Integrand size = 45, antiderivative size = 152 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {i A+B}{f (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}+\frac {(2 i A+B) \sqrt {c-i c \tan (e+f x)}}{3 c f (a+i a \tan (e+f x))^{3/2}}+\frac {(2 i A+B) \sqrt {c-i c \tan (e+f x)}}{3 a c f \sqrt {a+i a \tan (e+f x)}} \] Output:
-(I*A+B)/f/(a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(1/2)+1/3*(2*I*A+B) *(c-I*c*tan(f*x+e))^(1/2)/c/f/(a+I*a*tan(f*x+e))^(3/2)+1/3*(2*I*A+B)*(c-I* c*tan(f*x+e))^(1/2)/a/c/f/(a+I*a*tan(f*x+e))^(1/2)
Time = 1.97 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.64 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {A+i B+(-2 i A-B) \tan (e+f x)+(2 A-i B) \tan ^2(e+f x)}{3 a f (-i+\tan (e+f x)) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \] Input:
Integrate[(A + B*Tan[e + f*x])/((a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c - I*c* Tan[e + f*x]]),x]
Output:
(A + I*B + ((-2*I)*A - B)*Tan[e + f*x] + (2*A - I*B)*Tan[e + f*x]^2)/(3*a* f*(-I + Tan[e + f*x])*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x] ])
Time = 0.68 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3042, 4071, 87, 55, 48}
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.
| \(\displaystyle \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}dx\) |
| \(\Big \downarrow \) 4071 |
| \(\displaystyle \frac {a c \int \frac {A+B \tan (e+f x)}{(i \tan (e+f x) a+a)^{5/2} (c-i c \tan (e+f x))^{3/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {a c \left (\frac {(2 A-i B) \int \frac {1}{(i \tan (e+f x) a+a)^{5/2} \sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)}{c}-\frac {B+i A}{a c (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}\right )}{f}\) |
| \(\Big \downarrow \) 55 |
| \(\displaystyle \frac {a c \left (\frac {(2 A-i B) \left (\frac {\int \frac {1}{(i \tan (e+f x) a+a)^{3/2} \sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)}{3 a}+\frac {i \sqrt {c-i c \tan (e+f x)}}{3 a c (a+i a \tan (e+f x))^{3/2}}\right )}{c}-\frac {B+i A}{a c (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}\right )}{f}\) |
| \(\Big \downarrow \) 48 |
| \(\displaystyle \frac {a c \left (\frac {(2 A-i B) \left (\frac {i \sqrt {c-i c \tan (e+f x)}}{3 a^2 c \sqrt {a+i a \tan (e+f x)}}+\frac {i \sqrt {c-i c \tan (e+f x)}}{3 a c (a+i a \tan (e+f x))^{3/2}}\right )}{c}-\frac {B+i A}{a c (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}\right )}{f}\) |
Input:
Int[(A + B*Tan[e + f*x])/((a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c - I*c*Tan[e + f*x]]),x]
Output:
(a*c*(-((I*A + B)/(a*c*(a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c - I*c*Tan[e + f *x]])) + ((2*A - I*B)*(((I/3)*Sqrt[c - I*c*Tan[e + f*x]])/(a*c*(a + I*a*Ta n[e + f*x])^(3/2)) + ((I/3)*Sqrt[c - I*c*Tan[e + f*x]])/(a^2*c*Sqrt[a + I* a*Tan[e + f*x]])))/c))/f
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp [(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(S implify[m + n + 2]/((b*c - a*d)*(m + 1))) Int[(a + b*x)^Simplify[m + 1]*( c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 2], 0] && NeQ[m, -1] && !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] || !SumSimp lerQ[n, 1])
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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 = 0.79 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00
| method | result | size |
| derivativedivides | \(\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (2 i A \tan \left (f x +e \right )^{4}-i B \tan \left (f x +e \right )^{3}+B \tan \left (f x +e \right )^{4}+3 i A \tan \left (f x +e \right )^{2}+2 A \tan \left (f x +e \right )^{3}-i B \tan \left (f x +e \right )+i A +2 A \tan \left (f x +e \right )-B \right )}{3 f \,a^{2} c \left (i+\tan \left (f x +e \right )\right )^{2} \left (i-\tan \left (f x +e \right )\right )^{3}}\) | \(152\) |
| default | \(\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (2 i A \tan \left (f x +e \right )^{4}-i B \tan \left (f x +e \right )^{3}+B \tan \left (f x +e \right )^{4}+3 i A \tan \left (f x +e \right )^{2}+2 A \tan \left (f x +e \right )^{3}-i B \tan \left (f x +e \right )+i A +2 A \tan \left (f x +e \right )-B \right )}{3 f \,a^{2} c \left (i+\tan \left (f x +e \right )\right )^{2} \left (i-\tan \left (f x +e \right )\right )^{3}}\) | \(152\) |
| parts | \(\frac {A \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (2 i \tan \left (f x +e \right )^{3}-2 \tan \left (f x +e \right )^{4}+2 i \tan \left (f x +e \right )-3 \tan \left (f x +e \right )^{2}-1\right )}{3 f \,a^{2} c \left (i-\tan \left (f x +e \right )\right )^{3} \left (i+\tan \left (f x +e \right )\right )^{2}}-\frac {i B \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (i \tan \left (f x +e \right )^{3}-\tan \left (f x +e \right )^{4}+i \tan \left (f x +e \right )+1\right )}{3 f \,a^{2} c \left (i-\tan \left (f x +e \right )\right )^{3} \left (i+\tan \left (f x +e \right )\right )^{2}}\) | \(211\) |
Input:
int((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(1/2),x,m ethod=_RETURNVERBOSE)
