\(\int \frac {\sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{7/2}} \, dx\) [998]

Optimal result

Integrand size = 35, antiderivative size = 182 \[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{7/2}} \, dx=\frac {i \sqrt {c-i c \tan (e+f x)}}{7 f (a+i a \tan (e+f x))^{7/2}}+\frac {3 i \sqrt {c-i c \tan (e+f x)}}{35 a f (a+i a \tan (e+f x))^{5/2}}+\frac {2 i \sqrt {c-i c \tan (e+f x)}}{35 a^2 f (a+i a \tan (e+f x))^{3/2}}+\frac {2 i \sqrt {c-i c \tan (e+f x)}}{35 a^3 f \sqrt {a+i a \tan (e+f x)}} \] Output:

1/7*I*(c-I*c*tan(f*x+e))^(1/2)/f/(a+I*a*tan(f*x+e))^(7/2)+3/35*I*(c-I*c*ta 
n(f*x+e))^(1/2)/a/f/(a+I*a*tan(f*x+e))^(5/2)+2/35*I*(c-I*c*tan(f*x+e))^(1/ 
2)/a^2/f/(a+I*a*tan(f*x+e))^(3/2)+2/35*I*(c-I*c*tan(f*x+e))^(1/2)/a^3/f/(a 
+I*a*tan(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 1.20 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.49 \[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{7/2}} \, dx=\frac {\sqrt {c-i c \tan (e+f x)} \left (-12-13 i \tan (e+f x)+8 \tan ^2(e+f x)+2 i \tan ^3(e+f x)\right )}{35 a^3 f (-i+\tan (e+f x))^3 \sqrt {a+i a \tan (e+f x)}} \] Input:

Integrate[Sqrt[c - I*c*Tan[e + f*x]]/(a + I*a*Tan[e + f*x])^(7/2),x]
 

Output:

(Sqrt[c - I*c*Tan[e + f*x]]*(-12 - (13*I)*Tan[e + f*x] + 8*Tan[e + f*x]^2 
+ (2*I)*Tan[e + f*x]^3))/(35*a^3*f*(-I + Tan[e + f*x])^3*Sqrt[a + I*a*Tan[ 
e + f*x]])
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.14, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3042, 4006, 55, 55, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4006

\(\displaystyle \frac {a c \int \frac {1}{(i \tan (e+f x) a+a)^{9/2} \sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 55

\(\displaystyle \frac {a c \left (\frac {3 \int \frac {1}{(i \tan (e+f x) a+a)^{7/2} \sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)}{7 a}+\frac {i \sqrt {c-i c \tan (e+f x)}}{7 a c (a+i a \tan (e+f x))^{7/2}}\right )}{f}\)

\(\Big \downarrow \) 55

\(\displaystyle \frac {a c \left (\frac {3 \left (\frac {2 \int \frac {1}{(i \tan (e+f x) a+a)^{5/2} \sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)}{5 a}+\frac {i \sqrt {c-i c \tan (e+f x)}}{5 a c (a+i a \tan (e+f x))^{5/2}}\right )}{7 a}+\frac {i \sqrt {c-i c \tan (e+f x)}}{7 a c (a+i a \tan (e+f x))^{7/2}}\right )}{f}\)

\(\Big \downarrow \) 55

\(\displaystyle \frac {a c \left (\frac {3 \left (\frac {2 \left (\frac {\int \frac {1}{(i \tan (e+f x) a+a)^{3/2} \sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)}{3 a}+\frac {i \sqrt {c-i c \tan (e+f x)}}{3 a c (a+i a \tan (e+f x))^{3/2}}\right )}{5 a}+\frac {i \sqrt {c-i c \tan (e+f x)}}{5 a c (a+i a \tan (e+f x))^{5/2}}\right )}{7 a}+\frac {i \sqrt {c-i c \tan (e+f x)}}{7 a c (a+i a \tan (e+f x))^{7/2}}\right )}{f}\)

\(\Big \downarrow \) 48

\(\displaystyle \frac {a c \left (\frac {3 \left (\frac {2 \left (\frac {i \sqrt {c-i c \tan (e+f x)}}{3 a^2 c \sqrt {a+i a \tan (e+f x)}}+\frac {i \sqrt {c-i c \tan (e+f x)}}{3 a c (a+i a \tan (e+f x))^{3/2}}\right )}{5 a}+\frac {i \sqrt {c-i c \tan (e+f x)}}{5 a c (a+i a \tan (e+f x))^{5/2}}\right )}{7 a}+\frac {i \sqrt {c-i c \tan (e+f x)}}{7 a c (a+i a \tan (e+f x))^{7/2}}\right )}{f}\)

Input:

Int[Sqrt[c - I*c*Tan[e + f*x]]/(a + I*a*Tan[e + f*x])^(7/2),x]
 

Output:

(a*c*(((I/7)*Sqrt[c - I*c*Tan[e + f*x]])/(a*c*(a + I*a*Tan[e + f*x])^(7/2) 
) + (3*(((I/5)*Sqrt[c - I*c*Tan[e + f*x]])/(a*c*(a + I*a*Tan[e + f*x])^(5/ 
2)) + (2*(((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)))/(7*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 3042

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

rule 4006

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + ( 
f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(c/f)   Subst[Int[(a + b*x)^(m - 1)*( 
c + d*x)^(n - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n 
}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]
 

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.52

method	result	size
derivativedivides	\(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (10 i \tan \left (f x +e \right )^{3}-2 \tan \left (f x +e \right )^{4}-25 i \tan \left (f x +e \right )+21 \tan \left (f x +e \right )^{2}-12\right )}{35 f \,a^{4} \left (-\tan \left (f x +e \right )+i\right )^{5}}\)	\(95\)
default	\(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (10 i \tan \left (f x +e \right )^{3}-2 \tan \left (f x +e \right )^{4}-25 i \tan \left (f x +e \right )+21 \tan \left (f x +e \right )^{2}-12\right )}{35 f \,a^{4} \left (-\tan \left (f x +e \right )+i\right )^{5}}\)	\(95\)

Input:

int((c-I*c*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(7/2),x,method=_RETURNVERB 
OSE)
 

Output:

1/35/f*(-c*(I*tan(f*x+e)-1))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)/a^4*(10*I*ta 
n(f*x+e)^3-2*tan(f*x+e)^4-25*I*tan(f*x+e)+21*tan(f*x+e)^2-12)/(-tan(f*x+e) 
+I)^5
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.53 \[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{7/2}} \, dx=\frac {\sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (35 i \, e^{\left (8 i \, f x + 8 i \, e\right )} + 70 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 56 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 26 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 5 i\right )} e^{\left (-7 i \, f x - 7 i \, e\right )}}{280 \, a^{4} f} \] Input:

integrate((c-I*c*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(7/2),x, algorithm=" 
fricas")
 

Output:

1/280*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))* 
(35*I*e^(8*I*f*x + 8*I*e) + 70*I*e^(6*I*f*x + 6*I*e) + 56*I*e^(4*I*f*x + 4 
*I*e) + 26*I*e^(2*I*f*x + 2*I*e) + 5*I)*e^(-7*I*f*x - 7*I*e)/(a^4*f)
 

Sympy [F]

\[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{7/2}} \, dx=\int \frac {\sqrt {- i c \left (\tan {\left (e + f x \right )} + i\right )}}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {7}{2}}}\, dx \] Input:

integrate((c-I*c*tan(f*x+e))**(1/2)/(a+I*a*tan(f*x+e))**(7/2),x)
 

Output:

Integral(sqrt(-I*c*(tan(e + f*x) + I))/(I*a*(tan(e + f*x) - I))**(7/2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{7/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((c-I*c*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(7/2),x, algorithm=" 
maxima")
 

Output:

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

Giac [F]

\[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{7/2}} \, dx=\int { \frac {\sqrt {-i \, c \tan \left (f x + e\right ) + c}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {7}{2}}} \,d x } \] Input:

integrate((c-I*c*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(7/2),x, algorithm=" 
giac")
 

Output:

integrate(sqrt(-I*c*tan(f*x + e) + c)/(I*a*tan(f*x + e) + a)^(7/2), x)
 

Mupad [B] (verification not implemented)

Time = 3.77 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.01 \[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{7/2}} \, dx=\frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (\cos \left (2\,e+2\,f\,x\right )\,70{}\mathrm {i}+\cos \left (4\,e+4\,f\,x\right )\,56{}\mathrm {i}+\cos \left (6\,e+6\,f\,x\right )\,26{}\mathrm {i}+\cos \left (8\,e+8\,f\,x\right )\,5{}\mathrm {i}+70\,\sin \left (2\,e+2\,f\,x\right )+56\,\sin \left (4\,e+4\,f\,x\right )+26\,\sin \left (6\,e+6\,f\,x\right )+5\,\sin \left (8\,e+8\,f\,x\right )+35{}\mathrm {i}\right )}{560\,a^4\,f} \] Input:

int((c - c*tan(e + f*x)*1i)^(1/2)/(a + a*tan(e + f*x)*1i)^(7/2),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)*(cos(2*e + 2*f*x)*70i + cos(4*e + 4*f*x)*56i + cos(6*e + 6*f* 
x)*26i + cos(8*e + 8*f*x)*5i + 70*sin(2*e + 2*f*x) + 56*sin(4*e + 4*f*x) + 
 26*sin(6*e + 6*f*x) + 5*sin(8*e + 8*f*x) + 35i))/(560*a^4*f)
 

Reduce [F]

\[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{7/2}} \, dx=\frac {\sqrt {c}\, \sqrt {a}\, \left (\left (\int \frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )}{\tan \left (f x +e \right )^{5} i +3 \tan \left (f x +e \right )^{4}-2 \tan \left (f x +e \right )^{3} i +2 \tan \left (f x +e \right )^{2}-3 \tan \left (f x +e \right ) i -1}d x \right ) i -\left (\int \frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}}{\tan \left (f x +e \right )^{5} i +3 \tan \left (f x +e \right )^{4}-2 \tan \left (f x +e \right )^{3} i +2 \tan \left (f x +e \right )^{2}-3 \tan \left (f x +e \right ) i -1}d x \right )\right )}{a^{4}} \] Input:

int((c-I*c*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(7/2),x)
 

Output:

(sqrt(c)*sqrt(a)*(int((sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1 
)*tan(e + f*x))/(tan(e + f*x)**5*i + 3*tan(e + f*x)**4 - 2*tan(e + f*x)**3 
*i + 2*tan(e + f*x)**2 - 3*tan(e + f*x)*i - 1),x)*i - int((sqrt(tan(e + f* 
x)*i + 1)*sqrt( - tan(e + f*x)*i + 1))/(tan(e + f*x)**5*i + 3*tan(e + f*x) 
**4 - 2*tan(e + f*x)**3*i + 2*tan(e + f*x)**2 - 3*tan(e + f*x)*i - 1),x))) 
/a**4