\(\int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^3} \, dx\) [1185]

Optimal result

Integrand size = 28, antiderivative size = 264 \[ \int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^3} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (1,m,1+m,\frac {1}{2} (1+i \tan (e+f x))\right ) (a+i a \tan (e+f x))^m}{2 (i c+d)^3 f m}-\frac {d \left (2 i c d (3-m) m+c^2 \left (6-5 m+m^2\right )-d^2 \left (2-m+m^2\right )\right ) \operatorname {Hypergeometric2F1}\left (1,m,1+m,-\frac {d (1+i \tan (e+f x))}{i c-d}\right ) (a+i a \tan (e+f x))^m}{2 \left (c^2+d^2\right )^3 f m}-\frac {d (a+i a \tan (e+f x))^m}{2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^2}-\frac {d (c (4-m)+i d m) (a+i a \tan (e+f x))^m}{2 \left (c^2+d^2\right )^2 f (c+d \tan (e+f x))} \] Output:

-1/2*hypergeom([1, m],[1+m],1/2+1/2*I*tan(f*x+e))*(a+I*a*tan(f*x+e))^m/(I* 
c+d)^3/f/m-1/2*d*(2*I*c*d*(3-m)*m+c^2*(m^2-5*m+6)-d^2*(m^2-m+2))*hypergeom 
([1, m],[1+m],-d*(1+I*tan(f*x+e))/(I*c-d))*(a+I*a*tan(f*x+e))^m/(c^2+d^2)^ 
3/f/m-1/2*d*(a+I*a*tan(f*x+e))^m/(c^2+d^2)/f/(c+d*tan(f*x+e))^2-1/2*d*(c*( 
4-m)+I*d*m)*(a+I*a*tan(f*x+e))^m/(c^2+d^2)^2/f/(c+d*tan(f*x+e))
 

Mathematica [A] (verified)

Time = 2.24 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.81 \[ \int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^3} \, dx=\frac {(a+i a \tan (e+f x))^m \left (-2 d+\frac {2 (c+d \tan (e+f x)) \left ((c-i d) (c+i d) d m (c (-4+m)-i d m)-\left (i (c+i d)^3 \operatorname {Hypergeometric2F1}\left (1,m,1+m,\frac {1}{2} (1+i \tan (e+f x))\right )+d \left (-2 i c d (-3+m) m+d^2 \left (-2+m-m^2\right )+c^2 \left (6-5 m+m^2\right )\right ) \operatorname {Hypergeometric2F1}\left (1,m,1+m,\frac {d (1+i \tan (e+f x))}{-i c+d}\right )\right ) (c+d \tan (e+f x))\right )}{\left (c^2+d^2\right )^2 m}\right )}{4 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^2} \] Input:

Integrate[(a + I*a*Tan[e + f*x])^m/(c + d*Tan[e + f*x])^3,x]
 

Output:

((a + I*a*Tan[e + f*x])^m*(-2*d + (2*(c + d*Tan[e + f*x])*((c - I*d)*(c + 
I*d)*d*m*(c*(-4 + m) - I*d*m) - (I*(c + I*d)^3*Hypergeometric2F1[1, m, 1 + 
 m, (1 + I*Tan[e + f*x])/2] + d*((-2*I)*c*d*(-3 + m)*m + d^2*(-2 + m - m^2 
) + c^2*(6 - 5*m + m^2))*Hypergeometric2F1[1, m, 1 + m, (d*(1 + I*Tan[e + 
f*x]))/((-I)*c + d)])*(c + d*Tan[e + f*x])))/((c^2 + d^2)^2*m)))/(4*(c^2 + 
 d^2)*f*(c + d*Tan[e + f*x])^2)
 

Rubi [A] (verified)

Time = 1.60 (sec) , antiderivative size = 321, normalized size of antiderivative = 1.22, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.393, Rules used = {3042, 4044, 3042, 4081, 3042, 4083, 3042, 3962, 78, 4082, 78}

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.

