3.8.93 \(\int \frac {\sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx\) [793]

3.8.93.1 Optimal result

Integrand size = 45, antiderivative size = 155 \[ \int \frac {\sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {(i A+B) \sqrt {a+i a \tan (e+f x)}}{5 f (c-i c \tan (e+f x))^{5/2}}-\frac {(2 i A-3 B) \sqrt {a+i a \tan (e+f x)}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac {(2 i A-3 B) \sqrt {a+i a \tan (e+f x)}}{15 c^2 f \sqrt {c-i c \tan (e+f x)}} \]

output

-1/15*(2*I*A-3*B)*(a+I*a*tan(f*x+e))^(1/2)/c^2/f/(c-I*c*tan(f*x+e))^(1/2)- 
1/5*(I*A+B)*(a+I*a*tan(f*x+e))^(1/2)/f/(c-I*c*tan(f*x+e))^(5/2)-1/15*(2*I* 
A-3*B)*(a+I*a*tan(f*x+e))^(1/2)/c/f/(c-I*c*tan(f*x+e))^(3/2)
 

3.8.93.2 Mathematica [A] (verified)

Time = 6.20 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.71 \[ \int \frac {\sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=\frac {a (-i+\tan (e+f x)) \left (-7 A-3 i B+(6 i A-9 B) \tan (e+f x)+(2 A+3 i B) \tan ^2(e+f x)\right )}{15 c^2 f (i+\tan (e+f x))^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]

input

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

output

(a*(-I + Tan[e + f*x])*(-7*A - (3*I)*B + ((6*I)*A - 9*B)*Tan[e + f*x] + (2 
*A + (3*I)*B)*Tan[e + f*x]^2))/(15*c^2*f*(I + Tan[e + f*x])^2*Sqrt[a + I*a 
*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])
 

3.8.93.3 Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 167, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4071

\(\displaystyle \frac {a c \int \frac {A+B \tan (e+f x)}{\sqrt {i \tan (e+f x) a+a} (c-i c \tan (e+f x))^{7/2}}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {a c \left (\frac {(2 A+3 i B) \int \frac {1}{\sqrt {i \tan (e+f x) a+a} (c-i c \tan (e+f x))^{5/2}}d\tan (e+f x)}{5 c}-\frac {(B+i A) \sqrt {a+i a \tan (e+f x)}}{5 a c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\)

\(\Big \downarrow \) 55

\(\displaystyle \frac {a c \left (\frac {(2 A+3 i B) \left (\frac {\int \frac {1}{\sqrt {i \tan (e+f x) a+a} (c-i c \tan (e+f x))^{3/2}}d\tan (e+f x)}{3 c}-\frac {i \sqrt {a+i a \tan (e+f x)}}{3 a c (c-i c \tan (e+f x))^{3/2}}\right )}{5 c}-\frac {(B+i A) \sqrt {a+i a \tan (e+f x)}}{5 a c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\)

\(\Big \downarrow \) 48

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

input

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

output

(a*c*(-1/5*((I*A + B)*Sqrt[a + I*a*Tan[e + f*x]])/(a*c*(c - I*c*Tan[e + f* 
x])^(5/2)) + ((2*A + (3*I)*B)*(((-1/3*I)*Sqrt[a + I*a*Tan[e + f*x]])/(a*c* 
(c - I*c*Tan[e + f*x])^(3/2)) - ((I/3)*Sqrt[a + I*a*Tan[e + f*x]])/(a*c^2* 
Sqrt[c - I*c*Tan[e + f*x]])))/(5*c)))/f
 

3.8.93.3.1 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]
 

