Integrand size = 26, antiderivative size = 219 \[ \int \frac {\cos ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {63 i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{128 \sqrt {2} \sqrt {a} d}+\frac {63 i a^2}{160 d (a+i a \tan (c+d x))^{5/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {9 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac {21 i a}{64 d (a+i a \tan (c+d x))^{3/2}}+\frac {63 i}{128 d \sqrt {a+i a \tan (c+d x)}} \]
[Out]
Time = 0.15 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3568, 44, 53, 65, 212} \[ \int \frac {\cos ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {9 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac {63 i a^2}{160 d (a+i a \tan (c+d x))^{5/2}}-\frac {63 i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{128 \sqrt {2} \sqrt {a} d}+\frac {21 i a}{64 d (a+i a \tan (c+d x))^{3/2}}+\frac {63 i}{128 d \sqrt {a+i a \tan (c+d x)}} \]
[In]
[Out]
Rule 44
Rule 53
Rule 65
Rule 212
Rule 3568
Rubi steps \begin{align*} \text {integral}& = -\frac {\left (i a^5\right ) \text {Subst}\left (\int \frac {1}{(a-x)^3 (a+x)^{7/2}} \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = -\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {\left (9 i a^4\right ) \text {Subst}\left (\int \frac {1}{(a-x)^2 (a+x)^{7/2}} \, dx,x,i a \tan (c+d x)\right )}{8 d} \\ & = -\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {9 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}-\frac {\left (63 i a^3\right ) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{7/2}} \, dx,x,i a \tan (c+d x)\right )}{32 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 {9 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}-\frac {\left (63 i a^2\right ) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{5/2}} \, dx,x,i a \tan (c+d x)\right )}{64 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 {9 i a^3}{16 d (a-i a \tan (c+d x)) (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 a) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{3/2}} \, dx,x,i a \tan (c+d x)\right )}{128 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 {9 i a^3}{16 d (a-i a \tan (c+d x)) (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 {(63 i) \text {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,i a \tan (c+d x)\right )}{256 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 {9 i a^3}{16 d (a-i a \tan (c+d x)) (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 {(63 i) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{128 d} \\ & = -\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 {9 i a^3}{16 d (a-i a \tan (c+d x)) (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)}} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.20 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.24 \[ \int \frac {\cos ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {i a^2 \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},3,-\frac {3}{2},\frac {1}{2} (1+i \tan (c+d x))\right )}{20 d (a+i a \tan (c+d x))^{5/2}} \]
[In]
[Out]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 539 vs. \(2 (174 ) = 348\).
Time = 11.10 (sec) , antiderivative size = 540, normalized size of antiderivative = 2.47
method | result | size |
default | \(-\frac {32 i \left (\cos ^{5}\left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+32 i \left (\cos ^{4}\left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-288 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+84 i \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-288 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+84 i \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-420 \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-630 i \cos \left (d x +c \right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-315 i \cos \left (d x +c \right ) \arctan \left (\frac {\cos \left (d x +c \right )+1+i \sin \left (d x +c \right )}{2 \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 )-420 \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 )-630 i \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-315 i \arctan \left (\frac {\cos \left (d x +c \right )+1+i \sin \left (d x +c \right )}{2 \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 )+315 \arctan \left (\frac {\cos \left (d x +c \right )+1+i \sin \left (d x +c \right )}{2 \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 )}{1280 d \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\cos \left (d x +c \right )+1\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}\) | \(540\) |
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.34 \[ \int \frac {\cos ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {{\left (-315 i \, \sqrt {\frac {1}{2}} a d \sqrt {\frac {1}{a d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 315 i \, \sqrt {\frac {1}{2}} a d \sqrt {\frac {1}{a d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-10 i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 95 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 203 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 344 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 64 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 8 i\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{1280 \, a d} \]
[In]
[Out]
\[ \int \frac {\cos ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {\cos ^{4}{\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \]
[In]
[Out]
none
Time = 0.32 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.88 \[ \int \frac {\cos ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {i \, {\left (315 \, \sqrt {2} \sqrt {a} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (315 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a - 1050 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{2} + 672 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{3} + 192 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4} + 128 \, a^{5}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} - 4 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a + 4 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2}}\right )}}{2560 \, a d} \]
[In]
[Out]
\[ \int \frac {\cos ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int { \frac {\cos \left (d x + c\right )^{4}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}} \,d x } \]
[In]
[Out]
Timed out. \[ \int \frac {\cos ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^4}{\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \]
[In]
[Out]