Integrand size = 21, antiderivative size = 72 \[ \int \cos ^4(c+d x) (b \sec (c+d x))^{3/2} \, dx=\frac {6 b^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 b^3 \sin (c+d x)}{5 d (b \sec (c+d x))^{3/2}} \] Output:
6/5*b^2*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d/cos(d*x+c)^(1/2)/(b*sec(d* x+c))^(1/2)+2/5*b^3*sin(d*x+c)/d/(b*sec(d*x+c))^(3/2)
Time = 0.11 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.81 \[ \int \cos ^4(c+d x) (b \sec (c+d x))^{3/2} \, dx=\frac {b \sqrt {b \sec (c+d x)} \left (12 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\sin (c+d x)+\sin (3 (c+d x))\right )}{10 d} \] Input:
Integrate[Cos[c + d*x]^4*(b*Sec[c + d*x])^(3/2),x]
Output:
(b*Sqrt[b*Sec[c + d*x]]*(12*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + Sin[c + d*x] + Sin[3*(c + d*x)]))/(10*d)
Time = 0.38 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 2030, 4256, 3042, 4258, 3042, 3119}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \cos ^4(c+d x) (b \sec (c+d x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle b^4 \int \frac {1}{\left (b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4256 |
\(\displaystyle b^4 \left (\frac {3 \int \frac {1}{\sqrt {b \sec (c+d x)}}dx}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \sec (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^4 \left (\frac {3 \int \frac {1}{\sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \sec (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle b^4 \left (\frac {3 \int \sqrt {\cos (c+d x)}dx}{5 b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 b d (b \sec (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^4 \left (\frac {3 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 b d (b \sec (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle b^4 \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 b d (b \sec (c+d x))^{3/2}}\right )\) |
Input:
Int[Cos[c + d*x]^4*(b*Sec[c + d*x])^(3/2),x]
Output:
b^4*((6*EllipticE[(c + d*x)/2, 2])/(5*b^2*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[ c + d*x]]) + (2*Sin[c + d*x])/(5*b*d*(b*Sec[c + d*x])^(3/2)))
Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m Int[(b*v) ^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Csc[c + d*x])^(n + 1)/(b*d*n)), x] + Simp[(n + 1)/(b^2*n) Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2* n]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^n*Sin[c + d*x]^n Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
Result contains complex when optimal does not.
Time = 2.74 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.88
method | result | size |
default | \(\frac {2 \left (\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (\cos \left (d x +c \right )^{2}+\cos \left (d x +c \right )+3\right )+i \left (-3 \cos \left (d x +c \right )^{2}-6 \cos \left (d x +c \right )-3\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+i \left (3 \cos \left (d x +c \right )^{2}+6 \cos \left (d x +c \right )+3\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right )\right ) b \sqrt {b \sec \left (d x +c \right )}}{5 d \left (\cos \left (d x +c \right )+1\right )}\) | \(207\) |
Input:
int(cos(d*x+c)^4*(b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
Output:
2/5/d*(sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)^2+cos(d*x+c)+3)+I*(-3*cos(d*x+c)^ 2-6*cos(d*x+c)-3)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(cot(d*x+c )-csc(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)+I*(3*cos(d*x+c)^2+6*cos(d*x+c)+3 )*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*( cot(d*x+c)-csc(d*x+c)),I))*b*(b*sec(d*x+c))^(1/2)/(cos(d*x+c)+1)
Result contains complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.29 \[ \int \cos ^4(c+d x) (b \sec (c+d x))^{3/2} \, dx=\frac {2 \, b \sqrt {\frac {b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + 3 i \, \sqrt {2} b^{\frac {3}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} b^{\frac {3}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{5 \, d} \] Input:
integrate(cos(d*x+c)^4*(b*sec(d*x+c))^(3/2),x, algorithm="fricas")
Output:
1/5*(2*b*sqrt(b/cos(d*x + c))*cos(d*x + c)^2*sin(d*x + c) + 3*I*sqrt(2)*b^ (3/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*s in(d*x + c))) - 3*I*sqrt(2)*b^(3/2)*weierstrassZeta(-4, 0, weierstrassPInv erse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/d
Timed out. \[ \int \cos ^4(c+d x) (b \sec (c+d x))^{3/2} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**4*(b*sec(d*x+c))**(3/2),x)
Output:
Timed out
\[ \int \cos ^4(c+d x) (b \sec (c+d x))^{3/2} \, dx=\int { \left (b \sec \left (d x + c\right )\right )^{\frac {3}{2}} \cos \left (d x + c\right )^{4} \,d x } \] Input:
integrate(cos(d*x+c)^4*(b*sec(d*x+c))^(3/2),x, algorithm="maxima")
Output:
integrate((b*sec(d*x + c))^(3/2)*cos(d*x + c)^4, x)
\[ \int \cos ^4(c+d x) (b \sec (c+d x))^{3/2} \, dx=\int { \left (b \sec \left (d x + c\right )\right )^{\frac {3}{2}} \cos \left (d x + c\right )^{4} \,d x } \] Input:
integrate(cos(d*x+c)^4*(b*sec(d*x+c))^(3/2),x, algorithm="giac")
Output:
integrate((b*sec(d*x + c))^(3/2)*cos(d*x + c)^4, x)
Timed out. \[ \int \cos ^4(c+d x) (b \sec (c+d x))^{3/2} \, dx=\int {\cos \left (c+d\,x\right )}^4\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \] Input:
int(cos(c + d*x)^4*(b/cos(c + d*x))^(3/2),x)
Output:
int(cos(c + d*x)^4*(b/cos(c + d*x))^(3/2), x)
\[ \int \cos ^4(c+d x) (b \sec (c+d x))^{3/2} \, dx=\sqrt {b}\, \left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )d x \right ) b \] Input:
int(cos(d*x+c)^4*(b*sec(d*x+c))^(3/2),x)
Output:
sqrt(b)*int(sqrt(sec(c + d*x))*cos(c + d*x)**4*sec(c + d*x),x)*b