\(\int \frac {\sec ^3(x)}{\sqrt {\sin (2 x)}} \, dx\) [409]
Optimal result
Integrand size = 13, antiderivative size = 31 \[
\int \frac {\sec ^3(x)}{\sqrt {\sin (2 x)}} \, dx=\frac {4}{5} \sec (x) \sqrt {\sin (2 x)}+\frac {1}{5} \sec ^3(x) \sqrt {\sin (2 x)}
\]
[Out]
4/5*sec(x)*sin(2*x)^(1/2)+1/5*sec(x)^3*sin(2*x)^(1/2)
Rubi [A] (verified)
Time = 0.03 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number
of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {4384, 4376}
\[
\int \frac {\sec ^3(x)}{\sqrt {\sin (2 x)}} \, dx=\frac {1}{5} \sqrt {\sin (2 x)} \sec ^3(x)+\frac {4}{5} \sqrt {\sin (2 x)} \sec (x)
\]
[In]
Int[Sec[x]^3/Sqrt[Sin[2*x]],x]
[Out]
(4*Sec[x]*Sqrt[Sin[2*x]])/5 + (Sec[x]^3*Sqrt[Sin[2*x]])/5
Rule 4376
Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_.)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[(-(e*Cos[a +
b*x])^m)*((g*Sin[c + d*x])^(p + 1)/(b*g*m)), x] /; FreeQ[{a, b, c, d, e, g, m, p}, x] && EqQ[b*c - a*d, 0] &&
EqQ[d/b, 2] && !IntegerQ[p] && EqQ[m + 2*p + 2, 0]
Rule 4384
Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[(-(e*Cos[a +
b*x])^m)*((g*Sin[c + d*x])^(p + 1)/(2*b*g*(m + p + 1))), x] + Dist[(m + 2*p + 2)/(e^2*(m + p + 1)), Int[(e*Cos
[a + b*x])^(m + 2)*(g*Sin[c + d*x])^p, x], x] /; FreeQ[{a, b, c, d, e, g, p}, x] && EqQ[b*c - a*d, 0] && EqQ[d
/b, 2] && !IntegerQ[p] && LtQ[m, -1] && NeQ[m + 2*p + 2, 0] && NeQ[m + p + 1, 0] && IntegersQ[2*m, 2*p]
Rubi steps \begin{align*}
\text {integral}& = \frac {1}{5} \sec ^3(x) \sqrt {\sin (2 x)}+\frac {4}{5} \int \frac {\sec (x)}{\sqrt {\sin (2 x)}} \, dx \\ & = \frac {4}{5} \sec (x) \sqrt {\sin (2 x)}+\frac {1}{5} \sec ^3(x) \sqrt {\sin (2 x)} \\
\end{align*}
Mathematica [A] (verified)
Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.65
\[
\int \frac {\sec ^3(x)}{\sqrt {\sin (2 x)}} \, dx=\frac {1}{5} \sec (x) \left (4+\sec ^2(x)\right ) \sqrt {\sin (2 x)}
\]
[In]
Integrate[Sec[x]^3/Sqrt[Sin[2*x]],x]
[Out]
(Sec[x]*(4 + Sec[x]^2)*Sqrt[Sin[2*x]])/5
Maple [C] (verified)
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 4.94 (sec) , antiderivative size = 2946, normalized size of antiderivative =
95.03
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\(2946\) |
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[In]
int(1/cos(x)^3/sin(2*x)^(1/2),x,method=_RETURNVERBOSE)
[Out]
1/48*(-56*tan(1/2*x)-96*(1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticF((1+tan(1/2*
x))^(1/2),1/2*2^(1/2))*tan(1/2*x)^2+8*tan(1/2*x)^3-8*tan(1/2*x)^5+56*tan(1/2*x)^7-32*(1+tan(1/2*x))^(1/2)*(-2*
tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticF((1+tan(1/2*x))^(1/2),1/2*2^(1/2))+6*(1+tan(1/2*x))^(1/2)*(-2
*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticPi((1+tan(1/2*x))^(1/2),1/2-1/2*I,1/2*2^(1/2))+6*(1+tan(1/2*x
