Integrand size = 26, antiderivative size = 172 \[ \int \frac {\tan ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}-\frac {\tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {28 i \sqrt {a+i a \tan (c+d x)}}{5 a d}-\frac {7 i \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 a d}-\frac {23 i (a+i a \tan (c+d x))^{3/2}}{15 a^2 d} \]
-1/2*I*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/d*2^(1/2)/a^( 1/2)+28/5*I*(a+I*a*tan(d*x+c))^(1/2)/a/d-7/5*I*(a+I*a*tan(d*x+c))^(1/2)*ta n(d*x+c)^2/a/d-tan(d*x+c)^3/d/(a+I*a*tan(d*x+c))^(1/2)-23/15*I*(a+I*a*tan( d*x+c))^(3/2)/a^2/d
Time = 0.70 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.61 \[ \int \frac {\tan ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}+\frac {61 i-38 \tan (c+d x)+2 i \tan ^2(c+d x)+6 \tan ^3(c+d x)}{15 d \sqrt {a+i a \tan (c+d x)}} \]
((-I)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*Sqrt [a]*d) + (61*I - 38*Tan[c + d*x] + (2*I)*Tan[c + d*x]^2 + 6*Tan[c + d*x]^3 )/(15*d*Sqrt[a + I*a*Tan[c + d*x]])
Time = 0.90 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.06, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4041, 27, 3042, 4080, 27, 3042, 4075, 3042, 4010, 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.
| \(\displaystyle \int \frac {\tan ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^4}{\sqrt {a+i a \tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 4041 |
| \(\displaystyle -\frac {\int -\frac {1}{2} \tan ^2(c+d x) \sqrt {i \tan (c+d x) a+a} (6 a-7 i a \tan (c+d x))dx}{a^2}-\frac {\tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \tan ^2(c+d x) \sqrt {i \tan (c+d x) a+a} (6 a-7 i a \tan (c+d x))dx}{2 a^2}-\frac {\tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \tan (c+d x)^2 \sqrt {i \tan (c+d x) a+a} (6 a-7 i a \tan (c+d x))dx}{2 a^2}-\frac {\tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 4080 |
| \(\displaystyle \frac {\frac {2 \int \frac {1}{2} \tan (c+d x) \sqrt {i \tan (c+d x) a+a} \left (23 \tan (c+d x) a^2+28 i a^2\right )dx}{5 a}-\frac {14 i a \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}}{2 a^2}-\frac {\tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \tan (c+d x) \sqrt {i \tan (c+d x) a+a} \left (23 \tan (c+d x) a^2+28 i a^2\right )dx}{5 a}-\frac {14 i a \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}}{2 a^2}-\frac {\tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \tan (c+d x) \sqrt {i \tan (c+d x) a+a} \left (23 \tan (c+d x) a^2+28 i a^2\right )dx}{5 a}-\frac {14 i a \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}}{2 a^2}-\frac {\tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 4075 |
| \(\displaystyle \frac {\frac {\int \sqrt {i \tan (c+d x) a+a} \left (28 i a^2 \tan (c+d x)-23 a^2\right )dx-\frac {46 i a (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}-\frac {14 i a \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}}{2 a^2}-\frac {\tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \sqrt {i \tan (c+d x) a+a} \left (28 i a^2 \tan (c+d x)-23 a^2\right )dx-\frac {46 i a (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}-\frac {14 i a \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}}{2 a^2}-\frac {\tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 4010 |
| \(\displaystyle \frac {\frac {5 a^2 \int \sqrt {i \tan (c+d x) a+a}dx+\frac {56 i a^2 \sqrt {a+i a \tan (c+d x)}}{d}-\frac {46 i a (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}-\frac {14 i a \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}}{2 a^2}-\frac {\tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {5 a^2 \int \sqrt {i \tan (c+d x) a+a}dx+\frac {56 i a^2 \sqrt {a+i a \tan (c+d x)}}{d}-\frac {46 i a (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}-\frac {14 i a \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}}{2 a^2}-\frac {\tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 3961 |
| \(\displaystyle \frac {\frac {-\frac {10 i a^3 \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}+\frac {56 i a^2 \sqrt {a+i a \tan (c+d x)}}{d}-\frac {46 i a (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}-\frac {14 i a \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}}{2 a^2}-\frac {\tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {-\frac {5 i \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {56 i a^2 \sqrt {a+i a \tan (c+d x)}}{d}-\frac {46 i a (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}-\frac {14 i a \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}}{2 a^2}-\frac {\tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\) |
-(Tan[c + d*x]^3/(d*Sqrt[a + I*a*Tan[c + d*x]])) + ((((-14*I)/5)*a*Tan[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/d + (((-5*I)*Sqrt[2]*a^(5/2)*ArcTanh[S qrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + ((56*I)*a^2*Sqrt[a + I*a *Tan[c + d*x]])/d - (((46*I)/3)*a*(a + I*a*Tan[c + d*x])^(3/2))/d)/(5*a))/ (2*a^2)
3.2.9.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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[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]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Simp [(b*c + a*d)/b Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e , f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && !LtQ[m, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b*c - a*d))*(a + b*Tan[e + f*x])^m* ((c + d*Tan[e + f*x])^(n - 1)/(2*a*f*m)), x] + Simp[1/(2*a^2*m) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[c*(a*c*m + b*d*(n - 1)) - d*(b*c*m + a*d*(n - 1)) - d*(b*d*(m - n + 1) - a*c*(m + n - 1))*Ta n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, 0] && GtQ[n, 1] && (In tegerQ[m] || IntegersQ[2*m, 2*n])
