Integrand size = 26, antiderivative size = 170 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {15 i a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{32 \sqrt {2} d}+\frac {15 i a^2}{32 d \sqrt {a+i a \tan (c+d x)}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a^5}{16 d \sqrt {a+i a \tan (c+d x)} \left (a^3-i a^3 \tan (c+d x)\right )} \] Output:
-15/64*I*a^(3/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^( 1/2)/d+15/32*I*a^2/d/(a+I*a*tan(d*x+c))^(1/2)-1/4*I*a^4/d/(a-I*a*tan(d*x+c ))^2/(a+I*a*tan(d*x+c))^(1/2)-5/16*I*a^5/d/(a+I*a*tan(d*x+c))^(1/2)/(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 = 0.10 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.31 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {i a^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},3,\frac {1}{2},\frac {1}{2} (1+i \tan (c+d x))\right )}{4 d \sqrt {a+i a \tan (c+d x)}} \] Input:
Integrate[Cos[c + d*x]^4*(a + I*a*Tan[c + d*x])^(3/2),x]
Output:
((I/4)*a^2*Hypergeometric2F1[-1/2, 3, 1/2, (1 + I*Tan[c + d*x])/2])/(d*Sqr t[a + I*a*Tan[c + d*x]])
Time = 0.31 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.94, 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, 52, 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.
\(\displaystyle \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{3/2}}{\sec (c+d x)^4}dx\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle -\frac {i a^5 \int \frac {1}{(a-i a \tan (c+d x))^3 (i \tan (c+d x) a+a)^{3/2}}d(i a \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle -\frac {i a^5 \left (\frac {5 \int \frac {1}{(a-i a \tan (c+d x))^2 (i \tan (c+d x) a+a)^{3/2}}d(i a \tan (c+d x))}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}}\right )}{d}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle -\frac {i a^5 \left (\frac {5 \left (\frac {3 \int \frac {1}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{3/2}}d(i a \tan (c+d x))}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}}\right )}{d}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle -\frac {i a^5 \left (\frac {5 \left (\frac {3 \left (\frac {\int \frac {1}{(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{2 a}-\frac {1}{a \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}}\right )}{d}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {i a^5 \left (\frac {5 \left (\frac {3 \left (\frac {\int \frac {1}{a^2 \tan ^2(c+d x)+2 a}d\sqrt {i \tan (c+d x) a+a}}{a}-\frac {1}{a \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}}\right )}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {i a^5 \left (\frac {5 \left (\frac {3 \left (\frac {i \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2}}\right )}{\sqrt {2} a^{3/2}}-\frac {1}{a \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}}\right )}{d}\) |
Input:
Int[Cos[c + d*x]^4*(a + I*a*Tan[c + d*x])^(3/2),x]
Output:
((-I)*a^5*(1/(4*a*(a - I*a*Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]]) + ( 5*(1/(2*a*(a - I*a*Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]]) + (3*((I*ArcT an[(Sqrt[a]*Tan[c + d*x])/Sqrt[2]])/(Sqrt[2]*a^(3/2)) - 1/(a*Sqrt[a + I*a* Tan[c + d*x]])))/(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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 771 vs. \(2 (137 ) = 274\).
