Integrand size = 26, antiderivative size = 127 \[ \int \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {\sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {8 \sqrt {a+i a \tan (c+d x)}}{5 d}+\frac {2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {2 (a+i a \tan (c+d x))^{3/2}}{15 a d} \]
arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*a^(1/2)/d-8/ 5*(a+I*a*tan(d*x+c))^(1/2)/d+2/5*(a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^2/d-2 /15*(a+I*a*tan(d*x+c))^(3/2)/a/d
Time = 0.59 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.71 \[ \int \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {15 \sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )+2 \sqrt {a+i a \tan (c+d x)} \left (-13-i \tan (c+d x)+3 \tan ^2(c+d x)\right )}{15 d} \]
(15*Sqrt[2]*Sqrt[a]*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])] + 2*Sqrt[a + I*a*Tan[c + d*x]]*(-13 - I*Tan[c + d*x] + 3*Tan[c + d*x]^2))/ (15*d)
Time = 0.64 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.04, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3042, 4043, 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.
\(\displaystyle \int \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x)^3 \sqrt {a+i a \tan (c+d x)}dx\) |
\(\Big \downarrow \) 4043 |
\(\displaystyle \frac {2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {2 \int \frac {1}{2} \tan (c+d x) \sqrt {i \tan (c+d x) a+a} (i \tan (c+d x) a+4 a)dx}{5 a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {\int \tan (c+d x) \sqrt {i \tan (c+d x) a+a} (i \tan (c+d x) a+4 a)dx}{5 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {\int \tan (c+d x) \sqrt {i \tan (c+d x) a+a} (i \tan (c+d x) a+4 a)dx}{5 a}\) |
\(\Big \downarrow \) 4075 |
\(\displaystyle \frac {2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {\int \sqrt {i \tan (c+d x) a+a} (4 a \tan (c+d x)-i a)dx+\frac {2 (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {\int \sqrt {i \tan (c+d x) a+a} (4 a \tan (c+d x)-i a)dx+\frac {2 (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}\) |
\(\Big \downarrow \) 4010 |
\(\displaystyle \frac {2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {-5 i a \int \sqrt {i \tan (c+d x) a+a}dx+\frac {2 (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {8 a \sqrt {a+i a \tan (c+d x)}}{d}}{5 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {-5 i a \int \sqrt {i \tan (c+d x) a+a}dx+\frac {2 (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {8 a \sqrt {a+i a \tan (c+d x)}}{d}}{5 a}\) |
\(\Big \downarrow \) 3961 |
\(\displaystyle \frac {2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {-\frac {10 a^2 \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}+\frac {8 a \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {-\frac {5 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {8 a \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}\) |
(2*Tan[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(5*d) - ((-5*Sqrt[2]*a^(3/2) *ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + (8*a*Sqrt[a + I*a*Tan[c + d*x]])/d + (2*(a + I*a*Tan[c + d*x])^(3/2))/(3*d))/(5*a)
3.1.85.3.1 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[d*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Simp[1/(a*(m + n - 1)) Int[(a + b *Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 2)*Simp[d*(b*c*m + a*d*(-1 + n)) - a*c^2*(m + n - 1) + d*(b*d*m - a*c*(m + 2*n - 2))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2 , 0] && NeQ[c^2 + d^2, 0] && GtQ[n, 1] && NeQ[m + n - 1, 0] && (IntegerQ[n] || 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]
Time = 1.05 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.73
method | result | size |
derivativedivides | \(\frac {-\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}+a^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{a^{2} d}\) | \(93\) |
default | \(\frac {-\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}+a^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{a^{2} d}\) | \(93\) |
2/d/a^2*(-1/5*(a+I*a*tan(d*x+c))^(5/2)+1/3*a*(a+I*a*tan(d*x+c))^(3/2)-a^2* (a+I*a*tan(d*x+c))^(1/2)+1/2*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. 285 vs. \(2 (98) = 196\).
Time = 0.25 (sec) , antiderivative size = 285, normalized size of antiderivative = 2.24 \[ \int \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {15 \, \sqrt {2} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{d^{2}}} \log \left (4 \, {\left ({\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {a}{d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 15 \, \sqrt {2} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{d^{2}}} \log \left (-4 \, {\left ({\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {a}{d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 4 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (17 \, e^{\left (5 i \, d x + 5 i \, c\right )} + 20 \, e^{\left (3 i \, d x + 3 i \, c\right )} + 15 \, e^{\left (i \, d x + i \, c\right )}\right )}}{30 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
1/30*(15*sqrt(2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*sqr t(a/d^2)*log(4*((d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(a/d^2) + a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 15*sqrt(2)*(d*e^( 4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/d^2)*log(-4*((d*e^( 2*I*d*x + 2*I*c) + d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(a/d^2) - a*e^ (I*d*x + I*c))*e^(-I*d*x - I*c)) - 4*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(17*e^(5*I*d*x + 5*I*c) + 20*e^(3*I*d*x + 3*I*c) + 15*e^(I*d*x + I*c) ))/(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)
\[ \int \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\int \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )} \tan ^{3}{\left (c + d x \right )}\, dx \]
Time = 0.29 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.94 \[ \int \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {15 \, \sqrt {2} a^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + 12 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2} - 20 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{3} + 60 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{4}}{30 \, a^{4} d} \]
-1/30*(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))) + 12*(I*a*tan(d*x + c) + a)^(5/2)*a^2 - 20*(I*a*tan(d*x + c) + a)^(3/2)*a^3 + 60*sqrt(I*a*tan(d* x + c) + a)*a^4)/(a^4*d)
\[ \int \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\int { \sqrt {i \, a \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{3} \,d x } \]
Time = 5.26 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.79 \[ \int \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {2\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{d}+\frac {2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,a\,d}-\frac {2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{5\,a^2\,d}-\frac {\sqrt {2}\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {a}}\right )\,1{}\mathrm {i}}{d} \]