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[B *d*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f* x])^m*Simp[A*c - B*d + (B*c + A*d)*Tan[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && !LeQ[m, -1]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[B*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(f*(m + n))), x] + Simp[ 1/(a*(m + n)) Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Sim p[a*A*c*(m + n) - B*(b*c*m + a*d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*T an[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[n, 0]
Time = 0.80 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.66
| method | result | size |
| derivativedivides | \(\frac {2 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+2 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}+\frac {a^{3}}{2 \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {a^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4}\right )}{d \,a^{3}}\) | \(113\) |
| default | \(\frac {2 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+2 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}+\frac {a^{3}}{2 \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {a^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4}\right )}{d \,a^{3}}\) | \(113\) |
2*I/d/a^3*(1/5*(a+I*a*tan(d*x+c))^(5/2)-2/3*a*(a+I*a*tan(d*x+c))^(3/2)+2*a ^2*(a+I*a*tan(d*x+c))^(1/2)+1/2*a^3/(a+I*a*tan(d*x+c))^(1/2)-1/4*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. 343 vs. \(2 (131) = 262\).
Time = 0.24 (sec) , antiderivative size = 343, normalized size of antiderivative = 1.99 \[ \int \frac {\tan ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {15 \, \sqrt {2} {\left (i \, a d e^{\left (5 i \, d x + 5 i \, c\right )} + 2 i \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + i \, a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {1}{a d^{2}}} \log \left (4 \, {\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 15 \, \sqrt {2} {\left (-i \, a d e^{\left (5 i \, d x + 5 i \, c\right )} - 2 i \, a d e^{\left (3 i \, d x + 3 i \, c\right )} - i \, a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {1}{a d^{2}}} \log \left (-4 \, {\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 2 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-103 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 205 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 165 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 15 i\right )}}{60 \, {\left (a d e^{\left (5 i \, d x + 5 i \, c\right )} + 2 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )}} \]
-1/60*(15*sqrt(2)*(I*a*d*e^(5*I*d*x + 5*I*c) + 2*I*a*d*e^(3*I*d*x + 3*I*c) + I*a*d*e^(I*d*x + I*c))*sqrt(1/(a*d^2))*log(4*((a*d*e^(2*I*d*x + 2*I*c) + a*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a*d^2)) + a*e^(I*d*x + I* c))*e^(-I*d*x - I*c)) + 15*sqrt(2)*(-I*a*d*e^(5*I*d*x + 5*I*c) - 2*I*a*d*e ^(3*I*d*x + 3*I*c) - I*a*d*e^(I*d*x + I*c))*sqrt(1/(a*d^2))*log(-4*((a*d*e ^(2*I*d*x + 2*I*c) + a*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a*d^2) ) - a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + 2*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-103*I*e^(6*I*d*x + 6*I*c) - 205*I*e^(4*I*d*x + 4*I*c) - 165 *I*e^(2*I*d*x + 2*I*c) - 15*I))/(a*d*e^(5*I*d*x + 5*I*c) + 2*a*d*e^(3*I*d* x + 3*I*c) + a*d*e^(I*d*x + I*c))
\[ \int \frac {\tan ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {\tan ^{4}{\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \]
Time = 0.27 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.80 \[ \int \frac {\tan ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {i \, {\left (15 \, \sqrt {2} a^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + 24 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2} - 80 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{3} + 240 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{4} + \frac {60 \, a^{5}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}}{60 \, a^{5} d} \]
1/60*I*(15*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))) + 24*(I*a*tan(d*x + c ) + a)^(5/2)*a^2 - 80*(I*a*tan(d*x + c) + a)^(3/2)*a^3 + 240*sqrt(I*a*tan( d*x + c) + a)*a^4 + 60*a^5/sqrt(I*a*tan(d*x + c) + a))/(a^5*d)
\[ \int \frac {\tan ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int { \frac {\tan \left (d x + c\right )^{4}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}} \,d x } \]
Time = 4.36 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.75 \[ \int \frac {\tan ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {1{}\mathrm {i}}{d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}+\frac {\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,4{}\mathrm {i}}{a\,d}-\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,4{}\mathrm {i}}{3\,a^2\,d}+\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,2{}\mathrm {i}}{5\,a^3\,d}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{2\,\sqrt {-a}\,d} \]