Time = 4.34 (sec) , antiderivative size = 772, normalized size of antiderivative = 4.54
Input:
int(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
Output:
1/64/d*cos(d*x+c)*((-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(2^(1/2)*(cot (d*x+c)-csc(d*x+c))/(cot(d*x+c)^2-2*cot(d*x+c)*csc(d*x+c)+csc(d*x+c)^2-1)^ (1/2))*(-30*sin(d*x+c)^2-15*tan(d*x+c)*sin(d*x+c))+I*(-30*cos(d*x+c)^2-15* cos(d*x+c)+15)*tan(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(2^(1/ 2)*(cot(d*x+c)-csc(d*x+c))/(cot(d*x+c)^2-2*cot(d*x+c)*csc(d*x+c)+csc(d*x+c )^2-1)^(1/2))+I*(30*cos(d*x+c)+15)*sin(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)) ^(1/2)*arctanh(2^(1/2)*(cot(d*x+c)-csc(d*x+c))/(cot(d*x+c)^2-2*cot(d*x+c)* csc(d*x+c)+csc(d*x+c)^2-1)^(1/2))+(-30*cos(d*x+c)^2-15*cos(d*x+c)+15)*(-co s(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(2^(1/2)*(cot(d*x+c)-csc(d*x+c))/(co t(d*x+c)^2-2*cot(d*x+c)*csc(d*x+c)+csc(d*x+c)^2-1)^(1/2))+I*(-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))*(30*sin(d*x+c)^2+15*tan(d*x+c)*sin(d*x+c))+(-30*cos(d*x+c)^2-15*cos(d* x+c)+15)*tan(d*x+c)*(-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))+(30*cos(d*x+c)+15)*sin(d*x+c)*(-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*(30*cos(d*x+c)^2+15*cos(d*x+c)-15)*(-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))+40*I*si n(d*x+c)^2*cos(d*x+c)+sin(d*x+c)*(-24*cos(d*x+c)^2+30)+I*cos(d*x+c)*(24*co s(d*x+c)^2-30)+40*sin(d*x+c)*cos(d*x+c)^2)*a*(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. 287 vs. \(2 (127) = 254\).
Time = 0.09 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.69 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {{\left (15 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )} \log \left (-\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {-\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - a^{2} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{a}\right ) - 15 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )} \log \left (-\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {-\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - a^{2} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{a}\right ) - \sqrt {2} {\left (-2 i \, a e^{\left (6 i \, d x + 6 i \, c\right )} - 11 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} - i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 8 i \, a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{64 \, d} \] Input:
integrate(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")
Output:
-1/64*(15*sqrt(1/2)*sqrt(-a^3/d^2)*d*e^(I*d*x + I*c)*log(-4*(sqrt(2)*sqrt( 1/2)*(I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt(-a^3/d^2)*sqrt(a/(e^(2*I*d*x + 2 *I*c) + 1)) - a^2*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a) - 15*sqrt(1/2)*sqrt (-a^3/d^2)*d*e^(I*d*x + I*c)*log(-4*(sqrt(2)*sqrt(1/2)*(-I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt(-a^3/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)) - a^2*e^(I* d*x + I*c))*e^(-I*d*x - I*c)/a) - sqrt(2)*(-2*I*a*e^(6*I*d*x + 6*I*c) - 11 *I*a*e^(4*I*d*x + 4*I*c) - I*a*e^(2*I*d*x + 2*I*c) + 8*I*a)*sqrt(a/(e^(2*I *d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/d
Timed out. \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**4*(a+I*a*tan(d*x+c))**(3/2),x)
Output:
Timed out
Time = 0.14 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.93 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {i \, {\left (15 \, \sqrt {2} a^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (15 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{3} - 50 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4} + 32 \, a^{5}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} - 4 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a + 4 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{2}}\right )}}{128 \, a d} \] Input:
integrate(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")
Output:
1/128*I*(15*sqrt(2)*a^(5/2)*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*(15*(I*a*tan(d*x + c) + a)^2*a^3 - 50*(I*a*tan(d*x + c) + a)*a^4 + 32*a^5)/((I*a*tan(d*x + c) + a)^(5/2) - 4*(I*a*tan(d*x + c) + a)^(3/2)*a + 4*sqrt(I*a*tan(d*x + c ) + a)*a^2))/(a*d)
Exception generated. \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^(3/2),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. \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\int {\cos \left (c+d\,x\right )}^4\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \] Input:
int(cos(c + d*x)^4*(a + a*tan(c + d*x)*1i)^(3/2),x)
Output:
int(cos(c + d*x)^4*(a + a*tan(c + d*x)*1i)^(3/2), x)
\[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\sqrt {a}\, a \left (\left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{4} \tan \left (d x +c \right )d x \right ) i +\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{4}d x \right ) \] Input:
int(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^(3/2),x)
Output:
sqrt(a)*a*(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))