Integrand size = 26, antiderivative size = 310 \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=-\frac {195 i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{1024 \sqrt {2} a^{7/2} d}+\frac {195 i a^2}{352 d (a+i a \tan (c+d x))^{11/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{11/2}}+\frac {65 i a}{192 d (a+i a \tan (c+d x))^{9/2}}+\frac {195 i}{896 d (a+i a \tan (c+d x))^{7/2}}+\frac {39 i}{256 a d (a+i a \tan (c+d x))^{5/2}}+\frac {65 i}{512 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {195 i}{1024 a^3 d \sqrt {a+i a \tan (c+d x)}}-\frac {15 i a^5}{16 d (a+i a \tan (c+d x))^{11/2} \left (a^3-i a^3 \tan (c+d x)\right )} \] Output:
-195/2048*I*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/ a^(7/2)/d+195/352*I*a^2/d/(a+I*a*tan(d*x+c))^(11/2)-1/4*I*a^4/d/(a-I*a*tan (d*x+c))^2/(a+I*a*tan(d*x+c))^(11/2)+65/192*I*a/d/(a+I*a*tan(d*x+c))^(9/2) +195/896*I/d/(a+I*a*tan(d*x+c))^(7/2)+39/256*I/a/d/(a+I*a*tan(d*x+c))^(5/2 )+65/512*I/a^2/d/(a+I*a*tan(d*x+c))^(3/2)+195/1024*I/a^3/d/(a+I*a*tan(d*x+ c))^(1/2)-15/16*I*a^5/d/(a+I*a*tan(d*x+c))^(11/2)/(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 = 0.67 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.17 \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {i a^2 \operatorname {Hypergeometric2F1}\left (-\frac {11}{2},3,-\frac {9}{2},\frac {1}{2} (1+i \tan (c+d x))\right )}{44 d (a+i a \tan (c+d x))^{11/2}} \] Input:
Integrate[Cos[c + d*x]^4/(a + I*a*Tan[c + d*x])^(7/2),x]
Output:
((I/44)*a^2*Hypergeometric2F1[-11/2, 3, -9/2, (1 + I*Tan[c + d*x])/2])/(d* (a + I*a*Tan[c + d*x])^(11/2))
Time = 0.35 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.03, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {3042, 3968, 52, 52, 61, 61, 61, 61, 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.
| \(\displaystyle \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sec (c+d x)^4 (a+i a \tan (c+d x))^{7/2}}dx\) |
| \(\Big \downarrow \) 3968 |
| \(\displaystyle -\frac {i a^5 \int \frac {1}{(a-i a \tan (c+d x))^3 (i \tan (c+d x) a+a)^{13/2}}d(i a \tan (c+d x))}{d}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle -\frac {i a^5 \left (\frac {15 \int \frac {1}{(a-i a \tan (c+d x))^2 (i \tan (c+d x) a+a)^{13/2}}d(i a \tan (c+d x))}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{11/2}}\right )}{d}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle -\frac {i a^5 \left (\frac {15 \left (\frac {13 \int \frac {1}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{13/2}}d(i a \tan (c+d x))}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{11/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{11/2}}\right )}{d}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle -\frac {i a^5 \left (\frac {15 \left (\frac {13 \left (\frac {\int \frac {1}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{11/2}}d(i a \tan (c+d x))}{2 a}-\frac {1}{11 a (a+i a \tan (c+d x))^{11/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{11/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{11/2}}\right )}{d}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle -\frac {i a^5 \left (\frac {15 \left (\frac {13 \left (\frac {\frac {\int \frac {1}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{9/2}}d(i a \tan (c+d x))}{2 a}-\frac {1}{9 a (a+i a \tan (c+d x))^{9/2}}}{2 a}-\frac {1}{11 a (a+i a \tan (c+d x))^{11/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{11/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{11/2}}\right )}{d}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle -\frac {i a^5 \left (\frac {15 \left (\frac {13 \left (\frac {\frac {\frac {\int \frac {1}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{7/2}}d(i a \tan (c+d x))}{2 a}-\frac {1}{7 a (a+i a \tan (c+d x))^{7/2}}}{2 a}-\frac {1}{9 a (a+i a \tan (c+d x))^{9/2}}}{2 a}-\frac {1}{11 a (a+i a \tan (c+d x))^{11/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{11/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{11/2}}\right )}{d}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle -\frac {i a^5 \left (\frac {15 \left (\frac {13 \left (\frac {\frac {\frac {\frac {\int \frac {1}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{5/2}}d(i a \tan (c+d x))}{2 a}-\frac {1}{5 a (a+i a \tan (c+d x))^{5/2}}}{2 a}-\frac {1}{7 a (a+i a \tan (c+d x))^{7/2}}}{2 a}-\frac {1}{9 a (a+i a \tan (c+d x))^{9/2}}}{2 a}-\frac {1}{11 a (a+i a \tan (c+d x))^{11/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{11/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{11/2}}\right )}{d}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle -\frac {i a^5 \left (\frac {15 \left (\frac {13 \left (\frac {\frac {\frac {\frac {\frac {\int \frac {1}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{3/2}}d(i a \tan (c+d x))}{2 a}-\frac {1}{3 a (a+i a \tan (c+d x))^{3/2}}}{2 a}-\frac {1}{5 a (a+i a \tan (c+d x))^{5/2}}}{2 a}-\frac {1}{7 a (a+i a \tan (c+d x))^{7/2}}}{2 a}-\frac {1}{9 a (a+i a \tan (c+d x))^{9/2}}}{2 a}-\frac {1}{11 a (a+i a \tan (c+d x))^{11/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{11/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{11/2}}\right )}{d}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle -\frac {i a^5 \left (\frac {15 \left (\frac {13 \left (\frac {\frac {\frac {\frac {\frac {\frac {\int \frac {1}{(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{2 a}-\frac {1}{a \sqrt {a+i a \tan (c+d x)}}}{2 a}-\frac {1}{3 a (a+i a \tan (c+d x))^{3/2}}}{2 a}-\frac {1}{5 a (a+i a \tan (c+d x))^{5/2}}}{2 a}-\frac {1}{7 a (a+i a \tan (c+d x))^{7/2}}}{2 a}-\frac {1}{9 a (a+i a \tan (c+d x))^{9/2}}}{2 a}-\frac {1}{11 a (a+i a \tan (c+d x))^{11/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{11/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{11/2}}\right )}{d}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {i a^5 \left (\frac {15 \left (\frac {13 \left (\frac {\frac {\frac {\frac {\frac {\frac {\int \frac {1}{a^2 \tan ^2(c+d x)+2 a}d\sqrt {i \tan (c+d x) a+a}}{a}-\frac {1}{a \sqrt {a+i a \tan (c+d x)}}}{2 a}-\frac {1}{3 a (a+i a \tan (c+d x))^{3/2}}}{2 a}-\frac {1}{5 a (a+i a \tan (c+d x))^{5/2}}}{2 a}-\frac {1}{7 a (a+i a \tan (c+d x))^{7/2}}}{2 a}-\frac {1}{9 a (a+i a \tan (c+d x))^{9/2}}}{2 a}-\frac {1}{11 a (a+i a \tan (c+d x))^{11/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{11/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{11/2}}\right )}{d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {i a^5 \left (\frac {15 \left (\frac {13 \left (\frac {\frac {\frac {\frac {\frac {\frac {i \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2}}\right )}{\sqrt {2} a^{3/2}}-\frac {1}{a \sqrt {a+i a \tan (c+d x)}}}{2 a}-\frac {1}{3 a (a+i a \tan (c+d x))^{3/2}}}{2 a}-\frac {1}{5 a (a+i a \tan (c+d x))^{5/2}}}{2 a}-\frac {1}{7 a (a+i a \tan (c+d x))^{7/2}}}{2 a}-\frac {1}{9 a (a+i a \tan (c+d x))^{9/2}}}{2 a}-\frac {1}{11 a (a+i a \tan (c+d x))^{11/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{11/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{11/2}}\right )}{d}\) |
Input:
Int[Cos[c + d*x]^4/(a + I*a*Tan[c + d*x])^(7/2),x]
Output:
((-I)*a^5*(1/(4*a*(a - I*a*Tan[c + d*x])^2*(a + I*a*Tan[c + d*x])^(11/2)) + (15*(1/(2*a*(a - I*a*Tan[c + d*x])*(a + I*a*Tan[c + d*x])^(11/2)) + (13* (-1/11*1/(a*(a + I*a*Tan[c + d*x])^(11/2)) + (-1/9*1/(a*(a + I*a*Tan[c + d *x])^(9/2)) + (-1/7*1/(a*(a + I*a*Tan[c + d*x])^(7/2)) + (-1/5*1/(a*(a + I *a*Tan[c + d*x])^(5/2)) + (-1/3*1/(a*(a + I*a*Tan[c + d*x])^(3/2)) + ((I*A rcTan[(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))/(2*a))/(2*a))/(2*a))/(2*a)))/(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. 500 vs. \(2 (247 ) = 494\).
Time = 10.43 (sec) , antiderivative size = 501, normalized size of antiderivative = 1.62
| method | result | size |
| default | \(\frac {45045 \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2} \left (-1+8 \cos \left (d x +c \right )^{3}+4 \cos \left (d x +c \right )^{2}-4 \cos \left (d x +c \right )\right ) \operatorname {arctanh}\left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+i}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+45045 i \operatorname {arctanh}\left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+i}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \left (-4-8 \cos \left (d x +c \right )+8 \sec \left (d x +c \right )+3 \sec \left (d x +c \right )^{2}-\sec \left (d x +c \right )^{3}\right )+45045 i \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \tan \left (d x +c \right ) \left (-\sec \left (d x +c \right )^{2}+4\right )+45045 \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (-3 \sec \left (d x +c \right )^{2}+4\right )+14 i \tan \left (d x +c \right ) \sec \left (d x +c \right ) \left (-6435-5760 \cos \left (d x +c \right )^{6}-5760 \cos \left (d x +c \right )^{5}-11440 \cos \left (d x +c \right )^{4}-11440 \cos \left (d x +c \right )^{3}+39468 \cos \left (d x +c \right )^{2}+13728 \cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (147576-18816 \cos \left (d x +c \right )^{5}-18816 \cos \left (d x +c \right )^{4}-50960 \cos \left (d x +c \right )^{3}-50960 \cos \left (d x +c \right )^{2}+327756 \cos \left (d x +c \right )-165165 \sec \left (d x +c \right )-30030 \sec \left (d x +c \right )^{2}\right )}{473088 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 )}\, \left (-\tan \left (d x +c \right )+i\right )^{3} a^{3}}\) | \(501\) |
Input:
int(cos(d*x+c)^4/(a+I*a*tan(d*x+c))^(7/2),x,method=_RETURNVERBOSE)
Output:
1/473088/d/(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)/(-tan(d*x+c)+I)^3*(45045*tan(d*x+c)*sec(d*x+c)^2*(-1+8*cos(d *x+c)^3+4*cos(d*x+c)^2-4*cos(d*x+c))*arctanh(1/2/(-cos(d*x+c)/(cos(d*x+c)+ 1))^(1/2)*(cot(d*x+c)-csc(d*x+c)+I))+45045*I*arctanh(1/2/(-cos(d*x+c)/(cos (d*x+c)+1))^(1/2)*(cot(d*x+c)-csc(d*x+c)+I))*(-4-8*cos(d*x+c)+8*sec(d*x+c) +3*sec(d*x+c)^2-sec(d*x+c)^3)+45045*I*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1 ))^(1/2)*tan(d*x+c)*(-sec(d*x+c)^2+4)+45045*2^(1/2)*(-2*cos(d*x+c)/(cos(d* x+c)+1))^(1/2)*(-3*sec(d*x+c)^2+4)+14*I*tan(d*x+c)*sec(d*x+c)*(-6435-5760* cos(d*x+c)^6-5760*cos(d*x+c)^5-11440*cos(d*x+c)^4-11440*cos(d*x+c)^3+39468 *cos(d*x+c)^2+13728*cos(d*x+c))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+2*(-cos (d*x+c)/(cos(d*x+c)+1))^(1/2)*(147576-18816*cos(d*x+c)^5-18816*cos(d*x+c)^ 4-50960*cos(d*x+c)^3-50960*cos(d*x+c)^2+327756*cos(d*x+c)-165165*sec(d*x+c )-30030*sec(d*x+c)^2))/a^3
Time = 0.11 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.09 \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {{\left (-45045 i \, \sqrt {\frac {1}{2}} a^{4} d \sqrt {\frac {1}{a^{7} d^{2}}} e^{\left (11 i \, d x + 11 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{7} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 45045 i \, \sqrt {\frac {1}{2}} a^{4} d \sqrt {\frac {1}{a^{7} d^{2}}} e^{\left (11 i \, d x + 11 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{7} d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-462 i \, e^{\left (16 i \, d x + 16 i \, c\right )} - 7161 i \, e^{\left (14 i \, d x + 14 i \, c\right )} + 47413 i \, e^{\left (12 i \, d x + 12 i \, c\right )} + 78800 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 38512 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 19552 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 7184 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 1624 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 168 i\right )}\right )} e^{\left (-11 i \, d x - 11 i \, c\right )}}{473088 \, a^{4} d} \] Input:
integrate(cos(d*x+c)^4/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="fricas")
Output:
1/473088*(-45045*I*sqrt(1/2)*a^4*d*sqrt(1/(a^7*d^2))*e^(11*I*d*x + 11*I*c) *log(4*(sqrt(2)*sqrt(1/2)*(a^4*d*e^(2*I*d*x + 2*I*c) + a^4*d)*sqrt(a/(e^(2 *I*d*x + 2*I*c) + 1))*sqrt(1/(a^7*d^2)) + a*e^(I*d*x + I*c))*e^(-I*d*x - I *c)) + 45045*I*sqrt(1/2)*a^4*d*sqrt(1/(a^7*d^2))*e^(11*I*d*x + 11*I*c)*log (-4*(sqrt(2)*sqrt(1/2)*(a^4*d*e^(2*I*d*x + 2*I*c) + a^4*d)*sqrt(a/(e^(2*I* d*x + 2*I*c) + 1))*sqrt(1/(a^7*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))*(-462*I*e^(16*I*d*x + 16*I*c ) - 7161*I*e^(14*I*d*x + 14*I*c) + 47413*I*e^(12*I*d*x + 12*I*c) + 78800*I *e^(10*I*d*x + 10*I*c) + 38512*I*e^(8*I*d*x + 8*I*c) + 19552*I*e^(6*I*d*x + 6*I*c) + 7184*I*e^(4*I*d*x + 4*I*c) + 1624*I*e^(2*I*d*x + 2*I*c) + 168*I ))*e^(-11*I*d*x - 11*I*c)/(a^4*d)
Timed out. \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**4/(a+I*a*tan(d*x+c))**(7/2),x)
Output:
Timed out
Time = 0.12 (sec) , antiderivative size = 249, normalized size of antiderivative = 0.80 \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {i \, {\left (\frac {4 \, {\left (45045 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{7} - 150150 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{6} a + 96096 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} a^{2} + 27456 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a^{3} + 18304 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{4} + 16640 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{5} + 17920 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{6} + 21504 \, a^{7}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {15}{2}} a^{2} - 4 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {13}{2}} a^{3} + 4 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {11}{2}} a^{4}} + \frac {45045 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {5}{2}}}\right )}}{946176 \, a d} \] Input:
integrate(cos(d*x+c)^4/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="maxima")
Output:
1/946176*I*(4*(45045*(I*a*tan(d*x + c) + a)^7 - 150150*(I*a*tan(d*x + c) + a)^6*a + 96096*(I*a*tan(d*x + c) + a)^5*a^2 + 27456*(I*a*tan(d*x + c) + a )^4*a^3 + 18304*(I*a*tan(d*x + c) + a)^3*a^4 + 16640*(I*a*tan(d*x + c) + a )^2*a^5 + 17920*(I*a*tan(d*x + c) + a)*a^6 + 21504*a^7)/((I*a*tan(d*x + c) + a)^(15/2)*a^2 - 4*(I*a*tan(d*x + c) + a)^(13/2)*a^3 + 4*(I*a*tan(d*x + c) + a)^(11/2)*a^4) + 45045*sqrt(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)))/a^(5/2))/(a* d)
Exception generated. \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(cos(d*x+c)^4/(a+I*a*tan(d*x+c))^(7/2),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. \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^4}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2}} \,d x \] Input:
int(cos(c + d*x)^4/(a + a*tan(c + d*x)*1i)^(7/2),x)
Output:
int(cos(c + d*x)^4/(a + a*tan(c + d*x)*1i)^(7/2), x)
\[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{4} \tan \left (d x +c \right )}{\tan \left (d x +c \right )^{5} i +3 \tan \left (d x +c \right )^{4}-2 \tan \left (d x +c \right )^{3} i +2 \tan \left (d x +c \right )^{2}-3 \tan \left (d x +c \right ) i -1}d x \right ) i -\left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{4}}{\tan \left (d x +c \right )^{5} i +3 \tan \left (d x +c \right )^{4}-2 \tan \left (d x +c \right )^{3} i +2 \tan \left (d x +c \right )^{2}-3 \tan \left (d x +c \right ) i -1}d x \right )\right )}{a^{4}} \] Input:
int(cos(d*x+c)^4/(a+I*a*tan(d*x+c))^(7/2),x)
Output:
(sqrt(a)*(int((sqrt(tan(c + d*x)*i + 1)*cos(c + d*x)**4*tan(c + d*x))/(tan (c + d*x)**5*i + 3*tan(c + d*x)**4 - 2*tan(c + d*x)**3*i + 2*tan(c + d*x)* *2 - 3*tan(c + d*x)*i - 1),x)*i - int((sqrt(tan(c + d*x)*i + 1)*cos(c + d* x)**4)/(tan(c + d*x)**5*i + 3*tan(c + d*x)**4 - 2*tan(c + d*x)**3*i + 2*ta n(c + d*x)**2 - 3*tan(c + d*x)*i - 1),x)))/a**4