\(\int \frac {1}{(a+c x^2)^{7/2}} \, dx\) [62]

Optimal result

Integrand size = 11, antiderivative size = 58 \[ \int \frac {1}{\left (a+c x^2\right )^{7/2}} \, dx=\frac {x}{5 a \left (a+c x^2\right )^{5/2}}+\frac {4 x}{15 a^2 \left (a+c x^2\right )^{3/2}}+\frac {8 x}{15 a^3 \sqrt {a+c x^2}} \]

[Out]

1/5*x/a/(c*x^2+a)^(5/2)+4/15*x/a^2/(c*x^2+a)^(3/2)+8/15*x/a^3/(c*x^2+a)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {198, 197} \[ \int \frac {1}{\left (a+c x^2\right )^{7/2}} \, dx=\frac {8 x}{15 a^3 \sqrt {a+c x^2}}+\frac {4 x}{15 a^2 \left (a+c x^2\right )^{3/2}}+\frac {x}{5 a \left (a+c x^2\right )^{5/2}} \]

[In]

Int[(a + c*x^2)^(-7/2),x]

[Out]

x/(5*a*(a + c*x^2)^(5/2)) + (4*x)/(15*a^2*(a + c*x^2)^(3/2)) + (8*x)/(15*a^3*Sqrt[a + c*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}{5 a \left (a+c x^2\right )^{5/2}}+\frac {4 \int \frac {1}{\left (a+c x^2\right )^{5/2}} \, dx}{5 a} \\ & = \frac {x}{5 a \left (a+c x^2\right )^{5/2}}+\frac {4 x}{15 a^2 \left (a+c x^2\right )^{3/2}}+\frac {8 \int \frac {1}{\left (a+c x^2\right )^{3/2}} \, dx}{15 a^2} \\ & = \frac {x}{5 a \left (a+c x^2\right )^{5/2}}+\frac {4 x}{15 a^2 \left (a+c x^2\right )^{3/2}}+\frac {8 x}{15 a^3 \sqrt {a+c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.69 \[ \int \frac {1}{\left (a+c x^2\right )^{7/2}} \, dx=\frac {15 a^2 x+20 a c x^3+8 c^2 x^5}{15 a^3 \left (a+c x^2\right )^{5/2}} \]

[In]

Integrate[(a + c*x^2)^(-7/2),x]

[Out]

(15*a^2*x + 20*a*c*x^3 + 8*c^2*x^5)/(15*a^3*(a + c*x^2)^(5/2))

Maple [A] (verified)

Time = 1.98 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.64

method	result	size
gosper	\(\frac {x \left (8 x^{4} c^{2}+20 a c \,x^{2}+15 a^{2}\right )}{15 \left (c \,x^{2}+a \right )^{\frac {5}{2}} a^{3}}\)	\(37\)
trager	\(\frac {x \left (8 x^{4} c^{2}+20 a c \,x^{2}+15 a^{2}\right )}{15 \left (c \,x^{2}+a \right )^{\frac {5}{2}} a^{3}}\)	\(37\)
pseudoelliptic	\(\frac {x \left (8 x^{4} c^{2}+20 a c \,x^{2}+15 a^{2}\right )}{15 \left (c \,x^{2}+a \right )^{\frac {5}{2}} a^{3}}\)	\(37\)
default	\(\frac {x}{5 a \left (c \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 a \left (c \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {c \,x^{2}+a}}}{a}\)	\(53\)

[In]

int(1/(c*x^2+a)^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/15*x*(8*c^2*x^4+20*a*c*x^2+15*a^2)/(c*x^2+a)^(5/2)/a^3

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.19 \[ \int \frac {1}{\left (a+c x^2\right )^{7/2}} \, dx=\frac {{\left (8 \, c^{2} x^{5} + 20 \, a c x^{3} + 15 \, a^{2} x\right )} \sqrt {c x^{2} + a}}{15 \, {\left (a^{3} c^{3} x^{6} + 3 \, a^{4} c^{2} x^{4} + 3 \, a^{5} c x^{2} + a^{6}\right )}} \]

[In]

integrate(1/(c*x^2+a)^(7/2),x, algorithm="fricas")

[Out]

1/15*(8*c^2*x^5 + 20*a*c*x^3 + 15*a^2*x)*sqrt(c*x^2 + a)/(a^3*c^3*x^6 + 3*a^4*c^2*x^4 + 3*a^5*c*x^2 + a^6)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 413 vs. \(2 (51) = 102\).

