{\sqrt {e x +d}}+\frac {4 c d}{3 \left (e x +d \right )^{\frac {3}{2}}}}{e^{3}}\)	\(48\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.49 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.18 \[ \int \frac {a+c x^2}{(d+e x)^{7/2}} \, dx=-\frac {2 \, {\left (15 \, c e^{2} x^{2} + 20 \, c d e x + 8 \, c d^{2} + 3 \, a e^{2}\right )} \sqrt {e x + d}}{15 \, {\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}} \]

[In]

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

[Out]

-2/15*(15*c*e^2*x^2 + 20*c*d*e*x + 8*c*d^2 + 3*a*e^2)*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. 252 vs. \(2 (63) = 126\).

Time = 0.48 (sec) , antiderivative size = 252, normalized size of antiderivative = 4.13 \[ \int \frac {a+c x^2}{(d+e x)^{7/2}} \, dx=\begin {cases} - \frac {6 a e^{2}}{15 d^{2} e^{3} \sqrt {d + e x} + 30 d e^{4} x \sqrt {d + e x} + 15 e^{5} x^{2} \sqrt {d + e x}} - \frac {16 c d^{2}}{15 d^{2} e^{3} \sqrt {d + e x} + 30 d e^{4} x \sqrt {d + e x} + 15 e^{5} x^{2} \sqrt {d + e x}} - \frac {40 c d e x}{15 d^{2} e^{3} \sqrt {d + e x} + 30 d e^{4} x \sqrt {d + e x} + 15 e^{5} x^{2} \sqrt {d + e x}} - \frac {30 c e^{2} x^{2}}{15 d^{2} e^{3} \sqrt {d + e x} + 30 d e^{4} x \sqrt {d + e x} + 15 e^{5} x^{2} \sqrt {d + e x}} & \text {for}\: e \neq 0 \\\frac {a x + \frac {c x^{3}}{3}}{d^{\frac {7}{2}}} & \text {otherwise} \end {cases} \]

[In]

integrate((c*x**2+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)) - 1
6*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)) - 40*c*d*e*x/(1
5*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 + c*x**3/3)/d**
(7/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.72 \[ \int \frac {a+c x^2}{(d+e x)^{7/2}} \, dx=-\frac {2 \, {\left (15 \, {\left (e x + d\right )}^{2} c - 10 \, {\left (e x + d\right )} c d + 3 \, c d^{2} + 3 \, a e^{2}\right )}}{15 \, {\left (e x + d\right )}^{\frac {5}{2}} e^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.72 \[ \int \frac {a+c x^2}{(d+e x)^{7/2}} \, dx=-\frac {2 \, {\left (15 \, {\left (e x + d\right )}^{2} c - 10 \, {\left (e x + d\right )} c d + 3 \, c d^{2} + 3 \, a e^{2}\right )}}{15 \, {\left (e x + d\right )}^{\frac {5}{2}} e^{3}} \]

[In]

integrate((c*x^2+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 + 3*a*e^2)/((e*x + d)^(5/2)*e^3)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.72 \[ \int \frac {a+c x^2}{(d+e x)^{7/2}} \, dx=-\frac {30\,c\,{\left (d+e\,x\right )}^2+6\,a\,e^2+6\,c\,d^2-20\,c\,d\,\left (d+e\,x\right )}{15\,e^3\,{\left (d+e\,x\right )}^{5/2}} \]

[In]

int((a + c*x^2)/(d + e*x)^(7/2),x)

[Out]

-(30*c*(d + e*x)^2 + 6*a*e^2 + 6*c*d^2 - 20*c*d*(d + e*x))/(15*e^3*(d + e*x)^(5/2))