Integrand size = 26, antiderivative size = 150 \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {5 i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{8 \sqrt {2} \sqrt {a} d}+\frac {5 i a}{12 d (a+i a \tan (c+d x))^{3/2}}+\frac {5 i}{8 d \sqrt {a+i a \tan (c+d x)}}-\frac {i a^3}{2 d (a+i a \tan (c+d x))^{3/2} \left (a^2-i a^2 \tan (c+d x)\right )} \] Output:
-5/16*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+5/12*I*a/d/(a+I*a*tan(d*x+c))^(3/2)+5/8*I/d/(a+I*a*tan(d*x+c))^(1/2) -1/2*I*a^3/d/(a+I*a*tan(d*x+c))^(3/2)/(a^2-I*a^2*tan(d*x+c))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.12 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.34 \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {i a \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},2,-\frac {1}{2},\frac {1}{2} (1+i \tan (c+d x))\right )}{6 d (a+i a \tan (c+d x))^{3/2}} \] Input:
Integrate[Cos[c + d*x]^2/Sqrt[a + I*a*Tan[c + d*x]],x]
Output:
((I/6)*a*Hypergeometric2F1[-3/2, 2, -1/2, (1 + I*Tan[c + d*x])/2])/(d*(a + I*a*Tan[c + d*x])^(3/2))
Time = 0.29 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.96, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {3042, 3968, 52, 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.
\(\displaystyle \int \frac {\cos ^2(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sec (c+d x)^2 \sqrt {a+i a \tan (c+d x)}}dx\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle -\frac {i a^3 \int \frac {1}{(a-i a \tan (c+d x))^2 (i \tan (c+d x) a+a)^{5/2}}d(i a \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle -\frac {i a^3 \left (\frac {5 \int \frac {1}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{5/2}}d(i a \tan (c+d x))}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}\right )}{d}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle -\frac {i a^3 \left (\frac {5 \left (\frac {\int \frac {1}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{3/2}}d(i a \tan (c+d x))}{2 a}-\frac {1}{3 a (a+i a \tan (c+d x))^{3/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}\right )}{d}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle -\frac {i a^3 \left (\frac {5 \left (\frac {\frac {\int \frac {1}{(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{2 a}-\frac {1}{a \sqrt {a+i a \tan (c+d x)}}}{2 a}-\frac {1}{3 a (a+i a \tan (c+d x))^{3/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}\right )}{d}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {i a^3 \left (\frac {5 \left (\frac {\frac {\int \frac {1}{a^2 \tan ^2(c+d x)+2 a}d\sqrt {i \tan (c+d x) a+a}}{a}-\frac {1}{a \sqrt {a+i a \tan (c+d x)}}}{2 a}-\frac {1}{3 a (a+i a \tan (c+d x))^{3/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}\right )}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {i a^3 \left (\frac {5 \left (\frac {\frac {i \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2}}\right )}{\sqrt {2} a^{3/2}}-\frac {1}{a \sqrt {a+i a \tan (c+d x)}}}{2 a}-\frac {1}{3 a (a+i a \tan (c+d x))^{3/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}\right )}{d}\) |
Input:
Int[Cos[c + d*x]^2/Sqrt[a + I*a*Tan[c + d*x]],x]
Output:
((-I)*a^3*(1/(2*a*(a - I*a*Tan[c + d*x])*(a + I*a*Tan[c + d*x])^(3/2)) + ( 5*(-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) ))/(4*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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 303 vs. \(2 (118 ) = 236\).