\(\displaystyle \int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^3}dx\)

\(\Big \downarrow \) 4044

\(\displaystyle \frac {\int \frac {(i \tan (e+f x) a+a)^m (a (2 c+i d m)-a d (2-m) \tan (e+f x))}{(c+d \tan (e+f x))^2}dx}{2 a \left (c^2+d^2\right )}-\frac {d (a+i a \tan (e+f x))^m}{2 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {(i \tan (e+f x) a+a)^m (a (2 c+i d m)-a d (2-m) \tan (e+f x))}{(c+d \tan (e+f x))^2}dx}{2 a \left (c^2+d^2\right )}-\frac {d (a+i a \tan (e+f x))^m}{2 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^2}\)

\(\Big \downarrow \) 4081

\(\displaystyle \frac {\frac {\int \frac {(i \tan (e+f x) a+a)^m \left (a^2 \left (2 c^2+i d (5-m) m c-d^2 \left (m^2-m+2\right )\right )-a^2 d (1-m) (c (4-m)+i d m) \tan (e+f x)\right )}{c+d \tan (e+f x)}dx}{a \left (c^2+d^2\right )}-\frac {a d (c (4-m)+i d m) (a+i a \tan (e+f x))^m}{f \left (c^2+d^2\right ) (c+d \tan (e+f x))}}{2 a \left (c^2+d^2\right )}-\frac {d (a+i a \tan (e+f x))^m}{2 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {(i \tan (e+f x) a+a)^m \left (a^2 \left (2 c^2+i d (5-m) m c-d^2 \left (m^2-m+2\right )\right )-a^2 d (1-m) (c (4-m)+i d m) \tan (e+f x)\right )}{c+d \tan (e+f x)}dx}{a \left (c^2+d^2\right )}-\frac {a d (c (4-m)+i d m) (a+i a \tan (e+f x))^m}{f \left (c^2+d^2\right ) (c+d \tan (e+f x))}}{2 a \left (c^2+d^2\right )}-\frac {d (a+i a \tan (e+f x))^m}{2 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^2}\)

\(\Big \downarrow \) 4083

\(\displaystyle \frac {\frac {\frac {2 a^2 (c+i d)^2 \int (i \tan (e+f x) a+a)^mdx}{c-i d}+\frac {a d \left (c^2 \left (m^2-5 m+6\right )+2 i c d (3-m) m-d^2 \left (m^2-m+2\right )\right ) \int \frac {(a-i a \tan (e+f x)) (i \tan (e+f x) a+a)^m}{c+d \tan (e+f x)}dx}{d+i c}}{a \left (c^2+d^2\right )}-\frac {a d (c (4-m)+i d m) (a+i a \tan (e+f x))^m}{f \left (c^2+d^2\right ) (c+d \tan (e+f x))}}{2 a \left (c^2+d^2\right )}-\frac {d (a+i a \tan (e+f x))^m}{2 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {2 a^2 (c+i d)^2 \int (i \tan (e+f x) a+a)^mdx}{c-i d}+\frac {a d \left (c^2 \left (m^2-5 m+6\right )+2 i c d (3-m) m-d^2 \left (m^2-m+2\right )\right ) \int \frac {(a-i a \tan (e+f x)) (i \tan (e+f x) a+a)^m}{c+d \tan (e+f x)}dx}{d+i c}}{a \left (c^2+d^2\right )}-\frac {a d (c (4-m)+i d m) (a+i a \tan (e+f x))^m}{f \left (c^2+d^2\right ) (c+d \tan (e+f x))}}{2 a \left (c^2+d^2\right )}-\frac {d (a+i a \tan (e+f x))^m}{2 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^2}\)

\(\Big \downarrow \) 3962

\(\displaystyle \frac {\frac {\frac {a d \left (c^2 \left (m^2-5 m+6\right )+2 i c d (3-m) m-d^2 \left (m^2-m+2\right )\right ) \int \frac {(a-i a \tan (e+f x)) (i \tan (e+f x) a+a)^m}{c+d \tan (e+f x)}dx}{d+i c}-\frac {2 i a^3 (c+i d)^2 \int \frac {(i \tan (e+f x) a+a)^{m-1}}{a-i a \tan (e+f x)}d(i a \tan (e+f x))}{f (c-i d)}}{a \left (c^2+d^2\right )}-\frac {a d (c (4-m)+i d m) (a+i a \tan (e+f x))^m}{f \left (c^2+d^2\right ) (c+d \tan (e+f x))}}{2 a \left (c^2+d^2\right )}-\frac {d (a+i a \tan (e+f x))^m}{2 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^2}\)

