Integrand size = 17, antiderivative size = 130 \[ \int (a+i a \tan (c+d x))^{7/2} \, dx=-\frac {8 i \sqrt {2} a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {8 i a^3 \sqrt {a+i a \tan (c+d x)}}{d}+\frac {4 i a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 i a (a+i a \tan (c+d x))^{5/2}}{5 d} \]
-8*I*a^(7/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2) /d+8*I*a^3*(a+I*a*tan(d*x+c))^(1/2)/d+4/3*I*a^2*(a+I*a*tan(d*x+c))^(3/2)/d +2/5*I*a*(a+I*a*tan(d*x+c))^(5/2)/d
Time = 0.53 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.75 \[ \int (a+i a \tan (c+d x))^{7/2} \, dx=\frac {-120 i \sqrt {2} a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )+2 a^3 \sqrt {a+i a \tan (c+d x)} \left (73 i-16 \tan (c+d x)-3 i \tan ^2(c+d x)\right )}{15 d} \]
((-120*I)*Sqrt[2]*a^(7/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt [a])] + 2*a^3*Sqrt[a + I*a*Tan[c + d*x]]*(73*I - 16*Tan[c + d*x] - (3*I)*T an[c + d*x]^2))/(15*d)
Time = 0.47 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {3042, 3959, 3042, 3959, 3042, 3959, 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.
\(\displaystyle \int (a+i a \tan (c+d x))^{7/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+i a \tan (c+d x))^{7/2}dx\) |
\(\Big \downarrow \) 3959 |
\(\displaystyle 2 a \int (i \tan (c+d x) a+a)^{5/2}dx+\frac {2 i a (a+i a \tan (c+d x))^{5/2}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 a \int (i \tan (c+d x) a+a)^{5/2}dx+\frac {2 i a (a+i a \tan (c+d x))^{5/2}}{5 d}\) |
\(\Big \downarrow \) 3959 |
\(\displaystyle 2 a \left (2 a \int (i \tan (c+d x) a+a)^{3/2}dx+\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^{5/2}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 a \left (2 a \int (i \tan (c+d x) a+a)^{3/2}dx+\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^{5/2}}{5 d}\) |
\(\Big \downarrow \) 3959 |
\(\displaystyle 2 a \left (2 a \left (2 a \int \sqrt {i \tan (c+d x) a+a}dx+\frac {2 i a \sqrt {a+i a \tan (c+d x)}}{d}\right )+\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^{5/2}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 a \left (2 a \left (2 a \int \sqrt {i \tan (c+d x) a+a}dx+\frac {2 i a \sqrt {a+i a \tan (c+d x)}}{d}\right )+\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^{5/2}}{5 d}\) |
\(\Big \downarrow \) 3961 |
\(\displaystyle 2 a \left (2 a \left (\frac {2 i a \sqrt {a+i a \tan (c+d x)}}{d}-\frac {4 i a^2 \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}\right )+\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^{5/2}}{5 d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle 2 a \left (2 a \left (\frac {2 i a \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 i \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}\right )+\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^{5/2}}{5 d}\) |
(((2*I)/5)*a*(a + I*a*Tan[c + d*x])^(5/2))/d + 2*a*((((2*I)/3)*a*(a + I*a* Tan[c + d*x])^(3/2))/d + 2*a*(((-2*I)*Sqrt[2]*a^(3/2)*ArcTanh[Sqrt[a + I*a *Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + ((2*I)*a*Sqrt[a + I*a*Tan[c + d*x]] )/d))
3.2.7.3.1 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[b*((a + b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[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] && GtQ[n , 1]
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 = 0.92 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.71
method | result | size |
derivativedivides | \(\frac {2 i a \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+4 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}-4 a^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d}\) | \(92\) |
default | \(\frac {2 i a \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+4 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}-4 a^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d}\) | \(92\) |
2*I/d*a*(1/5*(a+I*a*tan(d*x+c))^(5/2)+2/3*a*(a+I*a*tan(d*x+c))^(3/2)+4*a^2 *(a+I*a*tan(d*x+c))^(1/2)-4*a^(5/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c)) ^(1/2)*2^(1/2)/a^(1/2)))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 321 vs. \(2 (95) = 190\).
Time = 0.24 (sec) , antiderivative size = 321, normalized size of antiderivative = 2.47 \[ \int (a+i a \tan (c+d x))^{7/2} \, dx=\frac {4 \, {\left (15 \, \sqrt {2} \sqrt {-\frac {a^{7}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {4 \, {\left (a^{4} e^{\left (i \, d x + i \, c\right )} + \sqrt {-\frac {a^{7}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{3}}\right ) - 15 \, \sqrt {2} \sqrt {-\frac {a^{7}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {4 \, {\left (a^{4} e^{\left (i \, d x + i \, c\right )} + \sqrt {-\frac {a^{7}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{3}}\right ) - 2 \, \sqrt {2} {\left (-23 i \, a^{3} e^{\left (5 i \, d x + 5 i \, c\right )} - 35 i \, a^{3} e^{\left (3 i \, d x + 3 i \, c\right )} - 15 i \, a^{3} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )}}{15 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
4/15*(15*sqrt(2)*sqrt(-a^7/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log(4*(a^4*e^(I*d*x + I*c) + sqrt(-a^7/d^2)*(I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^3) - 1 5*sqrt(2)*sqrt(-a^7/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log(4*(a^4*e^(I*d*x + I*c) + sqrt(-a^7/d^2)*(-I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^3) - 2*sqrt( 2)*(-23*I*a^3*e^(5*I*d*x + 5*I*c) - 35*I*a^3*e^(3*I*d*x + 3*I*c) - 15*I*a^ 3*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))/(d*e^(4*I*d*x + 4*I* c) + 2*d*e^(2*I*d*x + 2*I*c) + d)
\[ \int (a+i a \tan (c+d x))^{7/2} \, dx=\int \left (i a \tan {\left (c + d x \right )} + a\right )^{\frac {7}{2}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.92 \[ \int (a+i a \tan (c+d x))^{7/2} \, dx=\frac {2 i \, {\left (30 \, \sqrt {2} a^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + 3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2} + 10 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{3} + 60 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{4}\right )}}{15 \, a d} \]
2/15*I*(30*sqrt(2)*a^(9/2)*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))) + 3*(I*a*tan(d*x + c) + a)^(5/2)*a^2 + 10*(I*a*tan(d*x + c) + a)^(3/2)*a^3 + 60*sqrt(I*a*tan(d* x + c) + a)*a^4)/(a*d)
Timed out. \[ \int (a+i a \tan (c+d x))^{7/2} \, dx=\text {Timed out} \]
Time = 5.21 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.82 \[ \int (a+i a \tan (c+d x))^{7/2} \, dx=\frac {a^3\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,8{}\mathrm {i}}{d}+\frac {a^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,4{}\mathrm {i}}{3\,d}+\frac {a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,2{}\mathrm {i}}{5\,d}+\frac {\sqrt {2}\,{\left (-a\right )}^{7/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,8{}\mathrm {i}}{d} \]