Time = 9.30 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.03
method | result | size |
default | \(\frac {15 i \sin \left (d x +c \right ) \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )+i\right )}{2 \sqrt {\cot \left (d x +c \right )^{2}-2 \cot \left (d x +c \right ) \csc \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )+\left (15 \cos \left (d x +c \right )+15\right ) \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )+i\right )}{2 \sqrt {\cot \left (d x +c \right )^{2}-2 \cot \left (d x +c \right ) \csc \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )+15 i \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (20 \cos \left (d x +c \right )+20\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+i \cos \left (d x +c \right ) \left (-4 \cos \left (d x +c \right )^{2}-4 \cos \left (d x +c \right )+30\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}{48 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}\) | \(304\) |
Input:
int(cos(d*x+c)^2/(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
Output:
1/48/d*(15*I*sin(d*x+c)*arctanh(1/2*2^(1/2)*(cot(d*x+c)-csc(d*x+c)+I)/(cot (d*x+c)^2-2*cot(d*x+c)*csc(d*x+c)+csc(d*x+c)^2-1)^(1/2))+(15*cos(d*x+c)+15 )*arctanh(1/2*2^(1/2)*(cot(d*x+c)-csc(d*x+c)+I)/(cot(d*x+c)^2-2*cot(d*x+c) *csc(d*x+c)+csc(d*x+c)^2-1)^(1/2))+15*I*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c) +1))^(1/2)+sin(d*x+c)*cos(d*x+c)*(20*cos(d*x+c)+20)*(-cos(d*x+c)/(cos(d*x+ c)+1))^(1/2)+I*cos(d*x+c)*(-4*cos(d*x+c)^2-4*cos(d*x+c)+30)*(-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)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 271 vs. \(2 (109) = 218\).
Time = 0.09 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.81 \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {{\left (-15 i \, \sqrt {\frac {1}{2}} a d \sqrt {\frac {1}{a d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 15 i \, \sqrt {\frac {1}{2}} a d \sqrt {\frac {1}{a d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-3 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 11 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 16 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{48 \, a d} \] Input:
integrate(cos(d*x+c)^2/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")
Output:
1/48*(-15*I*sqrt(1/2)*a*d*sqrt(1/(a*d^2))*e^(3*I*d*x + 3*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)) + 15*I*sqrt(1/ 2)*a*d*sqrt(1/(a*d^2))*e^(3*I*d*x + 3*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))*(-3*I*e^(6*I*d*x + 6*I*c) + 11*I*e^(4*I*d*x + 4*I*c) + 16*I*e^ (2*I*d*x + 2*I*c) + 2*I))*e^(-3*I*d*x - 3*I*c)/(a*d)
\[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {\cos ^{2}{\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \] Input:
integrate(cos(d*x+c)**2/(a+I*a*tan(d*x+c))**(1/2),x)
Output:
Integral(cos(c + d*x)**2/sqrt(I*a*(tan(c + d*x) - I)), x)
Time = 0.12 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.92 \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {i \, {\left (15 \, \sqrt {2} \sqrt {a} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (15 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a - 20 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{2} - 8 \, a^{3}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} - 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a}\right )}}{96 \, a d} \] Input:
integrate(cos(d*x+c)^2/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")
Output:
1/96*I*(15*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*(15*(I*a*tan(d*x + c) + a)^2*a - 20*(I*a*tan(d*x + c) + a)*a^2 - 8*a^3)/((I*a*tan(d*x + c) + a)^(5/2) - 2*(I*a*tan(d*x + c) + a)^(3/2)*a))/(a*d)
Exception generated. \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(cos(d*x+c)^2/(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. \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2}{\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \] Input:
int(cos(c + d*x)^2/(a + a*tan(c + d*x)*1i)^(1/2),x)
Output:
int(cos(c + d*x)^2/(a + a*tan(c + d*x)*1i)^(1/2), x)
\[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {a}\, \left (-\left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{2} \tan \left (d x +c \right )}{\tan \left (d x +c \right )^{2}+1}d x \right ) i +\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{2}+1}d x \right )}{a} \] Input:
int(cos(d*x+c)^2/(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)**2*tan(c + d*x))/( tan(c + d*x)**2 + 1),x)*i + int((sqrt(tan(c + d*x)*i + 1)*cos(c + d*x)**2) /(tan(c + d*x)**2 + 1),x)))/a