Integrand size = 26, antiderivative size = 223 \[ \int \frac {\cos ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {63 i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{128 \sqrt {2} \sqrt {a} d}+\frac {63 i a^2}{160 d (a+i a \tan (c+d x))^{5/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}+\frac {21 i a}{64 d (a+i a \tan (c+d x))^{3/2}}+\frac {63 i}{128 d \sqrt {a+i a \tan (c+d x)}}-\frac {9 i a^5}{16 d (a+i a \tan (c+d x))^{5/2} \left (a^3-i a^3 \tan (c+d x)\right )} \] Output:
-63/256*I*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/a^ (1/2)/d+63/160*I*a^2/d/(a+I*a*tan(d*x+c))^(5/2)-1/4*I*a^4/d/(a-I*a*tan(d*x +c))^2/(a+I*a*tan(d*x+c))^(5/2)+21/64*I*a/d/(a+I*a*tan(d*x+c))^(3/2)+63/12 8*I/d/(a+I*a*tan(d*x+c))^(1/2)-9/16*I*a^5/d/(a+I*a*tan(d*x+c))^(5/2)/(a^3- I*a^3*tan(d*x+c))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.19 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.24 \[ \int \frac {\cos ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {i a^2 \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},3,-\frac {3}{2},\frac {1}{2} (1+i \tan (c+d x))\right )}{20 d (a+i a \tan (c+d x))^{5/2}} \] Input:
Integrate[Cos[c + d*x]^4/Sqrt[a + I*a*Tan[c + d*x]],x]
Output:
((I/20)*a^2*Hypergeometric2F1[-5/2, 3, -3/2, (1 + I*Tan[c + d*x])/2])/(d*( a + I*a*Tan[c + d*x])^(5/2))
Time = 0.34 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {3042, 3968, 52, 52, 61, 61, 61, 73, 219}
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.
| \(\displaystyle \int \frac {\cos ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sec (c+d x)^4 \sqrt {a+i a \tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 3968 |
| \(\displaystyle -\frac {i a^5 \int \frac {1}{(a-i a \tan (c+d x))^3 (i \tan (c+d x) a+a)^{7/2}}d(i a \tan (c+d x))}{d}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle -\frac {i a^5 \left (\frac {9 \int \frac {1}{(a-i a \tan (c+d x))^2 (i \tan (c+d x) a+a)^{7/2}}d(i a \tan (c+d x))}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}\right )}{d}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle -\frac {i a^5 \left (\frac {9 \left (\frac {7 \int \frac {1}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{7/2}}d(i a \tan (c+d x))}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}\right )}{d}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle -\frac {i a^5 \left (\frac {9 \left (\frac {7 \left (\frac {\int \frac {1}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{5/2}}d(i a \tan (c+d x))}{2 a}-\frac {1}{5 a (a+i a \tan (c+d x))^{5/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}\right )}{d}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle -\frac {i a^5 \left (\frac {9 \left (\frac {7 \left (\frac {\frac {\int \frac {1}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{3/2}}d(i a \tan (c+d x))}{2 a}-\frac {1}{3 a (a+i a \tan (c+d x))^{3/2}}}{2 a}-\frac {1}{5 a (a+i a \tan (c+d x))^{5/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}\right )}{d}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle -\frac {i a^5 \left (\frac {9 \left (\frac {7 \left (\frac {\frac {\frac {\int \frac {1}{(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{2 a}-\frac {1}{a \sqrt {a+i a \tan (c+d x)}}}{2 a}-\frac {1}{3 a (a+i a \tan (c+d x))^{3/2}}}{2 a}-\frac {1}{5 a (a+i a \tan (c+d x))^{5/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}\right )}{d}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {i a^5 \left (\frac {9 \left (\frac {7 \left (\frac {\frac {\frac {\int \frac {1}{a^2 \tan ^2(c+d x)+2 a}d\sqrt {i \tan (c+d x) a+a}}{a}-\frac {1}{a \sqrt {a+i a \tan (c+d x)}}}{2 a}-\frac {1}{3 a (a+i a \tan (c+d x))^{3/2}}}{2 a}-\frac {1}{5 a (a+i a \tan (c+d x))^{5/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}\right )}{d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {i a^5 \left (\frac {9 \left (\frac {7 \left (\frac {\frac {\frac {i \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2}}\right )}{\sqrt {2} a^{3/2}}-\frac {1}{a \sqrt {a+i a \tan (c+d x)}}}{2 a}-\frac {1}{3 a (a+i a \tan (c+d x))^{3/2}}}{2 a}-\frac {1}{5 a (a+i a \tan (c+d x))^{5/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}\right )}{d}\) |
Input:
Int[Cos[c + d*x]^4/Sqrt[a + I*a*Tan[c + d*x]],x]
Output:
((-I)*a^5*(1/(4*a*(a - I*a*Tan[c + d*x])^2*(a + I*a*Tan[c + d*x])^(5/2)) + (9*(1/(2*a*(a - I*a*Tan[c + d*x])*(a + I*a*Tan[c + d*x])^(5/2)) + (7*(-1/ 5*1/(a*(a + I*a*Tan[c + d*x])^(5/2)) + (-1/3*1/(a*(a + I*a*Tan[c + d*x])^( 3/2)) + ((I*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[2]])/(Sqrt[2]*a^(3/2)) - 1/ (a*Sqrt[a + I*a*Tan[c + d*x]]))/(2*a))/(2*a)))/(4*a)))/(8*a)))/d
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[1/(a^(m - 2)*b*f) Subst[Int[(a - x)^(m/2 - 1)*(a + x )^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]
Time = 7.74 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.40
| method | result | size |
