Integrand size = 26, antiderivative size = 166 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {15 i a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{32 \sqrt {2} d}+\frac {15 i a^2}{32 d \sqrt {a+i a \tan (c+d x)}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a^3}{16 d (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}} \]
-15/64*I*a^(3/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/d*2 ^(1/2)+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 ))^(1/2)/(a-I*a*tan(d*x+c))^2-5/16*I*a^3/d/(a+I*a*tan(d*x+c))^(1/2)/(a-I*a *tan(d*x+c))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.11 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.32 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {i a^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},3,\frac {1}{2},\frac {1}{2} (1+i \tan (c+d x))\right )}{4 d \sqrt {a+i a \tan (c+d x)}} \]
((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 = 0.30 (sec) , antiderivative size = 159, 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, 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 |
| \(\displaystyle -\frac {i a^5 \left (\frac {5 \left (\frac {3 \left (\frac {i \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2}}\right )}{\sqrt {2} a^{3/2}}-\frac {1}{a \sqrt {a+i a \tan (c+d x)}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}}\right )}{d}\) |
((-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
3.3.98.3.1 Defintions of rubi rules used
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. 658 vs. \(2 (133 ) = 266\).
Time = 14.37 (sec) , antiderivative size = 659, normalized size of antiderivative = 3.97
| method | result | size |
| default | \(-\frac {\left (-\tan \left (d x +c \right )+i\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \cos \left (d x +c \right ) \left (30 i \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \left (\cos ^{2}\left (d x +c \right )\right )+30 i \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )+15 i \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+15 i \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) \sin \left (d x +c \right )+40 i \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+30 \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )-30 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) \left (\cos ^{2}\left (d x +c \right )\right )-15 i \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+15 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sin \left (d x +c \right )-15 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) \cos \left (d x +c \right )-24 \left (\cos ^{3}\left (d x +c \right )\right )+15 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )+30 \cos \left (d x +c \right )\right )}{64 d}\) | \(659\) |
-1/64/d*(-tan(d*x+c)+I)*(a*(1+I*tan(d*x+c)))^(1/2)*a*cos(d*x+c)*(30*I*(-co s(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x +c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^2+30*I*(-cos(d*x+c)/(cos(d*x+c)+1))^ (1/2)*arctan((-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)*sin(d*x+c)+15* I*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-c os(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)+15*I*(-cos(d*x+c)/(cos(d*x+c)+ 1))^(1/2)*arctan((-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*sin(d*x+c)+40*I*cos(d *x+c)^2*sin(d*x+c)+30*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos( d*x+c)+1))^(1/2))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)*sin(d*x+c) -30*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan((-cos(d*x+c)/(cos(d*x+c)+1)) ^(1/2))*cos(d*x+c)^2-15*I*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d *x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+15*(-cos(d*x+c)/( cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d *x+c)+1))^(1/2))*sin(d*x+c)-15*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(( -cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)-24*cos(d*x+c)^3+15*(-cos(d*x +c)/(cos(d*x+c)+1))^(1/2)*arctan((-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+30*co s(d*x+c))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (123) = 246\).
Time = 0.26 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.73 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {{\left (15 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )} \log \left (-\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {-\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - a^{2} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{a}\right ) - 15 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )} \log \left (-\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {-\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - a^{2} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{a}\right ) - \sqrt {2} {\left (-2 i \, a e^{\left (6 i \, d x + 6 i \, c\right )} - 11 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} - i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 8 i \, a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{64 \, d} \]
-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. \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\text {Timed out} \]
Time = 0.31 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.95 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {i \, {\left (15 \, \sqrt {2} a^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (15 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{3} - 50 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4} + 32 \, a^{5}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} - 4 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a + 4 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{2}}\right )}}{128 \, a d} \]
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)
Timed out. \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\text {Timed out} \]
Timed out. \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\int {\cos \left (c+d\,x\right )}^4\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \]