Integrand size = 17, antiderivative size = 162 \[ \int \frac {1}{(a+i a \tan (c+d x))^{7/2}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{8 \sqrt {2} a^{7/2} d}+\frac {i}{7 d (a+i a \tan (c+d x))^{7/2}}+\frac {i}{10 a d (a+i a \tan (c+d x))^{5/2}}+\frac {i}{12 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {i}{8 a^3 d \sqrt {a+i a \tan (c+d x)}} \]
-1/16*I*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/a^(7/2)/d*2^ (1/2)+1/8*I/a^3/d/(a+I*a*tan(d*x+c))^(1/2)+1/7*I/d/(a+I*a*tan(d*x+c))^(7/2 )+1/10*I/a/d/(a+I*a*tan(d*x+c))^(5/2)+1/12*I/a^2/d/(a+I*a*tan(d*x+c))^(3/2 )
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.56 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.31 \[ \int \frac {1}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {i \operatorname {Hypergeometric2F1}\left (-\frac {7}{2},1,-\frac {5}{2},\frac {1}{2} (1+i \tan (c+d x))\right )}{7 d (a+i a \tan (c+d x))^{7/2}} \]
((I/7)*Hypergeometric2F1[-7/2, 1, -5/2, (1 + I*Tan[c + d*x])/2])/(d*(a + I *a*Tan[c + d*x])^(7/2))
Time = 0.59 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.07, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.647, Rules used = {3042, 3960, 3042, 3960, 3042, 3960, 3042, 3960, 3042, 3961, 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.
\(\displaystyle \int \frac {1}{(a+i a \tan (c+d x))^{7/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a+i a \tan (c+d x))^{7/2}}dx\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle \frac {\int \frac {1}{(i \tan (c+d x) a+a)^{5/2}}dx}{2 a}+\frac {i}{7 d (a+i a \tan (c+d x))^{7/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {1}{(i \tan (c+d x) a+a)^{5/2}}dx}{2 a}+\frac {i}{7 d (a+i a \tan (c+d x))^{7/2}}\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle \frac {\frac {\int \frac {1}{(i \tan (c+d x) a+a)^{3/2}}dx}{2 a}+\frac {i}{5 d (a+i a \tan (c+d x))^{5/2}}}{2 a}+\frac {i}{7 d (a+i a \tan (c+d x))^{7/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {1}{(i \tan (c+d x) a+a)^{3/2}}dx}{2 a}+\frac {i}{5 d (a+i a \tan (c+d x))^{5/2}}}{2 a}+\frac {i}{7 d (a+i a \tan (c+d x))^{7/2}}\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle \frac {\frac {\frac {\int \frac {1}{\sqrt {i \tan (c+d x) a+a}}dx}{2 a}+\frac {i}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a}+\frac {i}{5 d (a+i a \tan (c+d x))^{5/2}}}{2 a}+\frac {i}{7 d (a+i a \tan (c+d x))^{7/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {\int \frac {1}{\sqrt {i \tan (c+d x) a+a}}dx}{2 a}+\frac {i}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a}+\frac {i}{5 d (a+i a \tan (c+d x))^{5/2}}}{2 a}+\frac {i}{7 d (a+i a \tan (c+d x))^{7/2}}\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle \frac {\frac {\frac {\frac {\int \sqrt {i \tan (c+d x) a+a}dx}{2 a}+\frac {i}{d \sqrt {a+i a \tan (c+d x)}}}{2 a}+\frac {i}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a}+\frac {i}{5 d (a+i a \tan (c+d x))^{5/2}}}{2 a}+\frac {i}{7 d (a+i a \tan (c+d x))^{7/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {\frac {\int \sqrt {i \tan (c+d x) a+a}dx}{2 a}+\frac {i}{d \sqrt {a+i a \tan (c+d x)}}}{2 a}+\frac {i}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a}+\frac {i}{5 d (a+i a \tan (c+d x))^{5/2}}}{2 a}+\frac {i}{7 d (a+i a \tan (c+d x))^{7/2}}\) |
\(\Big \downarrow \) 3961 |
\(\displaystyle \frac {\frac {\frac {\frac {i}{d \sqrt {a+i a \tan (c+d x)}}-\frac {i \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}}{2 a}+\frac {i}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a}+\frac {i}{5 d (a+i a \tan (c+d x))^{5/2}}}{2 a}+\frac {i}{7 d (a+i a \tan (c+d x))^{7/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\frac {\frac {i}{d \sqrt {a+i a \tan (c+d x)}}-\frac {i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}}{2 a}+\frac {i}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a}+\frac {i}{5 d (a+i a \tan (c+d x))^{5/2}}}{2 a}+\frac {i}{7 d (a+i a \tan (c+d x))^{7/2}}\) |
(I/7)/(d*(a + I*a*Tan[c + d*x])^(7/2)) + ((I/5)/(d*(a + I*a*Tan[c + d*x])^ (5/2)) + ((I/3)/(d*(a + I*a*Tan[c + d*x])^(3/2)) + (((-I)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*Sqrt[a]*d) + I/(d*Sqrt[a + I*a*Tan[c + d*x]]))/(2*a))/(2*a))/(2*a)
3.2.34.3.1 Defintions of rubi rules used
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[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a*((a + b*Tan[c + d*x])^n/(2*b*d*n)), x] + Simp[1/(2*a) Int[(a + b*Tan[c + d*x])^ (n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && LtQ[n, 0]
Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*(b/d) Subst[Int[1/(2*a - x^2), x], x, Sqrt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a , b, c, d}, x] && EqQ[a^2 + b^2, 0]
Time = 0.90 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.72
method | result | size |
derivativedivides | \(\frac {2 i a \left (-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{32 a^{\frac {9}{2}}}+\frac {1}{16 a^{4} \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {1}{24 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {1}{20 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}+\frac {1}{14 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}\right )}{d}\) | \(116\) |
default | \(\frac {2 i a \left (-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{32 a^{\frac {9}{2}}}+\frac {1}{16 a^{4} \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {1}{24 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {1}{20 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}+\frac {1}{14 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}\right )}{d}\) | \(116\) |
2*I/d*a*(-1/32/a^(9/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2 )/a^(1/2))+1/16/a^4/(a+I*a*tan(d*x+c))^(1/2)+1/24/a^3/(a+I*a*tan(d*x+c))^( 3/2)+1/20/a^2/(a+I*a*tan(d*x+c))^(5/2)+1/14/a/(a+I*a*tan(d*x+c))^(7/2))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 294 vs. \(2 (115) = 230\).
Time = 0.25 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.81 \[ \int \frac {1}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {{\left (-105 i \, \sqrt {\frac {1}{2}} a^{4} d \sqrt {\frac {1}{a^{7} d^{2}}} e^{\left (7 i \, d x + 7 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{7} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 105 i \, \sqrt {\frac {1}{2}} a^{4} d \sqrt {\frac {1}{a^{7} d^{2}}} e^{\left (7 i \, d x + 7 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{7} d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (176 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 298 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 188 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 81 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 15 i\right )}\right )} e^{\left (-7 i \, d x - 7 i \, c\right )}}{1680 \, a^{4} d} \]
1/1680*(-105*I*sqrt(1/2)*a^4*d*sqrt(1/(a^7*d^2))*e^(7*I*d*x + 7*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)) + 105*I*sqrt(1/2)*a^4*d*sqrt(1/(a^7*d^2))*e^(7*I*d*x + 7*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))*(176*I*e^(8*I*d*x + 8*I*c) + 298*I*e^( 6*I*d*x + 6*I*c) + 188*I*e^(4*I*d*x + 4*I*c) + 81*I*e^(2*I*d*x + 2*I*c) + 15*I))*e^(-7*I*d*x - 7*I*c)/(a^4*d)
\[ \int \frac {1}{(a+i a \tan (c+d x))^{7/2}} \, dx=\int \frac {1}{\left (i a \tan {\left (c + d x \right )} + a\right )^{\frac {7}{2}}}\, dx \]
Time = 0.47 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.85 \[ \int \frac {1}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {i \, {\left (\frac {105 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {5}{2}}} + \frac {4 \, {\left (105 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} + 70 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a + 84 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{2} + 120 \, a^{3}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{2}}\right )}}{3360 \, a d} \]
1/3360*I*(105*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) + 4*(105*(I*a*tan( d*x + c) + a)^3 + 70*(I*a*tan(d*x + c) + a)^2*a + 84*(I*a*tan(d*x + c) + a )*a^2 + 120*a^3)/((I*a*tan(d*x + c) + a)^(7/2)*a^2))/(a*d)
\[ \int \frac {1}{(a+i a \tan (c+d x))^{7/2}} \, dx=\int { \frac {1}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}}} \,d x } \]
Time = 5.17 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.80 \[ \int \frac {1}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {\frac {1{}\mathrm {i}}{7\,d}+\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2\,1{}\mathrm {i}}{12\,a^2\,d}+\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{8\,a^3\,d}+\frac {\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{10\,a\,d}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2}}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{16\,{\left (-a\right )}^{7/2}\,d} \]