| default | \(-\frac {-315 i \sin \left (d x +c \right ) \operatorname {arctanh}\left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+i}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+315 \left (-\cos \left (d x +c \right )-1\right ) \operatorname {arctanh}\left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+i}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )-315 i \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+12 \sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-35-24 \cos \left (d x +c \right )^{3}-24 \cos \left (d x +c \right )^{2}-35 \cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+2 i \cos \left (d x +c \right ) \left (-315+16 \cos \left (d x +c \right )^{4}+16 \cos \left (d x +c \right )^{3}+42 \cos \left (d x +c \right )^{2}+42 \cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}{1280 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}\) | \(312\) |
Input:
int(cos(d*x+c)^4/(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/1280/d*(-315*I*sin(d*x+c)*arctanh(1/2/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2 )*(cot(d*x+c)-csc(d*x+c)+I))+315*(-cos(d*x+c)-1)*arctanh(1/2/(-cos(d*x+c)/ (cos(d*x+c)+1))^(1/2)*(cot(d*x+c)-csc(d*x+c)+I))-315*I*(-2*cos(d*x+c)/(cos (d*x+c)+1))^(1/2)*2^(1/2)+12*sin(d*x+c)*cos(d*x+c)*(-35-24*cos(d*x+c)^3-24 *cos(d*x+c)^2-35*cos(d*x+c))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+2*I*cos(d* x+c)*(-315+16*cos(d*x+c)^4+16*cos(d*x+c)^3+42*cos(d*x+c)^2+42*cos(d*x+c))* (-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c )+1))^(1/2)/(a*(1+I*tan(d*x+c)))^(1/2)
Time = 0.09 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.31 \[ \int \frac {\cos ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {{\left (-315 i \, \sqrt {\frac {1}{2}} a d \sqrt {\frac {1}{a d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 315 i \, \sqrt {\frac {1}{2}} a d \sqrt {\frac {1}{a d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-10 i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 95 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 203 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 344 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 64 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 8 i\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{1280 \, a d} \] Input:
integrate(cos(d*x+c)^4/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")
Output:
1/1280*(-315*I*sqrt(1/2)*a*d*sqrt(1/(a*d^2))*e^(5*I*d*x + 5*I*c)*log(4*(sq rt(2)*sqrt(1/2)*(a*d*e^(2*I*d*x + 2*I*c) + a*d)*sqrt(a/(e^(2*I*d*x + 2*I*c ) + 1))*sqrt(1/(a*d^2)) + a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + 315*I*sqr t(1/2)*a*d*sqrt(1/(a*d^2))*e^(5*I*d*x + 5*I*c)*log(-4*(sqrt(2)*sqrt(1/2)*( a*d*e^(2*I*d*x + 2*I*c) + a*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a *d^2)) - a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-10*I*e^(10*I*d*x + 10*I*c) - 95*I*e^(8*I*d*x + 8*I*c) + 203*I*e^(6*I*d*x + 6*I*c) + 344*I*e^(4*I*d*x + 4*I*c) + 64*I*e^(2*I*d*x + 2*I*c) + 8*I))*e^(-5*I*d*x - 5*I*c)/(a*d)
\[ \int \frac {\cos ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {\cos ^{4}{\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \] Input:
integrate(cos(d*x+c)**4/(a+I*a*tan(d*x+c))**(1/2),x)
Output:
Integral(cos(c + d*x)**4/sqrt(I*a*(tan(c + d*x) - I)), x)
Time = 0.13 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.86 \[ \int \frac {\cos ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {i \, {\left (315 \, \sqrt {2} \sqrt {a} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (315 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a - 1050 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{2} + 672 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{3} + 192 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4} + 128 \, a^{5}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} - 4 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a + 4 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2}}\right )}}{2560 \, a d} \] Input:
integrate(cos(d*x+c)^4/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")
Output:
1/2560*I*(315*sqrt(2)*sqrt(a)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c ) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(d*x + c) + a))) + 4*(315*(I*a*tan( d*x + c) + a)^4*a - 1050*(I*a*tan(d*x + c) + a)^3*a^2 + 672*(I*a*tan(d*x + c) + a)^2*a^3 + 192*(I*a*tan(d*x + c) + a)*a^4 + 128*a^5)/((I*a*tan(d*x + c) + a)^(9/2) - 4*(I*a*tan(d*x + c) + a)^(7/2)*a + 4*(I*a*tan(d*x + c) + a)^(5/2)*a^2))/(a*d)
Exception generated. \[ \int \frac {\cos ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(cos(d*x+c)^4/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad Argument TypeDone
Timed out. \[ \int \frac {\cos ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^4}{\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \] Input:
int(cos(c + d*x)^4/(a + a*tan(c + d*x)*1i)^(1/2),x)
Output:
int(cos(c + d*x)^4/(a + a*tan(c + d*x)*1i)^(1/2), x)
\[ \int \frac {\cos ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {a}\, \left (-\left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{4} \tan \left (d x +c \right )}{\tan \left (d x +c \right )^{2}+1}d x \right ) i +\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{4}}{\tan \left (d x +c \right )^{2}+1}d x \right )}{a} \] Input:
int(cos(d*x+c)^4/(a+I*a*tan(d*x+c))^(1/2),x)
Output:
(sqrt(a)*( - int((sqrt(tan(c + d*x)*i + 1)*cos(c + d*x)**4*tan(c + d*x))/( tan(c + d*x)**2 + 1),x)*i + int((sqrt(tan(c + d*x)*i + 1)*cos(c + d*x)**4) /(tan(c + d*x)**2 + 1),x)))/a