3.8.93.4 Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.63

method	result	size
risch	\(-\frac {\sqrt {\frac {a \,{\mathrm e}^{2 i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left (3 i A \,{\mathrm e}^{4 i \left (f x +e \right )}+3 B \,{\mathrm e}^{4 i \left (f x +e \right )}+10 i A \,{\mathrm e}^{2 i \left (f x +e \right )}+15 i A -15 B \right )}{60 c^{2} \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\)	\(98\)
derivativedivides	\(-\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (2 i A \tan \left (f x +e \right )^{3}-12 i B \tan \left (f x +e \right )^{2}-3 B \tan \left (f x +e \right )^{3}-13 i A \tan \left (f x +e \right )-8 A \tan \left (f x +e \right )^{2}+3 i B +12 B \tan \left (f x +e \right )+7 A \right )}{15 f \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{4}}\)	\(125\)
default	\(-\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (2 i A \tan \left (f x +e \right )^{3}-12 i B \tan \left (f x +e \right )^{2}-3 B \tan \left (f x +e \right )^{3}-13 i A \tan \left (f x +e \right )-8 A \tan \left (f x +e \right )^{2}+3 i B +12 B \tan \left (f x +e \right )+7 A \right )}{15 f \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{4}}\)	\(125\)
parts	\(\frac {A \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (8 i \tan \left (f x +e \right )^{2}+2 \tan \left (f x +e \right )^{3}-7 i-13 \tan \left (f x +e \right )\right )}{15 f \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{4}}+\frac {i B \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (4 i \tan \left (f x +e \right )^{2}+\tan \left (f x +e \right )^{3}-i-4 \tan \left (f x +e \right )\right )}{5 f \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{4}}\)	\(167\)

input

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

output

-1/60/c^2*(a*exp(2*I*(f*x+e))/(exp(2*I*(f*x+e))+1))^(1/2)/(c/(exp(2*I*(f*x 
+e))+1))^(1/2)*(3*I*A*exp(4*I*(f*x+e))+3*B*exp(4*I*(f*x+e))+10*I*A*exp(2*I 
*(f*x+e))+15*I*A-15*B)/f
 

3.8.93.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.73 \[ \int \frac {\sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {{\left (3 \, {\left (i \, A + B\right )} e^{\left (7 i \, f x + 7 i \, e\right )} - {\left (-13 i \, A - 3 \, B\right )} e^{\left (5 i \, f x + 5 i \, e\right )} + 5 \, {\left (5 i \, A - 3 \, B\right )} e^{\left (3 i \, f x + 3 i \, e\right )} + 15 \, {\left (i \, A - B\right )} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{60 \, c^{3} f} \]

input

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

output

-1/60*(3*(I*A + B)*e^(7*I*f*x + 7*I*e) - (-13*I*A - 3*B)*e^(5*I*f*x + 5*I* 
e) + 5*(5*I*A - 3*B)*e^(3*I*f*x + 3*I*e) + 15*(I*A - B)*e^(I*f*x + I*e))*s 
qrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(c^3*f)
 

3.8.93.6 Sympy [F]

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

input

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

output

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

3.8.93.7 Maxima [F(-2)]

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

input

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

output

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

3.8.93.8 Giac [F]

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

input

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

output

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

3.8.93.9 Mupad [B] (verification not implemented)

Time = 9.36 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.10 \[ \int \frac {\sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (A\,15{}\mathrm {i}-15\,B+A\,\cos \left (2\,e+2\,f\,x\right )\,10{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,3{}\mathrm {i}+3\,B\,\cos \left (4\,e+4\,f\,x\right )-10\,A\,\sin \left (2\,e+2\,f\,x\right )-3\,A\,\sin \left (4\,e+4\,f\,x\right )+B\,\sin \left (4\,e+4\,f\,x\right )\,3{}\mathrm {i}\right )}{60\,c^2\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]

input

int(((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^(1/2))/(c - c*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)*(A*15i - 15*B + A*cos(2*e + 2*f*x)*10i + A*cos(4*e + 4*f*x)*3i + 3 
*B*cos(4*e + 4*f*x) - 10*A*sin(2*e + 2*f*x) - 3*A*sin(4*e + 4*f*x) + B*sin 
(4*e + 4*f*x)*3i))/(60*c^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))