Output:
1/3*I/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)/a^2/c*(2*I* A*tan(f*x+e)^4-I*B*tan(f*x+e)^3+B*tan(f*x+e)^4+3*I*A*tan(f*x+e)^2+2*A*tan( f*x+e)^3-I*B*tan(f*x+e)+I*A+2*A*tan(f*x+e)-B)/(I+tan(f*x+e))^2/(I-tan(f*x+ e))^3
Time = 0.09 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.95 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {{\left (3 \, {\left (i \, A + B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + 4 \, {\left (i \, A - B\right )} e^{\left (5 i \, f x + 5 i \, e\right )} + 3 \, {\left (-i \, A + B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, {\left (i \, A - B\right )} e^{\left (3 i \, f x + 3 i \, e\right )} - {\left (7 i \, A - B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (-3 i \, f x - 3 i \, e\right )}}{12 \, a^{2} c f} \] Input:
integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(1/ 2),x, algorithm="fricas")
Output:
-1/12*(3*(I*A + B)*e^(6*I*f*x + 6*I*e) + 4*(I*A - B)*e^(5*I*f*x + 5*I*e) + 3*(-I*A + B)*e^(4*I*f*x + 4*I*e) + 4*(I*A - B)*e^(3*I*f*x + 3*I*e) - (7*I *A - B)*e^(2*I*f*x + 2*I*e) - I*A + B)*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*s qrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(-3*I*f*x - 3*I*e)/(a^2*c*f)
\[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}} \, dx=\int \frac {A + B \tan {\left (e + f x \right )}}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}} \sqrt {- i c \left (\tan {\left (e + f x \right )} + i\right )}}\, dx \] Input:
integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))**(3/2)/(c-I*c*tan(f*x+e))**( 1/2),x)
Output:
Integral((A + B*tan(e + f*x))/((I*a*(tan(e + f*x) - I))**(3/2)*sqrt(-I*c*( tan(e + f*x) + I))), x)
Exception generated. \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(1/ 2),x, algorithm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un defined.
\[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}} \, dx=\int { \frac {B \tan \left (f x + e\right ) + A}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \sqrt {-i \, c \tan \left (f x + e\right ) + c}} \,d x } \] Input:
integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(1/ 2),x, algorithm="giac")
Output:
integrate((B*tan(f*x + e) + A)/((I*a*tan(f*x + e) + a)^(3/2)*sqrt(-I*c*tan (f*x + e) + c)), x)
Time = 6.27 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.12 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (6\,A\,\sin \left (2\,e+2\,f\,x\right )-3\,B+A\,\cos \left (2\,e+2\,f\,x\right )\,6{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}-B\,\cos \left (4\,e+4\,f\,x\right )-A\,3{}\mathrm {i}+A\,\sin \left (4\,e+4\,f\,x\right )+B\,\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}\right )}{12\,a^2\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \] Input:
int((A + B*tan(e + f*x))/((a + a*tan(e + f*x)*1i)^(3/2)*(c - c*tan(e + f*x )*1i)^(1/2)),x)
Output:
(((a*(cos(2*e + 2*f*x) + sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1)) ^(1/2)*(A*cos(2*e + 2*f*x)*6i - 3*B - A*3i + A*cos(4*e + 4*f*x)*1i - B*cos (4*e + 4*f*x) + 6*A*sin(2*e + 2*f*x) + A*sin(4*e + 4*f*x) + B*sin(4*e + 4* f*x)*1i))/(12*a^2*f*((c*(cos(2*e + 2*f*x) - sin(2*e + 2*f*x)*1i + 1))/(cos (2*e + 2*f*x) + 1))^(1/2))
\[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {\sqrt {c}\, \sqrt {a}\, \left (-\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right ) b i -\left (\int \frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}}{\tan \left (f x +e \right )^{3}-\tan \left (f x +e \right )^{2} i +\tan \left (f x +e \right )-i}d x \right ) \tan \left (f x +e \right )^{2} a f i +\left (\int \frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}}{\tan \left (f x +e \right )^{3}-\tan \left (f x +e \right )^{2} i +\tan \left (f x +e \right )-i}d x \right ) \tan \left (f x +e \right )^{2} b f -\left (\int \frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}}{\tan \left (f x +e \right )^{3}-\tan \left (f x +e \right )^{2} i +\tan \left (f x +e \right )-i}d x \right ) a f i +\left (\int \frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}}{\tan \left (f x +e \right )^{3}-\tan \left (f x +e \right )^{2} i +\tan \left (f x +e \right )-i}d x \right ) b f \right )}{a^{2} c f \left (\tan \left (f x +e \right )^{2}+1\right )} \] Input:
int((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(1/2),x)
Output:
(sqrt(c)*sqrt(a)*( - sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)* tan(e + f*x)*b*i - int((sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1))/(tan(e + f*x)**3 - tan(e + f*x)**2*i + tan(e + f*x) - i),x)*tan(e + f* x)**2*a*f*i + int((sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1))/( tan(e + f*x)**3 - tan(e + f*x)**2*i + tan(e + f*x) - i),x)*tan(e + f*x)**2 *b*f - int((sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1))/(tan(e + f*x)**3 - tan(e + f*x)**2*i + tan(e + f*x) - i),x)*a*f*i + int((sqrt(tan( e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1))/(tan(e + f*x)**3 - tan(e + f* x)**2*i + tan(e + f*x) - i),x)*b*f))/(a**2*c*f*(tan(e + f*x)**2 + 1))