))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticPi((1+tan(1/2*x))^(1/2),1/2+1/2*I,1/2*2^(1/2))+12
*ln(1/tan(1/2*x)*(tan(1/2*x)^2+2*(tan(1/2*x)^3-tan(1/2*x))^(1/2)+2*tan(1/2*x)-1))*(tan(1/2*x)^3-tan(1/2*x))^(3
/2)-24*arctan(((tan(1/2*x)^3-tan(1/2*x))^(1/2)+tan(1/2*x))/tan(1/2*x))*(tan(1/2*x)^3-tan(1/2*x))^(3/2)-12*ln(-
1/tan(1/2*x)*(-tan(1/2*x)^2+2*(tan(1/2*x)^3-tan(1/2*x))^(1/2)-2*tan(1/2*x)+1))*(tan(1/2*x)^3-tan(1/2*x))^(3/2)
-24*arctan(((tan(1/2*x)^3-tan(1/2*x))^(1/2)-tan(1/2*x))/tan(1/2*x))*(tan(1/2*x)^3-tan(1/2*x))^(3/2)-3*ln(1/tan
(1/2*x)*(tan(1/2*x)^2+2*(tan(1/2*x)^3-tan(1/2*x))^(1/2)+2*tan(1/2*x)-1))*(tan(1/2*x)^3-tan(1/2*x))^(1/2)+6*arc
tan(((tan(1/2*x)^3-tan(1/2*x))^(1/2)+tan(1/2*x))/tan(1/2*x))*(tan(1/2*x)^3-tan(1/2*x))^(1/2)+3*ln(-1/tan(1/2*x
)*(-tan(1/2*x)^2+2*(tan(1/2*x)^3-tan(1/2*x))^(1/2)-2*tan(1/2*x)+1))*(tan(1/2*x)^3-tan(1/2*x))^(1/2)+6*arctan((
(tan(1/2*x)^3-tan(1/2*x))^(1/2)-tan(1/2*x))/tan(1/2*x))*(tan(1/2*x)^3-tan(1/2*x))^(1/2)+6*I*(1+tan(1/2*x))^(1/
2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticPi((1+tan(1/2*x))^(1/2),1/2-1/2*I,1/2*2^(1/2))-6*I*(1+t
an(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticPi((1+tan(1/2*x))^(1/2),1/2+1/2*I,1/2*2^(
1/2))+6*(1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticPi((1+tan(1/2*x))^(1/2),1/2-1
/2*I,1/2*2^(1/2))*tan(1/2*x)^6+6*(1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticPi((
1+tan(1/2*x))^(1/2),1/2+1/2*I,1/2*2^(1/2))*tan(1/2*x)^6+18*(1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(
1/2*x))^(1/2)*EllipticPi((1+tan(1/2*x))^(1/2),1/2-1/2*I,1/2*2^(1/2))*tan(1/2*x)^4+18*(1+tan(1/2*x))^(1/2)*(-2*
tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticPi((1+tan(1/2*x))^(1/2),1/2+1/2*I,1/2*2^(1/2))*tan(1/2*x)^4+18
*(1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticPi((1+tan(1/2*x))^(1/2),1/2-1/2*I,1/
2*2^(1/2))*tan(1/2*x)^2+18*(1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticPi((1+tan(
1/2*x))^(1/2),1/2+1/2*I,1/2*2^(1/2))*tan(1/2*x)^2-18*I*(1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*
x))^(1/2)*EllipticPi((1+tan(1/2*x))^(1/2),1/2+1/2*I,1/2*2^(1/2))*tan(1/2*x)^2+6*I*(1+tan(1/2*x))^(1/2)*(-2*tan
(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticPi((1+tan(1/2*x))^(1/2),1/2-1/2*I,1/2*2^(1/2))*tan(1/2*x)^6-6*I*(
1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticPi((1+tan(1/2*x))^(1/2),1/2+1/2*I,1/2*
2^(1/2))*tan(1/2*x)^6+18*I*(1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticPi((1+tan(
1/2*x))^(1/2),1/2-1/2*I,1/2*2^(1/2))*tan(1/2*x)^4-18*I*(1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*
x))^(1/2)*EllipticPi((1+tan(1/2*x))^(1/2),1/2+1/2*I,1/2*2^(1/2))*tan(1/2*x)^4+18*I*(1+tan(1/2*x))^(1/2)*(-2*ta
