Integrand size = 45, antiderivative size = 157 \[ \int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=-\frac {i a^{3/2} A c^{3/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}+\frac {a A c \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}+\frac {B (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 f} \]
-I*a^(3/2)*A*c^(3/2)*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))/f+1/2*a*A*c*(a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e ))^(1/2)*tan(f*x+e)/f+1/3*B*(a+I*a*tan(f*x+e))^(3/2)*(c-I*c*tan(f*x+e))^(3 /2)/f
Time = 5.46 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.15 \[ \int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=\frac {a^{3/2} c^2 (i+\tan (e+f x)) \left (-6 A \arcsin \left (\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \sqrt {a+i a \tan (e+f x)}+\sqrt {a} \sqrt {1-i \tan (e+f x)} (-i+\tan (e+f x)) \left (2 B+3 A \tan (e+f x)+2 B \tan ^2(e+f x)\right )\right )}{6 f \sqrt {1-i \tan (e+f x)} \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]
(a^(3/2)*c^2*(I + Tan[e + f*x])*(-6*A*ArcSin[Sqrt[a + I*a*Tan[e + f*x]]/(S qrt[2]*Sqrt[a])]*Sqrt[a + I*a*Tan[e + f*x]] + Sqrt[a]*Sqrt[1 - I*Tan[e + f *x]]*(-I + Tan[e + f*x])*(2*B + 3*A*Tan[e + f*x] + 2*B*Tan[e + f*x]^2)))/( 6*f*Sqrt[1 - I*Tan[e + f*x]]*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])
Time = 0.40 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3042, 4071, 90, 40, 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.
\(\displaystyle \int (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2} (A+B \tan (e+f x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2} (A+B \tan (e+f x))dx\) |
\(\Big \downarrow \) 4071 |
\(\displaystyle \frac {a c \int \sqrt {i \tan (e+f x) a+a} (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {a c \left (A \int \sqrt {i \tan (e+f x) a+a} \sqrt {c-i c \tan (e+f x)}d\tan (e+f x)+\frac {B (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 a c}\right )}{f}\) |
\(\Big \downarrow \) 40 |
\(\displaystyle \frac {a c \left (A \left (\frac {1}{2} a c \int \frac {1}{\sqrt {i \tan (e+f x) a+a} \sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)+\frac {1}{2} \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}\right )+\frac {B (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 a c}\right )}{f}\) |
\(\Big \downarrow \) 45 |
\(\displaystyle \frac {a c \left (A \left (a c \int \frac {1}{i a+\frac {i c (i \tan (e+f x) a+a)}{c-i c \tan (e+f x)}}d\frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c-i c \tan (e+f x)}}+\frac {1}{2} \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}\right )+\frac {B (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 a c}\right )}{f}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {a c \left (A \left (\frac {1}{2} \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}-i \sqrt {a} \sqrt {c} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )\right )+\frac {B (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 a c}\right )}{f}\) |
(a*c*((B*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(3/2))/(3*a*c ) + A*((-I)*Sqrt[a]*Sqrt[c]*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(S qrt[a]*Sqrt[c - I*c*Tan[e + f*x]])] + (Tan[e + f*x]*Sqrt[a + I*a*Tan[e + f *x]]*Sqrt[c - I*c*Tan[e + f*x]])/2)))/f
3.8.97.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[x* (a + b*x)^m*((c + d*x)^m/(2*m + 1)), x] + Simp[2*a*c*(m/(2*m + 1)) Int[(a + b*x)^(m - 1)*(c + d*x)^(m - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[ b*c + a*d, 0] && IGtQ[m + 1/2, 0]
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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
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_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Si mp[a*(c/f) Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1)*(A + B*x), x], x , Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]
Time = 0.40 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.14
method | result | size |
parts | \(-\frac {A \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a c \left (\tan \left (f x +e \right ) \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}+a c \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right )\right )}{2 f \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}-\frac {B \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a c \left (1+\tan \left (f x +e \right )^{2}\right )}{3 f}\) | \(179\) |
derivativedivides | \(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a c \left (2 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{2}+3 A \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c +3 A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+2 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{6 f \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}\) | \(186\) |
default | \(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a c \left (2 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{2}+3 A \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c +3 A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+2 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{6 f \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}\) | \(186\) |
int((a+I*a*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/2),x,m ethod=_RETURNVERBOSE)
-1/2*A/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a*c*(tan(f *x+e)*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)+a*c*ln((a*c*tan(f*x+e)+(a*c )^(1/2)*(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)-1/3*B/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1) )^(1/2)*a*c*(1+tan(f*x+e)^2)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 433 vs. \(2 (119) = 238\).
