\(\int \frac {1}{(c+d x^2)^{5/2}} \, dx\) [100]

Optimal result

Integrand size = 11, antiderivative size = 39 \[ \int \frac {1}{\left (c+d x^2\right )^{5/2}} \, dx=\frac {x}{3 c \left (c+d x^2\right )^{3/2}}+\frac {2 x}{3 c^2 \sqrt {c+d x^2}} \]

[Out]

1/3*x/c/(d*x^2+c)^(3/2)+2/3*x/c^2/(d*x^2+c)^(1/2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {198, 197} \[ \int \frac {1}{\left (c+d x^2\right )^{5/2}} \, dx=\frac {2 x}{3 c^2 \sqrt {c+d x^2}}+\frac {x}{3 c \left (c+d x^2\right )^{3/2}} \]

[In]

Int[(c + d*x^2)^(-5/2),x]

[Out]

x/(3*c*(c + d*x^2)^(3/2)) + (2*x)/(3*c^2*Sqrt[c + d*x^2])

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 198

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p +
1], 0] && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {x}{3 c \left (c+d x^2\right )^{3/2}}+\frac {2 \int \frac {1}{\left (c+d x^2\right )^{3/2}} \, dx}{3 c} \\ & = \frac {x}{3 c \left (c+d x^2\right )^{3/2}}+\frac {2 x}{3 c^2 \sqrt {c+d x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.74 \[ \int \frac {1}{\left (c+d x^2\right )^{5/2}} \, dx=\frac {3 c x+2 d x^3}{3 c^2 \left (c+d x^2\right )^{3/2}} \]

[In]

Integrate[(c + d*x^2)^(-5/2),x]

[Out]

(3*c*x + 2*d*x^3)/(3*c^2*(c + d*x^2)^(3/2))

Maple [A] (verified)

Time = 2.31 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.67

method	result	size
gosper	\(\frac {x \left (2 d \,x^{2}+3 c \right )}{3 \left (d \,x^{2}+c \right )^{\frac {3}{2}} c^{2}}\)	\(26\)
trager	\(\frac {x \left (2 d \,x^{2}+3 c \right )}{3 \left (d \,x^{2}+c \right )^{\frac {3}{2}} c^{2}}\)	\(26\)
pseudoelliptic	\(\frac {x \left (2 d \,x^{2}+3 c \right )}{3 \left (d \,x^{2}+c \right )^{\frac {3}{2}} c^{2}}\)	\(26\)
default	\(\frac {x}{3 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{2} \sqrt {d \,x^{2}+c}}\)	\(32\)

[In]

int(1/(d*x^2+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/3*x*(2*d*x^2+3*c)/(d*x^2+c)^(3/2)/c^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.21 \[ \int \frac {1}{\left (c+d x^2\right )^{5/2}} \, dx=\frac {{\left (2 \, d x^{3} + 3 \, c x\right )} \sqrt {d x^{2} + c}}{3 \, {\left (c^{2} d^{2} x^{4} + 2 \, c^{3} d x^{2} + c^{4}\right )}} \]

[In]

integrate(1/(d*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

1/3*(2*d*x^3 + 3*c*x)*sqrt(d*x^2 + c)/(c^2*d^2*x^4 + 2*c^3*d*x^2 + c^4)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (32) = 64\).

Time = 0.52 (sec) , antiderivative size = 95, normalized size of antiderivative = 2.44 \[ \int \frac {1}{\left (c+d x^2\right )^{5/2}} \, dx=\frac {3 c x}{3 c^{\frac {7}{2}} \sqrt {1 + \frac {d x^{2}}{c}} + 3 c^{\frac {5}{2}} d x^{2} \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {2 d x^{3}}{3 c^{\frac {7}{2}} \sqrt {1 + \frac {d x^{2}}{c}} + 3 c^{\frac {5}{2}} d x^{2} \sqrt {1 + \frac {d x^{2}}{c}}} \]

[In]

integrate(1/(d*x**2+c)**(5/2),x)

[Out]

3*c*x/(3*c**(7/2)*sqrt(1 + d*x**2/c) + 3*c**(5/2)*d*x**2*sqrt(1 + d*x**2/c)) + 2*d*x**3/(3*c**(7/2)*sqrt(1 + d
*x**2/c) + 3*c**(5/2)*d*x**2*sqrt(1 + d*x**2/c))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.79 \[ \int \frac {1}{\left (c+d x^2\right )^{5/2}} \, dx=\frac {2 \, x}{3 \, \sqrt {d x^{2} + c} c^{2}} + \frac {x}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c} \]

[In]

integrate(1/(d*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

2/3*x/(sqrt(d*x^2 + c)*c^2) + 1/3*x/((d*x^2 + c)^(3/2)*c)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.69 \[ \int \frac {1}{\left (c+d x^2\right )^{5/2}} \, dx=\frac {x {\left (\frac {2 \, d x^{2}}{c^{2}} + \frac {3}{c}\right )}}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}}} \]

[In]

integrate(1/(d*x^2+c)^(5/2),x, algorithm="giac")

[Out]

1/3*x*(2*d*x^2/c^2 + 3/c)/(d*x^2 + c)^(3/2)

Mupad [B] (verification not implemented)

Time = 4.55 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.72 \[ \int \frac {1}{\left (c+d x^2\right )^{5/2}} \, dx=\frac {2\,x\,\left (d\,x^2+c\right )+c\,x}{3\,c^2\,{\left (d\,x^2+c\right )}^{3/2}} \]

[In]

int(1/(c + d*x^2)^(5/2),x)

[Out]

(2*x*(c + d*x^2) + c*x)/(3*c^2*(c + d*x^2)^(3/2))