n(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticPi((1+tan(1/2*x))^(1/2),1/2-1/2*I,1/2*2^(1/2))*tan(1/2*x)^2-32*(
1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticF((1+tan(1/2*x))^(1/2),1/2*2^(1/2))*ta
n(1/2*x)^6-96*(1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticF((1+tan(1/2*x))^(1/2),
1/2*2^(1/2))*tan(1/2*x)^4-12*ln(1/tan(1/2*x)*(tan(1/2*x)^2+2*(tan(1/2*x)^3-tan(1/2*x))^(1/2)+2*tan(1/2*x)-1))*
(tan(1/2*x)^3-tan(1/2*x))^(1/2)*tan(1/2*x)^5+24*arctan(((tan(1/2*x)^3-tan(1/2*x))^(1/2)+tan(1/2*x))/tan(1/2*x)
)*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*tan(1/2*x)^5+12*ln(-1/tan(1/2*x)*(-tan(1/2*x)^2+2*(tan(1/2*x)^3-tan(1/2*x))^
(1/2)-2*tan(1/2*x)+1))*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*tan(1/2*x)^5+24*arctan(((tan(1/2*x)^3-tan(1/2*x))^(1/2)
-tan(1/2*x))/tan(1/2*x))*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*tan(1/2*x)^5-3*ln(1/tan(1/2*x)*(tan(1/2*x)^2+2*(tan(1
/2*x)^3-tan(1/2*x))^(1/2)+2*tan(1/2*x)-1))*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*tan(1/2*x)^6+6*arctan(((tan(1/2*x)^
3-tan(1/2*x))^(1/2)+tan(1/2*x))/tan(1/2*x))*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*tan(1/2*x)^6+3*ln(-1/tan(1/2*x)*(-
tan(1/2*x)^2+2*(tan(1/2*x)^3-tan(1/2*x))^(1/2)-2*tan(1/2*x)+1))*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*tan(1/2*x)^6+6
*arctan(((tan(1/2*x)^3-tan(1/2*x))^(1/2)-tan(1/2*x))/tan(1/2*x))*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*tan(1/2*x)^6+
12*ln(1/tan(1/2*x)*(tan(1/2*x)^2+2*(tan(1/2*x)^3-tan(1/2*x))^(1/2)+2*tan(1/2*x)-1))*(tan(1/2*x)^3-tan(1/2*x))^
(3/2)*tan(1/2*x)^2-9*ln(1/tan(1/2*x)*(tan(1/2*x)^2+2*(tan(1/2*x)^3-tan(1/2*x))^(1/2)+2*tan(1/2*x)-1))*(tan(1/2
*x)^3-tan(1/2*x))^(1/2)*tan(1/2*x)^4-24*arctan(((tan(1/2*x)^3-tan(1/2*x))^(1/2)+tan(1/2*x))/tan(1/2*x))*(tan(1
/2*x)^3-tan(1/2*x))^(3/2)*tan(1/2*x)^2+18*arctan(((tan(1/2*x)^3-tan(1/2*x))^(1/2)+tan(1/2*x))/tan(1/2*x))*(tan
(1/2*x)^3-tan(1/2*x))^(1/2)*tan(1/2*x)^4-12*ln(-1/tan(1/2*x)*(-tan(1/2*x)^2+2*(tan(1/2*x)^3-tan(1/2*x))^(1/2)-
2*tan(1/2*x)+1))*(tan(1/2*x)^3-tan(1/2*x))^(3/2)*tan(1/2*x)^2+9*ln(-1/tan(1/2*x)*(-tan(1/2*x)^2+2*(tan(1/2*x)^
3-tan(1/2*x))^(1/2)-2*tan(1/2*x)+1))*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*tan(1/2*x)^4-24*arctan(((tan(1/2*x)^3-tan
(1/2*x))^(1/2)-tan(1/2*x))/tan(1/2*x))*(tan(1/2*x)^3-tan(1/2*x))^(3/2)*tan(1/2*x)^2+18*arctan(((tan(1/2*x)^3-t
an(1/2*x))^(1/2)-tan(1/2*x))/tan(1/2*x))*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*tan(1/2*x)^4-9*ln(1/tan(1/2*x)*(tan(1
/2*x)^2+2*(tan(1/2*x)^3-tan(1/2*x))^(1/2)+2*tan(1/2*x)-1))*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*tan(1/2*x)^2+18*arc
