Integrand size = 26, antiderivative size = 141 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {3 i a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{16 \sqrt {2} d}-\frac {i a^4 \sqrt {a+i a \tan (c+d x)}}{4 d (a-i a \tan (c+d x))^2}-\frac {3 i a^5 \sqrt {a+i a \tan (c+d x)}}{16 d \left (a^3-i a^3 \tan (c+d x)\right )} \] Output:
-3/32*I*a^(5/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1 /2)/d-1/4*I*a^4*(a+I*a*tan(d*x+c))^(1/2)/d/(a-I*a*tan(d*x+c))^2-3/16*I*a^5 *(a+I*a*tan(d*x+c))^(1/2)/d/(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 = 0.08 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.38 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {i a^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3,\frac {3}{2},\frac {1}{2} (1+i \tan (c+d x))\right ) \sqrt {a+i a \tan (c+d x)}}{4 d} \] Input:
Integrate[Cos[c + d*x]^4*(a + I*a*Tan[c + d*x])^(5/2),x]
Output:
((-1/4*I)*a^2*Hypergeometric2F1[1/2, 3, 3/2, (1 + I*Tan[c + d*x])/2]*Sqrt[ a + I*a*Tan[c + d*x]])/d
Time = 0.30 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.93, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3042, 3968, 52, 52, 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.
\(\displaystyle \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{5/2}}{\sec (c+d x)^4}dx\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle -\frac {i a^5 \int \frac {1}{(a-i a \tan (c+d x))^3 \sqrt {i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle -\frac {i a^5 \left (\frac {3 \int \frac {1}{(a-i a \tan (c+d x))^2 \sqrt {i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{8 a}+\frac {\sqrt {a+i a \tan (c+d x)}}{4 a (a-i a \tan (c+d x))^2}\right )}{d}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle -\frac {i a^5 \left (\frac {3 \left (\frac {\int \frac {1}{(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{4 a}+\frac {\sqrt {a+i a \tan (c+d x)}}{2 a (a-i a \tan (c+d x))}\right )}{8 a}+\frac {\sqrt {a+i a \tan (c+d x)}}{4 a (a-i a \tan (c+d x))^2}\right )}{d}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {i a^5 \left (\frac {3 \left (\frac {\int \frac {1}{a^2 \tan ^2(c+d x)+2 a}d\sqrt {i \tan (c+d x) a+a}}{2 a}+\frac {\sqrt {a+i a \tan (c+d x)}}{2 a (a-i a \tan (c+d x))}\right )}{8 a}+\frac {\sqrt {a+i a \tan (c+d x)}}{4 a (a-i a \tan (c+d x))^2}\right )}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {i a^5 \left (\frac {3 \left (\frac {i \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2}}\right )}{2 \sqrt {2} a^{3/2}}+\frac {\sqrt {a+i a \tan (c+d x)}}{2 a (a-i a \tan (c+d x))}\right )}{8 a}+\frac {\sqrt {a+i a \tan (c+d x)}}{4 a (a-i a \tan (c+d x))^2}\right )}{d}\) |
Input:
Int[Cos[c + d*x]^4*(a + I*a*Tan[c + d*x])^(5/2),x]
Output:
((-I)*a^5*(Sqrt[a + I*a*Tan[c + d*x]]/(4*a*(a - I*a*Tan[c + d*x])^2) + (3* (((I/2)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[2]])/(Sqrt[2]*a^(3/2)) + Sqrt[a + I*a*Tan[c + d*x]]/(2*a*(a - I*a*Tan[c + d*x]))))/(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] :> 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. 471 vs. \(2 (114 ) = 228\).
