\(\int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{5/2}} \, dx\) [847]

Optimal result

Integrand size = 45, antiderivative size = 157 \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {(i A-B) \sqrt {c-i c \tan (e+f x)}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {(2 i A+3 B) \sqrt {c-i c \tan (e+f x)}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(2 i A+3 B) \sqrt {c-i c \tan (e+f x)}}{15 a^2 f \sqrt {a+i a \tan (e+f x)}} \] Output:

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)+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)
 

Mathematica [A] (verified)

Time = 2.11 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.63 \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {\sqrt {c-i c \tan (e+f x)} \left (-7 i A-3 B+(6 A-9 i B) \tan (e+f x)+(2 i A+3 B) \tan ^2(e+f x)\right )}{15 a^2 f (-i+\tan (e+f x))^2 \sqrt {a+i a \tan (e+f x)}} \] Input:

Integrate[((A + B*Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]])/(a + I*a*Tan[e 
 + f*x])^(5/2),x]
 

Output:

(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]])
 

Rubi [A] (verified)

Time = 0.67 (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

\(\displaystyle \frac {a c \left (\frac {(2 A-3 i B) \left (\frac {i \sqrt {c-i c \tan (e+f x)}}{3 a^2 c \sqrt {a+i a \tan (e+f x)}}+\frac {i \sqrt {c-i c \tan (e+f x)}}{3 a c (a+i a \tan (e+f x))^{3/2}}\right )}{5 a}+\frac {(-B+i A) \sqrt {c-i c \tan (e+f x)}}{5 a c (a+i a \tan (e+f x))^{5/2}}\right )}{f}\)

Input:

Int[((A + B*Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]])/(a + I*a*Tan[e + f*x 
])^(5/2),x]
 

Output:

(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
 

Defintions of rubi rules used

rule 48

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]
 

rule 55

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])
 

rule 87

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]))))
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4071

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]
 

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.81

method	result	size
derivativedivides	\(-\frac {i \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (2 i A \tan \left (f x +e \right )^{3}-12 i B \tan \left (f x +e \right )^{2}+3 B \tan \left (f x +e \right )^{3}-13 i \tan \left (f x +e \right ) A +8 A \tan \left (f x +e \right )^{2}+3 i B -12 B \tan \left (f x +e \right )-7 A \right )}{15 f \,a^{3} \left (i-\tan \left (f x +e \right )\right )^{4}}\)	\(127\)
default	\(-\frac {i \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (2 i A \tan \left (f x +e \right )^{3}-12 i B \tan \left (f x +e \right )^{2}+3 B \tan \left (f x +e \right )^{3}-13 i \tan \left (f x +e \right ) A +8 A \tan \left (f x +e \right )^{2}+3 i B -12 B \tan \left (f x +e \right )-7 A \right )}{15 f \,a^{3} \left (i-\tan \left (f x +e \right )\right )^{4}}\)	\(127\)
parts	\(-\frac {A \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (8 i \tan \left (f x +e \right )^{2}-2 \tan \left (f x +e \right )^{3}-7 i+13 \tan \left (f x +e \right )\right )}{15 f \,a^{3} \left (i-\tan \left (f x +e \right )\right )^{4}}+\frac {i B \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (4 i \tan \left (f x +e \right )^{2}-\tan \left (f x +e \right )^{3}-i+4 \tan \left (f x +e \right )\right )}{5 f \,a^{3} \left (i-\tan \left (f x +e \right )\right )^{4}}\)	\(173\)

Input:

int((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(5/2),x,m 
ethod=_RETURNVERBOSE)
 

Output:

-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
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.71 \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{5/2}} \, dx=-\frac {{\left (15 \, {\left (-i \, A - B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + 5 \, {\left (-5 i \, A - 3 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} - {\left (13 i \, A - 3 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - 3 i \, A + 3 \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (-5 i \, f x - 5 i \, e\right )}}{60 \, a^{3} f} \] Input:

integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(5/ 
2),x, algorithm="fricas")
 

Output:

-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)
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\int \frac {\sqrt {- i c \left (\tan {\left (e + f x \right )} + i\right )} \left (A + B \tan {\left (e + f x \right )}\right )}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**(1/2)/(a+I*a*tan(f*x+e))**( 
5/2),x)
 

Output:

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)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(5/ 
2),x, algorithm="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(5/ 
2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeError: Bad Argument TypeError: Bad Argument TypeDone
 

Mupad [B] (verification not implemented)

Time = 7.66 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.57 \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (A\,15{}\mathrm {i}+15\,B+A\,\cos \left (2\,e+2\,f\,x\right )\,25{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,13{}\mathrm {i}+A\,\cos \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}+15\,B\,\cos \left (2\,e+2\,f\,x\right )-3\,B\,\cos \left (4\,e+4\,f\,x\right )-3\,B\,\cos \left (6\,e+6\,f\,x\right )+25\,A\,\sin \left (2\,e+2\,f\,x\right )+13\,A\,\sin \left (4\,e+4\,f\,x\right )+3\,A\,\sin \left (6\,e+6\,f\,x\right )-B\,\sin \left (2\,e+2\,f\,x\right )\,15{}\mathrm {i}+B\,\sin \left (4\,e+4\,f\,x\right )\,3{}\mathrm {i}+B\,\sin \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}\right )}{120\,a^3\,f} \] Input:

int(((A + B*tan(e + f*x))*(c - c*tan(e + f*x)*1i)^(1/2))/(a + a*tan(e + f* 
x)*1i)^(5/2),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)*((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)
 

Reduce [F]

\[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{5/2}} \, dx=-\frac {\sqrt {c}\, \left (\left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}}{\sqrt {\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}-2 \sqrt {\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right ) i -\sqrt {\tan \left (f x +e \right ) i +1}}d x \right ) a +\left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )}{\sqrt {\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}-2 \sqrt {\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right ) i -\sqrt {\tan \left (f x +e \right ) i +1}}d x \right ) b \right )}{\sqrt {a}\, a^{2}} \] Input:

int((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(5/2),x)
 

Output:

( - sqrt(c)*(int(sqrt( - tan(e + f*x)*i + 1)/(sqrt(tan(e + f*x)*i + 1)*tan 
(e + f*x)**2 - 2*sqrt(tan(e + f*x)*i + 1)*tan(e + f*x)*i - sqrt(tan(e + f* 
x)*i + 1)),x)*a + int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x))/(sqrt(tan 
(e + f*x)*i + 1)*tan(e + f*x)**2 - 2*sqrt(tan(e + f*x)*i + 1)*tan(e + f*x) 
*i - sqrt(tan(e + f*x)*i + 1)),x)*b))/(sqrt(a)*a**2)