\(\int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx\) [211]

Optimal result

Integrand size = 34, antiderivative size = 386 \[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=-\frac {\left (3-4 m+m^2\right ) \left (i B \left (1-m^2\right )-A \left (1-4 m+m^2\right )\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(c+d x)\right ) \tan ^{1+m}(c+d x)}{48 a^4 d (1+m)}-\frac {\left (i B \left (1+3 m-m^2\right )-A \left (13-7 m+m^2\right )\right ) \tan ^{1+m}(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}-\frac {(2-m) \left (i B \left (2+2 m-m^2\right )-A \left (8-6 m+m^2\right )\right ) \tan ^{1+m}(c+d x)}{48 a^4 d (1+i \tan (c+d x))}+\frac {(2-m) m \left (B \left (2+2 m-m^2\right )+i A \left (8-6 m+m^2\right )\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\tan ^2(c+d x)\right ) \tan ^{2+m}(c+d x)}{48 a^4 d (2+m)}+\frac {(A+i B) \tan ^{1+m}(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(i B (1-m)+A (5-m)) \tan ^{1+m}(c+d x)}{24 a d (a+i a \tan (c+d x))^3} \]

[Out]

-1/48*(m^2-4*m+3)*(I*B*(-m^2+1)-A*(m^2-4*m+1))*hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],-tan(d*x+c)^2)*tan(d*x+c)^
(1+m)/a^4/d/(1+m)-1/48*(I*B*(-m^2+3*m+1)-A*(m^2-7*m+13))*tan(d*x+c)^(1+m)/a^4/d/(1+I*tan(d*x+c))^2-1/48*(2-m)*
(I*B*(-m^2+2*m+2)-A*(m^2-6*m+8))*tan(d*x+c)^(1+m)/a^4/d/(1+I*tan(d*x+c))+1/48*(2-m)*m*(B*(-m^2+2*m+2)+I*A*(m^2
-6*m+8))*hypergeom([1, 1+1/2*m],[2+1/2*m],-tan(d*x+c)^2)*tan(d*x+c)^(2+m)/a^4/d/(2+m)+1/8*(A+I*B)*tan(d*x+c)^(
1+m)/d/(a+I*a*tan(d*x+c))^4+1/24*(I*B*(1-m)+A*(5-m))*tan(d*x+c)^(1+m)/a/d/(a+I*a*tan(d*x+c))^3

Rubi [A] (verified)

Time = 1.59 (sec) , antiderivative size = 386, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3677, 3619, 3557, 371} \[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=-\frac {\left (m^2-4 m+3\right ) \left (-A \left (m^2-4 m+1\right )+i B \left (1-m^2\right )\right ) \tan ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\tan ^2(c+d x)\right )}{48 a^4 d (m+1)}+\frac {(2-m) m \left (B \left (-m^2+2 m+2\right )+i A \left (m^2-6 m+8\right )\right ) \tan ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},-\tan ^2(c+d x)\right )}{48 a^4 d (m+2)}-\frac {(2-m) \left (-A \left (m^2-6 m+8\right )+i B \left (-m^2+2 m+2\right )\right ) \tan ^{m+1}(c+d x)}{48 a^4 d (1+i \tan (c+d x))}-\frac {\left (-A \left (m^2-7 m+13\right )+i B \left (-m^2+3 m+1\right )\right ) \tan ^{m+1}(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}+\frac {(A (5-m)+i B (1-m)) \tan ^{m+1}(c+d x)}{24 a d (a+i a \tan (c+d x))^3}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{8 d (a+i a \tan (c+d x))^4} \]

[In]

Int[(Tan[c + d*x]^m*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^4,x]

[Out]

-1/48*((3 - 4*m + m^2)*(I*B*(1 - m^2) - A*(1 - 4*m + m^2))*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[c +
d*x]^2]*Tan[c + d*x]^(1 + m))/(a^4*d*(1 + m)) - ((I*B*(1 + 3*m - m^2) - A*(13 - 7*m + m^2))*Tan[c + d*x]^(1 +
m))/(48*a^4*d*(1 + I*Tan[c + d*x])^2) - ((2 - m)*(I*B*(2 + 2*m - m^2) - A*(8 - 6*m + m^2))*Tan[c + d*x]^(1 +
m))/(48*a^4*d*(1 + I*Tan[c + d*x])) + ((2 - m)*m*(B*(2 + 2*m - m^2) + I*A*(8 - 6*m + m^2))*Hypergeometric2F1[1
, (2 + m)/2, (4 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x]^(2 + m))/(48*a^4*d*(2 + m)) + ((A + I*B)*Tan[c + d*x]^(1
+ m))/(8*d*(a + I*a*Tan[c + d*x])^4) + ((I*B*(1 - m) + A*(5 - m))*Tan[c + d*x]^(1 + m))/(24*a*d*(a + I*a*Tan[
c + d*x])^3)

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 3557

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 3619

Int[((b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*T
an[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Tan[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x] && NeQ
[c^2 + d^2, 0] &&  !IntegerQ[2*m]