Time = 0.80 (sec) , antiderivative size = 413, normalized size of antiderivative = 7.12 \[ \int \frac {1}{\left (a+c x^2\right )^{7/2}} \, dx=\frac {15 a^{5} x}{15 a^{\frac {17}{2}} \sqrt {1 + \frac {c x^{2}}{a}} + 45 a^{\frac {15}{2}} c x^{2} \sqrt {1 + \frac {c x^{2}}{a}} + 45 a^{\frac {13}{2}} c^{2} x^{4} \sqrt {1 + \frac {c x^{2}}{a}} + 15 a^{\frac {11}{2}} c^{3} x^{6} \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {35 a^{4} c x^{3}}{15 a^{\frac {17}{2}} \sqrt {1 + \frac {c x^{2}}{a}} + 45 a^{\frac {15}{2}} c x^{2} \sqrt {1 + \frac {c x^{2}}{a}} + 45 a^{\frac {13}{2}} c^{2} x^{4} \sqrt {1 + \frac {c x^{2}}{a}} + 15 a^{\frac {11}{2}} c^{3} x^{6} \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {28 a^{3} c^{2} x^{5}}{15 a^{\frac {17}{2}} \sqrt {1 + \frac {c x^{2}}{a}} + 45 a^{\frac {15}{2}} c x^{2} \sqrt {1 + \frac {c x^{2}}{a}} + 45 a^{\frac {13}{2}} c^{2} x^{4} \sqrt {1 + \frac {c x^{2}}{a}} + 15 a^{\frac {11}{2}} c^{3} x^{6} \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {8 a^{2} c^{3} x^{7}}{15 a^{\frac {17}{2}} \sqrt {1 + \frac {c x^{2}}{a}} + 45 a^{\frac {15}{2}} c x^{2} \sqrt {1 + \frac {c x^{2}}{a}} + 45 a^{\frac {13}{2}} c^{2} x^{4} \sqrt {1 + \frac {c x^{2}}{a}} + 15 a^{\frac {11}{2}} c^{3} x^{6} \sqrt {1 + \frac {c x^{2}}{a}}} \]

[In]

integrate(1/(c*x**2+a)**(7/2),x)

[Out]

15*a**5*x/(15*a**(17/2)*sqrt(1 + c*x**2/a) + 45*a**(15/2)*c*x**2*sqrt(1 + c*x**2/a) + 45*a**(13/2)*c**2*x**4*s
qrt(1 + c*x**2/a) + 15*a**(11/2)*c**3*x**6*sqrt(1 + c*x**2/a)) + 35*a**4*c*x**3/(15*a**(17/2)*sqrt(1 + c*x**2/
a) + 45*a**(15/2)*c*x**2*sqrt(1 + c*x**2/a) + 45*a**(13/2)*c**2*x**4*sqrt(1 + c*x**2/a) + 15*a**(11/2)*c**3*x*
*6*sqrt(1 + c*x**2/a)) + 28*a**3*c**2*x**5/(15*a**(17/2)*sqrt(1 + c*x**2/a) + 45*a**(15/2)*c*x**2*sqrt(1 + c*x
**2/a) + 45*a**(13/2)*c**2*x**4*sqrt(1 + c*x**2/a) + 15*a**(11/2)*c**3*x**6*sqrt(1 + c*x**2/a)) + 8*a**2*c**3*
x**7/(15*a**(17/2)*sqrt(1 + c*x**2/a) + 45*a**(15/2)*c*x**2*sqrt(1 + c*x**2/a) + 45*a**(13/2)*c**2*x**4*sqrt(1
+ c*x**2/a) + 15*a**(11/2)*c**3*x**6*sqrt(1 + c*x**2/a))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.79 \[ \int \frac {1}{\left (a+c x^2\right )^{7/2}} \, dx=\frac {8 \, x}{15 \, \sqrt {c x^{2} + a} a^{3}} + \frac {4 \, x}{15 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} a^{2}} + \frac {x}{5 \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} a} \]

[In]

integrate(1/(c*x^2+a)^(7/2),x, algorithm="maxima")

[Out]

8/15*x/(sqrt(c*x^2 + a)*a^3) + 4/15*x/((c*x^2 + a)^(3/2)*a^2) + 1/5*x/((c*x^2 + a)^(5/2)*a)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.71 \[ \int \frac {1}{\left (a+c x^2\right )^{7/2}} \, dx=\frac {{\left (4 \, x^{2} {\left (\frac {2 \, c^{2} x^{2}}{a^{3}} + \frac {5 \, c}{a^{2}}\right )} + \frac {15}{a}\right )} x}{15 \, {\left (c x^{2} + a\right )}^{\frac {5}{2}}} \]

[In]

integrate(1/(c*x^2+a)^(7/2),x, algorithm="giac")

[Out]

1/15*(4*x^2*(2*c^2*x^2/a^3 + 5*c/a^2) + 15/a)*x/(c*x^2 + a)^(5/2)

Mupad [B] (verification not implemented)

Time = 9.04 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.76 \[ \int \frac {1}{\left (a+c x^2\right )^{7/2}} \, dx=\frac {8\,x\,{\left (c\,x^2+a\right )}^2+3\,a^2\,x+4\,a\,x\,\left (c\,x^2+a\right )}{15\,a^3\,{\left (c\,x^2+a\right )}^{5/2}} \]

[In]

int(1/(a + c*x^2)^(7/2),x)

[Out]

(8*x*(a + c*x^2)^2 + 3*a^2*x + 4*a*x*(a + c*x^2))/(15*a^3*(a + c*x^2)^(5/2))