Time = 0.28 (sec) , antiderivative size = 433, normalized size of antiderivative = 2.76 \[ \int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=\frac {3 \, \sqrt {\frac {A^{2} a^{3} c^{3}}{f^{2}}} {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {4 \, {\left (2 \, {\left (A a c e^{\left (3 i \, f x + 3 i \, e\right )} + A a c e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - \sqrt {\frac {A^{2} a^{3} c^{3}}{f^{2}}} {\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, f\right )}\right )}}{A a c e^{\left (2 i \, f x + 2 i \, e\right )} + A a c}\right ) - 3 \, \sqrt {\frac {A^{2} a^{3} c^{3}}{f^{2}}} {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {4 \, {\left (2 \, {\left (A a c e^{\left (3 i \, f x + 3 i \, e\right )} + A a c e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - \sqrt {\frac {A^{2} a^{3} c^{3}}{f^{2}}} {\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, f\right )}\right )}}{A a c e^{\left (2 i \, f x + 2 i \, e\right )} + A a c}\right ) - 4 \, {\left (3 i \, A a c e^{\left (5 i \, f x + 5 i \, e\right )} - 8 \, B a c e^{\left (3 i \, f x + 3 i \, e\right )} - 3 i \, A a c e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{12 \, {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]
integrate((a+I*a*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/ 2),x, algorithm="fricas")
1/12*(3*sqrt(A^2*a^3*c^3/f^2)*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2* I*e) + f)*log(4*(2*(A*a*c*e^(3*I*f*x + 3*I*e) + A*a*c*e^(I*f*x + I*e))*sqr t(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) - sqrt(A^ 2*a^3*c^3/f^2)*(I*f*e^(2*I*f*x + 2*I*e) - I*f))/(A*a*c*e^(2*I*f*x + 2*I*e) + A*a*c)) - 3*sqrt(A^2*a^3*c^3/f^2)*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f *x + 2*I*e) + f)*log(4*(2*(A*a*c*e^(3*I*f*x + 3*I*e) + A*a*c*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^2*a^3*c^3/f^2)*(-I*f*e^(2*I*f*x + 2*I*e) + I*f))/(A*a*c*e^(2*I*f*x + 2*I*e) + A*a*c)) - 4*(3*I*A*a*c*e^(5*I*f*x + 5*I*e) - 8*B*a*c*e^(3*I*f*x + 3*I*e) - 3*I*A*a*c*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*s qrt(c/(e^(2*I*f*x + 2*I*e) + 1)))/(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)
\[ \int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=\int \left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}} \left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {3}{2}} \left (A + B \tan {\left (e + f x \right )}\right )\, dx \]
Integral((I*a*(tan(e + f*x) - I))**(3/2)*(-I*c*(tan(e + f*x) + I))**(3/2)* (A + B*tan(e + f*x)), x)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 853 vs. \(2 (119) = 238\).
Time = 0.45 (sec) , antiderivative size = 853, normalized size of antiderivative = 5.43 \[ \int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=\text {Too large to display} \]
integrate((a+I*a*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/ 2),x, algorithm="maxima")
-(12*A*a*c*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 32*I*B*a *c*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 12*A*a*c*cos(1/2 *arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 12*I*A*a*c*sin(5/2*arctan2 (sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 32*B*a*c*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 12*I*A*a*c*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 6*(A*a*c*cos(6*f*x + 6*e) + 3*A*a*c*cos(4*f*x + 4*e) + 3*A*a*c*cos(2*f*x + 2*e) + I*A*a*c*sin(6*f*x + 6*e) + 3*I*A*a*c*sin(4*f *x + 4*e) + 3*I*A*a*c*sin(2*f*x + 2*e) + A*a*c)*arctan2(cos(1/2*arctan2(si n(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*a*c*cos(6*f*x + 6*e) + 3*A*a*c*cos(4*f*x + 4*e) + 3*A*a*c*cos(2*f*x + 2*e) + I*A*a*c*sin(6*f*x + 6*e) + 3*I*A*a*c*sin(4*f *x + 4*e) + 3*I*A*a*c*sin(2*f*x + 2*e) + A*a*c)*arctan2(cos(1/2*arctan2(si n(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) + 3*(I*A*a*c*cos(6*f*x + 6*e) + 3*I*A*a*c*cos(4*f*x + 4*e) + 3*I*A*a*c*cos(2*f*x + 2*e) - A*a*c*sin(6*f*x + 6*e) - 3*A*a*c*sin( 4*f*x + 4*e) - 3*A*a*c*sin(2*f*x + 2*e) + I*A*a*c)*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), co s(2*f*x + 2*e)))^2 + 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)) ) + 1) + 3*(-I*A*a*c*cos(6*f*x + 6*e) - 3*I*A*a*c*cos(4*f*x + 4*e) - 3*I*A *a*c*cos(2*f*x + 2*e) + A*a*c*sin(6*f*x + 6*e) + 3*A*a*c*sin(4*f*x + 4*...
\[ \int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=\int { {\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} \,d x } \]
integrate((a+I*a*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/ 2),x, algorithm="giac")
Timed out. \[ \int (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=\int \left (A+B\,\mathrm {tan}\left (e+f\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \]