Integrand size = 45, antiderivative size = 157 \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {(i A-B) \sqrt {c-i c \tan (e+f x)}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {(2 i A+3 B) \sqrt {c-i c \tan (e+f x)}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(2 i A+3 B) \sqrt {c-i c \tan (e+f x)}}{15 a^2 f \sqrt {a+i a \tan (e+f x)}} \]
1/15*(2*I*A+3*B)*(c-I*c*tan(f*x+e))^(1/2)/a^2/f/(a+I*a*tan(f*x+e))^(1/2)+1 /5*(I*A-B)*(c-I*c*tan(f*x+e))^(1/2)/f/(a+I*a*tan(f*x+e))^(5/2)+1/15*(2*I*A +3*B)*(c-I*c*tan(f*x+e))^(1/2)/a/f/(a+I*a*tan(f*x+e))^(3/2)
Time = 3.53 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.63 \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {\sqrt {c-i c \tan (e+f x)} \left (-7 i A-3 B+(6 A-9 i B) \tan (e+f x)+(2 i A+3 B) \tan ^2(e+f x)\right )}{15 a^2 f (-i+\tan (e+f x))^2 \sqrt {a+i a \tan (e+f x)}} \]
(Sqrt[c - I*c*Tan[e + f*x]]*((-7*I)*A - 3*B + (6*A - (9*I)*B)*Tan[e + f*x] + ((2*I)*A + 3*B)*Tan[e + f*x]^2))/(15*a^2*f*(-I + Tan[e + f*x])^2*Sqrt[a + I*a*Tan[e + f*x]])
Time = 0.40 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.08, 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.
\(\displaystyle \int \frac {\sqrt {c-i c \tan (e+f x)} (A+B \tan (e+f x))}{(a+i a \tan (e+f x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {c-i c \tan (e+f x)} (A+B \tan (e+f x))}{(a+i a \tan (e+f x))^{5/2}}dx\) |
\(\Big \downarrow \) 4071 |
\(\displaystyle \frac {a c \int \frac {A+B \tan (e+f x)}{(i \tan (e+f x) a+a)^{7/2} \sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {a c \left (\frac {(2 A-3 i B) \int \frac {1}{(i \tan (e+f x) a+a)^{5/2} \sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)}{5 a}+\frac {(-B+i A) \sqrt {c-i c \tan (e+f x)}}{5 a c (a+i a \tan (e+f x))^{5/2}}\right )}{f}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {a c \left (\frac {(2 A-3 i B) \left (\frac {\int \frac {1}{(i \tan (e+f x) a+a)^{3/2} \sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)}{3 a}+\frac {i \sqrt {c-i c \tan (e+f x)}}{3 a c (a+i a \tan (e+f x))^{3/2}}\right )}{5 a}+\frac {(-B+i A) \sqrt {c-i c \tan (e+f x)}}{5 a c (a+i a \tan (e+f x))^{5/2}}\right )}{f}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {a c \left (\frac {(2 A-3 i B) \left (\frac {i \sqrt {c-i c \tan (e+f x)}}{3 a^2 c \sqrt {a+i a \tan (e+f x)}}+\frac {i \sqrt {c-i c \tan (e+f x)}}{3 a c (a+i a \tan (e+f x))^{3/2}}\right )}{5 a}+\frac {(-B+i A) \sqrt {c-i c \tan (e+f x)}}{5 a c (a+i a \tan (e+f x))^{5/2}}\right )}{f}\) |
(a*c*(((I*A - B)*Sqrt[c - I*c*Tan[e + f*x]])/(5*a*c*(a + I*a*Tan[e + f*x]) ^(5/2)) + ((2*A - (3*I)*B)*(((I/3)*Sqrt[c - I*c*Tan[e + f*x]])/(a*c*(a + I *a*Tan[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]])))/(5*a)))/f
3.9.47.3.1 Defintions of rubi rules used
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 = 0.43 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.81
method | result | size |
derivativedivides | \(-\frac {i \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (2 i A \tan \left (f x +e \right )^{3}-12 i B \tan \left (f x +e \right )^{2}+3 B \tan \left (f x +e \right )^{3}-13 i A \tan \left (f x +e \right )+8 A \tan \left (f x +e \right )^{2}+3 i B -12 B \tan \left (f x +e \right )-7 A \right )}{15 f \,a^{3} \left (i-\tan \left (f x +e \right )\right )^{4}}\) | \(127\) |
default | \(-\frac {i \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (2 i A \tan \left (f x +e \right )^{3}-12 i B \tan \left (f x +e \right )^{2}+3 B \tan \left (f x +e \right )^{3}-13 i A \tan \left (f x +e \right )+8 A \tan \left (f x +e \right )^{2}+3 i B -12 B \tan \left (f x +e \right )-7 A \right )}{15 f \,a^{3} \left (i-\tan \left (f x +e \right )\right )^{4}}\) | \(127\) |
parts | \(-\frac {A \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (8 i \tan \left (f x +e \right )^{2}-2 \tan \left (f x +e \right )^{3}-7 i+13 \tan \left (f x +e \right )\right )}{15 f \,a^{3} \left (i-\tan \left (f x +e \right )\right )^{4}}+\frac {i B \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (4 i \tan \left (f x +e \right )^{2}-\tan \left (f x +e \right )^{3}-i+4 \tan \left (f x +e \right )\right )}{5 f \,a^{3} \left (i-\tan \left (f x +e \right )\right )^{4}}\) | \(173\) |
int((c-I*c*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^(5/2),x,m ethod=_RETURNVERBOSE)
-1/15*I/f*(-c*(I*tan(f*x+e)-1))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)/a^3*(2*I* A*tan(f*x+e)^3-12*I*B*tan(f*x+e)^2+3*B*tan(f*x+e)^3-13*I*A*tan(f*x+e)+8*A* tan(f*x+e)^2+3*I*B-12*B*tan(f*x+e)-7*A)/(I-tan(f*x+e))^4
Time = 0.25 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.71 \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{5/2}} \, dx=-\frac {{\left (15 \, {\left (-i \, A - B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + 5 \, {\left (-5 i \, A - 3 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} - {\left (13 i \, A - 3 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - 3 i \, A + 3 \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (-5 i \, f x - 5 i \, e\right )}}{60 \, a^{3} f} \]
integrate((c-I*c*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^(5/ 2),x, algorithm="fricas")
-1/60*(15*(-I*A - B)*e^(6*I*f*x + 6*I*e) + 5*(-5*I*A - 3*B)*e^(4*I*f*x + 4 *I*e) - (13*I*A - 3*B)*e^(2*I*f*x + 2*I*e) - 3*I*A + 3*B)*sqrt(a/(e^(2*I*f *x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(-5*I*f*x - 5*I*e)/( a^3*f)
\[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\int \frac {\sqrt {- i c \left (\tan {\left (e + f x \right )} + i\right )} \left (A + B \tan {\left (e + f x \right )}\right )}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]
Integral(sqrt(-I*c*(tan(e + f*x) + I))*(A + B*tan(e + f*x))/(I*a*(tan(e + f*x) - I))**(5/2), x)
Exception generated. \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]
integrate((c-I*c*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^(5/ 2),x, algorithm="maxima")
\[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\int { \frac {{\left (B \tan \left (f x + e\right ) + A\right )} \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
integrate((c-I*c*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^(5/ 2),x, algorithm="giac")
Time = 10.32 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.57 \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (A\,15{}\mathrm {i}+15\,B+A\,\cos \left (2\,e+2\,f\,x\right )\,25{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,13{}\mathrm {i}+A\,\cos \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}+15\,B\,\cos \left (2\,e+2\,f\,x\right )-3\,B\,\cos \left (4\,e+4\,f\,x\right )-3\,B\,\cos \left (6\,e+6\,f\,x\right )+25\,A\,\sin \left (2\,e+2\,f\,x\right )+13\,A\,\sin \left (4\,e+4\,f\,x\right )+3\,A\,\sin \left (6\,e+6\,f\,x\right )-B\,\sin \left (2\,e+2\,f\,x\right )\,15{}\mathrm {i}+B\,\sin \left (4\,e+4\,f\,x\right )\,3{}\mathrm {i}+B\,\sin \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}\right )}{120\,a^3\,f} \]
(((a*(cos(2*e + 2*f*x) + sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1)) ^(1/2)*((c*(cos(2*e + 2*f*x) - sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2)*(A*15i + 15*B + A*cos(2*e + 2*f*x)*25i + A*cos(4*e + 4*f*x)*1 3i + A*cos(6*e + 6*f*x)*3i + 15*B*cos(2*e + 2*f*x) - 3*B*cos(4*e + 4*f*x) - 3*B*cos(6*e + 6*f*x) + 25*A*sin(2*e + 2*f*x) + 13*A*sin(4*e + 4*f*x) + 3 *A*sin(6*e + 6*f*x) - B*sin(2*e + 2*f*x)*15i + B*sin(4*e + 4*f*x)*3i + B*s in(6*e + 6*f*x)*3i))/(120*a^3*f)