Integrand size = 35, antiderivative size = 154 \[ \int (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)} \, dx=-\frac {3 i a^{5/2} \sqrt {c} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}+\frac {3 i a^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}+\frac {i a (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 f} \]
-3*I*a^(5/2)*arctan(c^(1/2)*(a+I*a*tan(f*x+e))^(1/2)/a^(1/2)/(c-I*c*tan(f* x+e))^(1/2))*c^(1/2)/f+3/2*I*a^2*(a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e ))^(1/2)/f+1/2*I*a*(c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^(3/2)/f
Time = 3.24 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.24 \[ \int (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)} \, dx=-\frac {8 i a^{5/2} \sqrt {c} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}+\frac {\sqrt {c-i c \tan (e+f x)} \left (10 i a^{5/2} \arcsin \left (\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {2} \sqrt {a}}\right )-a^2 \sqrt {1-i \tan (e+f x)} (-4 i+\tan (e+f x)) \sqrt {a+i a \tan (e+f x)}\right )}{2 f \sqrt {1-i \tan (e+f x)}} \]
((-8*I)*a^(5/2)*Sqrt[c]*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[ a]*Sqrt[c - I*c*Tan[e + f*x]])])/f + (Sqrt[c - I*c*Tan[e + f*x]]*((10*I)*a ^(5/2)*ArcSin[Sqrt[a + I*a*Tan[e + f*x]]/(Sqrt[2]*Sqrt[a])] - a^2*Sqrt[1 - I*Tan[e + f*x]]*(-4*I + Tan[e + f*x])*Sqrt[a + I*a*Tan[e + f*x]]))/(2*f*S qrt[1 - I*Tan[e + f*x]])
Time = 0.33 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3042, 4006, 60, 60, 45, 218}
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.
\(\displaystyle \int (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}dx\) |
\(\Big \downarrow \) 4006 |
\(\displaystyle \frac {a c \int \frac {(i \tan (e+f x) a+a)^{3/2}}{\sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {a c \left (\frac {3}{2} a \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)+\frac {i (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c}\right )}{f}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {a c \left (\frac {3}{2} a \left (a \int \frac {1}{\sqrt {i \tan (e+f x) a+a} \sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)+\frac {i \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c}\right )+\frac {i (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c}\right )}{f}\) |
\(\Big \downarrow \) 45 |
\(\displaystyle \frac {a c \left (\frac {3}{2} a \left (2 a \int \frac {1}{i a+\frac {i c (i \tan (e+f x) a+a)}{c-i c \tan (e+f x)}}d\frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c-i c \tan (e+f x)}}+\frac {i \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c}\right )+\frac {i (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c}\right )}{f}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {a c \left (\frac {3}{2} a \left (\frac {i \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c}-\frac {2 i \sqrt {a} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {c}}\right )+\frac {i (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c}\right )}{f}\) |
(a*c*(((I/2)*(a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c - I*c*Tan[e + f*x]])/c + (3*a*(((-2*I)*Sqrt[a]*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a] *Sqrt[c - I*c*Tan[e + f*x]])])/Sqrt[c] + (I*Sqrt[a + I*a*Tan[e + f*x]]*Sqr t[c - I*c*Tan[e + f*x]])/c))/2))/f
3.10.92.3.1 Defintions of rubi rules used
Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 2 Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre eQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0] && !GtQ[c, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(c/f) Subst[Int[(a + b*x)^(m - 1)*( c + d*x)^(n - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n }, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]
Time = 1.17 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00
method | result | size |
derivativedivides | \(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, a^{2} \left (4 i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}+3 a c \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right )-\tan \left (f x +e \right ) \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\right )}{2 f \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}\) | \(154\) |
default | \(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, a^{2} \left (4 i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}+3 a c \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right )-\tan \left (f x +e \right ) \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\right )}{2 f \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}\) | \(154\) |
1/2/f*(-c*(I*tan(f*x+e)-1))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*a^2*(4*I*(a*c *(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)+3*a*c*ln((c*a*tan(f*x+e)+(a*c*(1+tan( f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2))-tan(f*x+e)*(a*c*(1+tan(f*x+e)^2 ))^(1/2)*(a*c)^(1/2))/(a*c*(1+tan(f*x+e)^2))^(1/2)/(a*c)^(1/2)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 361 vs. \(2 (112) = 224\).