\(\Big \downarrow \) 78

\(\displaystyle \frac {\frac {\frac {a d \left (c^2 \left (m^2-5 m+6\right )+2 i c d (3-m) m-d^2 \left (m^2-m+2\right )\right ) \int \frac {(a-i a \tan (e+f x)) (i \tan (e+f x) a+a)^m}{c+d \tan (e+f x)}dx}{d+i c}-\frac {i a^2 (c+i d)^2 (a+i a \tan (e+f x))^m \operatorname {Hypergeometric2F1}\left (1,m,m+1,\frac {i \tan (e+f x) a+a}{2 a}\right )}{f m (c-i d)}}{a \left (c^2+d^2\right )}-\frac {a d (c (4-m)+i d m) (a+i a \tan (e+f x))^m}{f \left (c^2+d^2\right ) (c+d \tan (e+f x))}}{2 a \left (c^2+d^2\right )}-\frac {d (a+i a \tan (e+f x))^m}{2 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^2}\)

\(\Big \downarrow \) 4082

\(\displaystyle \frac {\frac {\frac {a^3 d \left (c^2 \left (m^2-5 m+6\right )+2 i c d (3-m) m-d^2 \left (m^2-m+2\right )\right ) \int \frac {(i \tan (e+f x) a+a)^{m-1}}{c+d \tan (e+f x)}d\tan (e+f x)}{f (d+i c)}-\frac {i a^2 (c+i d)^2 (a+i a \tan (e+f x))^m \operatorname {Hypergeometric2F1}\left (1,m,m+1,\frac {i \tan (e+f x) a+a}{2 a}\right )}{f m (c-i d)}}{a \left (c^2+d^2\right )}-\frac {a d (c (4-m)+i d m) (a+i a \tan (e+f x))^m}{f \left (c^2+d^2\right ) (c+d \tan (e+f x))}}{2 a \left (c^2+d^2\right )}-\frac {d (a+i a \tan (e+f x))^m}{2 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^2}\)

\(\Big \downarrow \) 78

\(\displaystyle \frac {\frac {\frac {a^2 d \left (c^2 \left (m^2-5 m+6\right )+2 i c d (3-m) m-d^2 \left (m^2-m+2\right )\right ) (a+i a \tan (e+f x))^m \operatorname {Hypergeometric2F1}\left (1,m,m+1,-\frac {d (i \tan (e+f x)+1)}{i c-d}\right )}{f m (-d+i c) (d+i c)}-\frac {i a^2 (c+i d)^2 (a+i a \tan (e+f x))^m \operatorname {Hypergeometric2F1}\left (1,m,m+1,\frac {i \tan (e+f x) a+a}{2 a}\right )}{f m (c-i d)}}{a \left (c^2+d^2\right )}-\frac {a d (c (4-m)+i d m) (a+i a \tan (e+f x))^m}{f \left (c^2+d^2\right ) (c+d \tan (e+f x))}}{2 a \left (c^2+d^2\right )}-\frac {d (a+i a \tan (e+f x))^m}{2 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^2}\)

Input:

Int[(a + I*a*Tan[e + f*x])^m/(c + d*Tan[e + f*x])^3,x]
 

Output:

-1/2*(d*(a + I*a*Tan[e + f*x])^m)/((c^2 + d^2)*f*(c + d*Tan[e + f*x])^2) + 
 (-((a*d*(c*(4 - m) + I*d*m)*(a + I*a*Tan[e + f*x])^m)/((c^2 + d^2)*f*(c + 
 d*Tan[e + f*x]))) + ((a^2*d*((2*I)*c*d*(3 - m)*m + c^2*(6 - 5*m + m^2) - 
d^2*(2 - m + m^2))*Hypergeometric2F1[1, m, 1 + m, -((d*(1 + I*Tan[e + f*x] 
))/(I*c - d))]*(a + I*a*Tan[e + f*x])^m)/((I*c - d)*(I*c + d)*f*m) - (I*a^ 
2*(c + I*d)^2*Hypergeometric2F1[1, m, 1 + m, (a + I*a*Tan[e + f*x])/(2*a)] 
*(a + I*a*Tan[e + f*x])^m)/((c - I*d)*f*m))/(a*(c^2 + d^2)))/(2*a*(c^2 + d 
^2))
 