Time = 67.12 (sec) , antiderivative size = 472, normalized size of antiderivative = 3.35
method | result | size |
default | \(-\frac {\cos \left (d x +c \right )^{2} \left (\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (6 \cos \left (d x +c \right )+6\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ) \sqrt {2}}{\sqrt {\cot \left (d x +c \right )^{2}-2 \cot \left (d x +c \right ) \csc \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )+i \left (6 \cos \left (d x +c \right )^{3}+6 \cos \left (d x +c \right )^{2}-3 \cos \left (d x +c \right )-3\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ) \sqrt {2}}{\sqrt {\cot \left (d x +c \right )^{2}-2 \cot \left (d x +c \right ) \csc \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )+i \sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (6 \cos \left (d x +c \right )+6\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {2}}{2}\right )+\left (-6 \cos \left (d x +c \right )^{3}-6 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+3\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {2}}{2}\right )+i \sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (4 \cos \left (d x +c \right )-3\right )+\cos \left (d x +c \right ) \left (4 \cos \left (d x +c \right )^{2}+7 \cos \left (d x +c \right )+3\right )\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (-\tan \left (d x +c \right )+i\right )^{2}}{16 d \left (-\sin \left (d x +c \right )+i \cos \left (d x +c \right )+i\right )}\) | \(472\) |
Input:
int(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/16/d*cos(d*x+c)^2*(sin(d*x+c)*cos(d*x+c)*(6*cos(d*x+c)+6)*(-cos(d*x+c)/ (cos(d*x+c)+1))^(1/2)*arctanh(1/(cot(d*x+c)^2-2*cot(d*x+c)*csc(d*x+c)+csc( d*x+c)^2-1)^(1/2)*(csc(d*x+c)-cot(d*x+c))*2^(1/2))+I*(6*cos(d*x+c)^3+6*cos (d*x+c)^2-3*cos(d*x+c)-3)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(1/(co t(d*x+c)^2-2*cot(d*x+c)*csc(d*x+c)+csc(d*x+c)^2-1)^(1/2)*(csc(d*x+c)-cot(d *x+c))*2^(1/2))+I*sin(d*x+c)*cos(d*x+c)*(6*cos(d*x+c)+6)*(-cos(d*x+c)/(cos (d*x+c)+1))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*2^(1/2)) +(-6*cos(d*x+c)^3-6*cos(d*x+c)^2+3*cos(d*x+c)+3)*(-cos(d*x+c)/(cos(d*x+c)+ 1))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*2^(1/2))+I*sin(d *x+c)*cos(d*x+c)*(4*cos(d*x+c)-3)+cos(d*x+c)*(4*cos(d*x+c)^2+7*cos(d*x+c)+ 3))*(a*(1+I*tan(d*x+c)))^(1/2)*a^2*(-tan(d*x+c)+I)^2/(-sin(d*x+c)+I*cos(d* x+c)+I)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 263 vs. \(2 (106) = 212\).
Time = 0.09 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.87 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {3 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{5}}{d^{2}}} d \log \left (\frac {4 \, {\left (a^{3} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{5}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2}}\right ) - 3 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{5}}{d^{2}}} d \log \left (\frac {4 \, {\left (a^{3} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{5}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2}}\right ) - \sqrt {2} {\left (-2 i \, a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} - 7 i \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} - 5 i \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{32 \, d} \] Input:
integrate(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")
Output:
-1/32*(3*sqrt(1/2)*sqrt(-a^5/d^2)*d*log(4*(a^3*e^(I*d*x + I*c) - sqrt(2)*s qrt(1/2)*sqrt(-a^5/d^2)*(I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^2) - 3*sqrt(1/2)*sqrt(-a^5/d^2)*d*log( 4*(a^3*e^(I*d*x + I*c) - sqrt(2)*sqrt(1/2)*sqrt(-a^5/d^2)*(-I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^2) - sqrt(2)*(-2*I*a^2*e^(5*I*d*x + 5*I*c) - 7*I*a^2*e^(3*I*d*x + 3*I*c) - 5* I*a^2*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))/d
Timed out. \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**4*(a+I*a*tan(d*x+c))**(5/2),x)
Output:
Timed out
Time = 0.12 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.99 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {i \, {\left (3 \, \sqrt {2} a^{\frac {7}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{4} - 10 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{5}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} - 4 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 4 \, a^{2}}\right )}}{64 \, a d} \] Input:
integrate(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")
Output:
1/64*I*(3*sqrt(2)*a^(7/2)*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*(3*(I*a*tan(d*x + c) + a)^(3/2)*a^4 - 10*sqrt(I*a*tan(d*x + c) + a)*a^5)/((I*a*tan(d*x + c) + a)^2 - 4*(I*a*tan(d*x + c) + a)*a + 4*a^2))/(a*d)
Exception generated. \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^(5/2),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. \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\int {\cos \left (c+d\,x\right )}^4\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2} \,d x \] Input:
int(cos(c + d*x)^4*(a + a*tan(c + d*x)*1i)^(5/2),x)
Output:
int(cos(c + d*x)^4*(a + a*tan(c + d*x)*1i)^(5/2), x)
\[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\sqrt {a}\, a^{2} \left (-\left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{4} \tan \left (d x +c \right )^{2}d x \right )+2 \left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{4} \tan \left (d x +c \right )d x \right ) i +\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{4}d x \right ) \] Input:
int(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^(5/2),x)
Output:
sqrt(a)*a**2*( - int(sqrt(tan(c + d*x)*i + 1)*cos(c + d*x)**4*tan(c + d*x) **2,x) + 2*int(sqrt(tan(c + d*x)*i + 1)*cos(c + d*x)**4*tan(c + d*x),x)*i + int(sqrt(tan(c + d*x)*i + 1)*cos(c + d*x)**4,x))