Time = 0.28 (sec) , antiderivative size = 361, normalized size of antiderivative = 2.34 \[ \int (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)} \, dx=\frac {3 \, \sqrt {\frac {a^{5} c}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {4 \, {\left (2 \, {\left (a^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - \sqrt {\frac {a^{5} c}{f^{2}}} {\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, f\right )}\right )}}{a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2}}\right ) - 3 \, \sqrt {\frac {a^{5} c}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {4 \, {\left (2 \, {\left (a^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - \sqrt {\frac {a^{5} c}{f^{2}}} {\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, f\right )}\right )}}{a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2}}\right ) - 4 \, {\left (-5 i \, a^{2} e^{\left (3 i \, f x + 3 i \, e\right )} - 3 i \, a^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{4 \, {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]
1/4*(3*sqrt(a^5*c/f^2)*(f*e^(2*I*f*x + 2*I*e) + f)*log(4*(2*(a^2*e^(3*I*f* x + 3*I*e) + a^2*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c /(e^(2*I*f*x + 2*I*e) + 1)) - sqrt(a^5*c/f^2)*(I*f*e^(2*I*f*x + 2*I*e) - I *f))/(a^2*e^(2*I*f*x + 2*I*e) + a^2)) - 3*sqrt(a^5*c/f^2)*(f*e^(2*I*f*x + 2*I*e) + f)*log(4*(2*(a^2*e^(3*I*f*x + 3*I*e) + a^2*e^(I*f*x + I*e))*sqrt( a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) - sqrt(a^5* c/f^2)*(-I*f*e^(2*I*f*x + 2*I*e) + I*f))/(a^2*e^(2*I*f*x + 2*I*e) + a^2)) - 4*(-5*I*a^2*e^(3*I*f*x + 3*I*e) - 3*I*a^2*e^(I*f*x + I*e))*sqrt(a/(e^(2* I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))/(f*e^(2*I*f*x + 2* I*e) + f)
\[ \int (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)} \, dx=\int \left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}} \sqrt {- i c \left (\tan {\left (e + f x \right )} + i\right )}\, dx \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 662 vs. \(2 (112) = 224\).
Time = 0.37 (sec) , antiderivative size = 662, normalized size of antiderivative = 4.30 \[ \int (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)} \, dx=\text {Too large to display} \]
(20*a^2*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 12*a^2*cos( 1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 20*I*a^2*sin(3/2*arctan 2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 12*I*a^2*sin(1/2*arctan2(sin(2*f* x + 2*e), cos(2*f*x + 2*e))) - 6*(a^2*cos(4*f*x + 4*e) + 2*a^2*cos(2*f*x + 2*e) + I*a^2*sin(4*f*x + 4*e) + 2*I*a^2*sin(2*f*x + 2*e) + a^2)*arctan2(c os(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), sin(1/2*arctan2(sin(2 *f*x + 2*e), cos(2*f*x + 2*e))) + 1) - 6*(a^2*cos(4*f*x + 4*e) + 2*a^2*cos (2*f*x + 2*e) + I*a^2*sin(4*f*x + 4*e) + 2*I*a^2*sin(2*f*x + 2*e) + a^2)*a rctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), -sin(1/2*arct an2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) - 3*(I*a^2*cos(4*f*x + 4*e) + 2*I*a^2*cos(2*f*x + 2*e) - a^2*sin(4*f*x + 4*e) - 2*a^2*sin(2*f*x + 2*e) + I*a^2)*log(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin (1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 2*sin(1/2*arctan2(si n(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) - 3*(-I*a^2*cos(4*f*x + 4*e) - 2*I *a^2*cos(2*f*x + 2*e) + a^2*sin(4*f*x + 4*e) + 2*a^2*sin(2*f*x + 2*e) - I* a^2)*log(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2* arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 - 2*sin(1/2*arctan2(sin(2*f *x + 2*e), cos(2*f*x + 2*e))) + 1))*sqrt(a)*sqrt(c)/(f*(-4*I*cos(4*f*x + 4 *e) - 8*I*cos(2*f*x + 2*e) + 4*sin(4*f*x + 4*e) + 8*sin(2*f*x + 2*e) - 4*I ))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 311 vs. \(2 (112) = 224\).