tan(((tan(1/2*x)^3-tan(1/2*x))^(1/2)+tan(1/2*x))/tan(1/2*x))*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*tan(1/2*x)^2+9*ln
(-1/tan(1/2*x)*(-tan(1/2*x)^2+2*(tan(1/2*x)^3-tan(1/2*x))^(1/2)-2*tan(1/2*x)+1))*(tan(1/2*x)^3-tan(1/2*x))^(1/
2)*tan(1/2*x)^2+18*arctan(((tan(1/2*x)^3-tan(1/2*x))^(1/2)-tan(1/2*x))/tan(1/2*x))*(tan(1/2*x)^3-tan(1/2*x))^(
1/2)*tan(1/2*x)^2+12*ln(1/tan(1/2*x)*(tan(1/2*x)^2+2*(tan(1/2*x)^3-tan(1/2*x))^(1/2)+2*tan(1/2*x)-1))*(tan(1/2
*x)^3-tan(1/2*x))^(1/2)*tan(1/2*x)-24*arctan(((tan(1/2*x)^3-tan(1/2*x))^(1/2)+tan(1/2*x))/tan(1/2*x))*(tan(1/2
*x)^3-tan(1/2*x))^(1/2)*tan(1/2*x)-12*ln(-1/tan(1/2*x)*(-tan(1/2*x)^2+2*(tan(1/2*x)^3-tan(1/2*x))^(1/2)-2*tan(
1/2*x)+1))*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*tan(1/2*x)-24*arctan(((tan(1/2*x)^3-tan(1/2*x))^(1/2)-tan(1/2*x))/t
an(1/2*x))*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*tan(1/2*x))*(tan(1/2*x)^2-1)*(-tan(1/2*x)/(tan(1/2*x)^2-1))^(1/2)/(
-tan(1/2*x)^2+2*(tan(1/2*x)^3-tan(1/2*x))^(1/2)-2*tan(1/2*x)+1)/(tan(1/2*x)^2+2*(tan(1/2*x)^3-tan(1/2*x))^(1/2
)+2*tan(1/2*x)-1)/(1+tan(1/2*x)^2)/(tan(1/2*x)^3-tan(1/2*x))^(1/2)/(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)
Fricas [A] (verification not implemented)
none
Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03
\[
\int \frac {\sec ^3(x)}{\sqrt {\sin (2 x)}} \, dx=\frac {4 \, \cos \left (x\right )^{3} + \sqrt {2} {\left (4 \, \cos \left (x\right )^{2} + 1\right )} \sqrt {\cos \left (x\right ) \sin \left (x\right )}}{5 \, \cos \left (x\right )^{3}}
\]
[In]
integrate(1/cos(x)^3/sin(2*x)^(1/2),x, algorithm="fricas")
[Out]
1/5*(4*cos(x)^3 + sqrt(2)*(4*cos(x)^2 + 1)*sqrt(cos(x)*sin(x)))/cos(x)^3
Sympy [F(-1)]
Timed out. \[
\int \frac {\sec ^3(x)}{\sqrt {\sin (2 x)}} \, dx=\text {Timed out}
\]
[In]
integrate(1/cos(x)**3/sin(2*x)**(1/2),x)
[Out]
Timed out
Maxima [F]
\[
\int \frac {\sec ^3(x)}{\sqrt {\sin (2 x)}} \, dx=\int { \frac {1}{\cos \left (x\right )^{3} \sqrt {\sin \left (2 \, x\right )}} \,d x }
\]
[In]
integrate(1/cos(x)^3/sin(2*x)^(1/2),x, algorithm="maxima")
[Out]
integrate(1/(cos(x)^3*sqrt(sin(2*x))), x)
Giac [F]
\[
\int \frac {\sec ^3(x)}{\sqrt {\sin (2 x)}} \, dx=\int { \frac {1}{\cos \left (x\right )^{3} \sqrt {\sin \left (2 \, x\right )}} \,d x }
\]
[In]
integrate(1/cos(x)^3/sin(2*x)^(1/2),x, algorithm="giac")
[Out]
integrate(1/(cos(x)^3*sqrt(sin(2*x))), x)
Mupad [B] (verification not implemented)
Time = 0.43 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.65
\[
\int \frac {\sec ^3(x)}{\sqrt {\sin (2 x)}} \, dx=\frac {\sqrt {\sin \left (2\,x\right )}\,\left (2\,\cos \left (2\,x\right )+3\right )}{5\,{\cos \left (x\right )}^3}
\]
[In]
int(1/(sin(2*x)^(1/2)*cos(x)^3),x)
[Out]
(sin(2*x)^(1/2)*(2*cos(2*x) + 3))/(5*cos(x)^3)