Defintions of rubi rules used

rule 78

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b 
*c - a*d)^n*((a + b*x)^(m + 1)/(b^(n + 1)*(m + 1)))*Hypergeometric2F1[-n, m 
 + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m}, x] 
 &&  !IntegerQ[m] && IntegerQ[n]
 

rule 3042

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

rule 3962

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

rule 4044

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

rule 4081

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(A*d - B*c)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*(n + 
1)*(c^2 + d^2))), x] - Simp[1/(a*(n + 1)*(c^2 + d^2))   Int[(a + b*Tan[e + 
f*x])^m*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c* 
m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Fr 
eeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 
0] && LtQ[n, -1]
 

rule 4082

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

rule 4083

Int[(((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
 (f_.)*(x_)]))/((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[( 
A*b + a*B)/(b*c + a*d)   Int[(a + b*Tan[e + f*x])^m, x], x] - Simp[(B*c - A 
*d)/(b*c + a*d)   Int[(a + b*Tan[e + f*x])^m*((a - b*Tan[e + f*x])/(c + d*T 
an[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - 
 a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]
 

Maple [F]

\[\int \frac {\left (a +i a \tan \left (f x +e \right )\right )^{m}}{\left (c +d \tan \left (f x +e \right )\right )^{3}}d x\]

Input:

int((a+I*a*tan(f*x+e))^m/(c+d*tan(f*x+e))^3,x)
 

Output:

int((a+I*a*tan(f*x+e))^m/(c+d*tan(f*x+e))^3,x)
 

Fricas [F]

\[ \int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^3} \, dx=\int { \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{3}} \,d x } \] Input:

integrate((a+I*a*tan(f*x+e))^m/(c+d*tan(f*x+e))^3,x, algorithm="fricas")
 

Output:

integral((2*a*e^(2*I*f*x + 2*I*e)/(e^(2*I*f*x + 2*I*e) + 1))^m*(-I*e^(6*I* 
f*x + 6*I*e) - 3*I*e^(4*I*f*x + 4*I*e) - 3*I*e^(2*I*f*x + 2*I*e) - I)/(-I* 
c^3 + 3*c^2*d + 3*I*c*d^2 - d^3 + (-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*e^( 
6*I*f*x + 6*I*e) - 3*(I*c^3 + c^2*d + I*c*d^2 + d^3)*e^(4*I*f*x + 4*I*e) - 
 3*(I*c^3 - c^2*d + I*c*d^2 - d^3)*e^(2*I*f*x + 2*I*e)), x)
 

Sympy [F]

\[ \int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^3} \, dx=\int \frac {\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{m}}{\left (c + d \tan {\left (e + f x \right )}\right )^{3}}\, dx \] Input:

integrate((a+I*a*tan(f*x+e))**m/(c+d*tan(f*x+e))**3,x)
 

Output:

Integral((I*a*(tan(e + f*x) - I))**m/(c + d*tan(e + f*x))**3, x)
 

Maxima [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^3} \, dx=\text {Timed out} \] Input:

integrate((a+I*a*tan(f*x+e))^m/(c+d*tan(f*x+e))^3,x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

\[ \int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^3} \, dx=\int { \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{3}} \,d x } \] Input:

integrate((a+I*a*tan(f*x+e))^m/(c+d*tan(f*x+e))^3,x, algorithm="giac")
 

Output:

integrate((I*a*tan(f*x + e) + a)^m/(d*tan(f*x + e) + c)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^3} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^m}{{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^3} \,d x \] Input:

int((a + a*tan(e + f*x)*1i)^m/(c + d*tan(e + f*x))^3,x)
 

Output:

int((a + a*tan(e + f*x)*1i)^m/(c + d*tan(e + f*x))^3, x)
 

Reduce [F]

\[ \int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^3} \, dx=\int \frac {\left (a +i a \tan \left (f x +e \right )\right )^{m}}{\left (d \tan \left (f x +e \right )+c \right )^{3}}d x \] Input:

int((a+I*a*tan(f*x+e))^m/(c+d*tan(f*x+e))^3,x)
 

Output:

int((a+I*a*tan(f*x+e))^m/(c+d*tan(f*x+e))^3,x)