Time = 1.86 (sec) , antiderivative size = 311, normalized size of antiderivative = 2.02 \[ \int (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)} \, dx=\frac {15 \, {\left (a^{3} c - a^{2} c\right )} \sqrt {-a c} e^{\left (9 i \, f x + 9 i \, e\right )} + 74 \, {\left (a^{3} c - a^{2} c\right )} \sqrt {-a c} e^{\left (7 i \, f x + 7 i \, e\right )} + 132 \, {\left (a^{3} c - a^{2} c\right )} \sqrt {-a c} e^{\left (5 i \, f x + 5 i \, e\right )} + 102 \, {\left (a^{3} c - a^{2} c\right )} \sqrt {-a c} e^{\left (3 i \, f x + 3 i \, e\right )} + 29 \, {\left (a^{3} c - a^{2} c\right )} \sqrt {-a c} e^{\left (i \, f x + i \, e\right )}}{4 \, {\left ({\left (a - 1\right )} c f e^{\left (10 i \, f x + 10 i \, e\right )} + 5 \, {\left (a - 1\right )} c f e^{\left (8 i \, f x + 8 i \, e\right )} + 10 \, {\left (a - 1\right )} c f e^{\left (6 i \, f x + 6 i \, e\right )} + 10 \, {\left (a - 1\right )} c f e^{\left (4 i \, f x + 4 i \, e\right )} + 5 \, {\left (a - 1\right )} c f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (a - 1\right )} c f\right )}} - \frac {12 i \, a^{\frac {5}{2}} \sqrt {c} \arctan \left (e^{\left (i \, f x + i \, e\right )}\right ) - \frac {i \, {\left (5 \, a^{\frac {5}{2}} \sqrt {c} e^{\left (3 i \, f x + 3 i \, e\right )} + 7 \, a^{\frac {5}{2}} \sqrt {c} e^{\left (i \, f x + i \, e\right )}\right )}}{{\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}^{2}}}{4 \, f} \]
1/4*(15*(a^3*c - a^2*c)*sqrt(-a*c)*e^(9*I*f*x + 9*I*e) + 74*(a^3*c - a^2*c )*sqrt(-a*c)*e^(7*I*f*x + 7*I*e) + 132*(a^3*c - a^2*c)*sqrt(-a*c)*e^(5*I*f *x + 5*I*e) + 102*(a^3*c - a^2*c)*sqrt(-a*c)*e^(3*I*f*x + 3*I*e) + 29*(a^3 *c - a^2*c)*sqrt(-a*c)*e^(I*f*x + I*e))/((a - 1)*c*f*e^(10*I*f*x + 10*I*e) + 5*(a - 1)*c*f*e^(8*I*f*x + 8*I*e) + 10*(a - 1)*c*f*e^(6*I*f*x + 6*I*e) + 10*(a - 1)*c*f*e^(4*I*f*x + 4*I*e) + 5*(a - 1)*c*f*e^(2*I*f*x + 2*I*e) + (a - 1)*c*f) - 1/4*(12*I*a^(5/2)*sqrt(c)*arctan(e^(I*f*x + I*e)) - I*(5*a ^(5/2)*sqrt(c)*e^(3*I*f*x + 3*I*e) + 7*a^(5/2)*sqrt(c)*e^(I*f*x + I*e))/(e ^(2*I*f*x + 2*I*e) + 1)^2)/f
Timed out. \[ \int (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)} \, dx=\int {\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}} \,d x \]