Rule 3677

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

Rubi steps \begin{align*} \text {integral}& = \frac {(A+i B) \tan ^{1+m}(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {\int \frac {\tan ^m(c+d x) (a (A (7-m)-i B (1+m))-a (i A-B) (3-m) \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx}{8 a^2} \\ & = \frac {(A+i B) \tan ^{1+m}(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(i B (1-m)+A (5-m)) \tan ^{1+m}(c+d x)}{24 a d (a+i a \tan (c+d x))^3}+\frac {\int \frac {\tan ^m(c+d x) \left (-2 a^2 \left (i B \left (4+3 m-m^2\right )-A \left (16-7 m+m^2\right )\right )+2 a^2 (B (1-m)-i A (5-m)) (2-m) \tan (c+d x)\right )}{(a+i a \tan (c+d x))^2} \, dx}{48 a^4} \\ & = -\frac {\left (i B \left (1+3 m-m^2\right )-A \left (13-7 m+m^2\right )\right ) \tan ^{1+m}(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}+\frac {(A+i B) \tan ^{1+m}(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(i B (1-m)+A (5-m)) \tan ^{1+m}(c+d x)}{24 a d (a+i a \tan (c+d x))^3}+\frac {\int \frac {\tan ^m(c+d x) \left (4 a^3 \left (A \left (19-20 m+8 m^2-m^3\right )-i B \left (7+2 m-4 m^2+m^3\right )\right )-4 a^3 (1-m) \left (B \left (1+3 m-m^2\right )+i A \left (13-7 m+m^2\right )\right ) \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{192 a^6} \\ & = -\frac {\left (i B \left (1+3 m-m^2\right )-A \left (13-7 m+m^2\right )\right ) \tan ^{1+m}(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}+\frac {(A+i B) \tan ^{1+m}(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(i B (1-m)+A (5-m)) \tan ^{1+m}(c+d x)}{24 a d (a+i a \tan (c+d x))^3}-\frac {(2-m) \left (i B \left (2+2 m-m^2\right )-A \left (8-6 m+m^2\right )\right ) \tan ^{1+m}(c+d x)}{48 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {\int \tan ^m(c+d x) \left (-8 a^4 \left (3-4 m+m^2\right ) \left (i B \left (1-m^2\right )-A \left (1-4 m+m^2\right )\right )+8 a^4 (2-m) m \left (B \left (2+2 m-m^2\right )+i A \left (8-6 m+m^2\right )\right ) \tan (c+d x)\right ) \, dx}{384 a^8} \\ & = -\frac {\left (i B \left (1+3 m-m^2\right )-A \left (13-7 m+m^2\right )\right ) \tan ^{1+m}(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}+\frac {(A+i B) \tan ^{1+m}(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(i B (1-m)+A (5-m)) \tan ^{1+m}(c+d x)}{24 a d (a+i a \tan (c+d x))^3}-\frac {(2-m) \left (i B \left (2+2 m-m^2\right )-A \left (8-6 m+m^2\right )\right ) \tan ^{1+m}(c+d x)}{48 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {\left ((2-m) m \left (B \left (2+2 m-m^2\right )+i A \left (8-6 m+m^2\right )\right )\right ) \int \tan ^{1+m}(c+d x) \, dx}{48 a^4}-\frac {\left (\left (3-4 m+m^2\right ) \left (i B \left (1-m^2\right )-A \left (1-4 m+m^2\right )\right )\right ) \int \tan ^m(c+d x) \, dx}{48 a^4} \\ & = -\frac {\left (i B \left (1+3 m-m^2\right )-A \left (13-7 m+m^2\right )\right ) \tan ^{1+m}(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}+\frac {(A+i B) \tan ^{1+m}(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(i B (1-m)+A (5-m)) \tan ^{1+m}(c+d x)}{24 a d (a+i a \tan (c+d x))^3}-\frac {(2-m) \left (i B \left (2+2 m-m^2\right )-A \left (8-6 m+m^2\right )\right ) \tan ^{1+m}(c+d x)}{48 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {\left ((2-m) m \left (B \left (2+2 m-m^2\right )+i A \left (8-6 m+m^2\right )\right )\right ) \text {Subst}\left (\int \frac {x^{1+m}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{48 a^4 d}-\frac {\left (\left (3-4 m+m^2\right ) \left (i B \left (1-m^2\right )-A \left (1-4 m+m^2\right )\right )\right ) \text {Subst}\left (\int \frac {x^m}{1+x^2} \, dx,x,\tan (c+d x)\right )}{48 a^4 d} \\ & = -\frac {\left (3-4 m+m^2\right ) \left (i B \left (1-m^2\right )-A \left (1-4 m+m^2\right )\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(c+d x)\right ) \tan ^{1+m}(c+d x)}{48 a^4 d (1+m)}-\frac {\left (i B \left (1+3 m-m^2\right )-A \left (13-7 m+m^2\right )\right ) \tan ^{1+m}(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}+\frac {(2-m) m \left (B \left (2+2 m-m^2\right )+i A \left (8-6 m+m^2\right )\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\tan ^2(c+d x)\right ) \tan ^{2+m}(c+d x)}{48 a^4 d (2+m)}+\frac {(A+i B) \tan ^{1+m}(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(i B (1-m)+A (5-m)) \tan ^{1+m}(c+d x)}{24 a d (a+i a \tan (c+d x))^3}-\frac {(2-m) \left (i B \left (2+2 m-m^2\right )-A \left (8-6 m+m^2\right )\right ) \tan ^{1+m}(c+d x)}{48 d \left (a^4+i a^4 \tan (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.56 (sec) , antiderivative size = 277, normalized size of antiderivative = 0.72 \[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=\frac {\tan ^{1+m}(c+d x) \left (\frac {\left (3-4 m+m^2\right ) \left (i B \left (-1+m^2\right )+A \left (1-4 m+m^2\right )\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(c+d x)\right )}{1+m}+\frac {(-2+m) m \left (-i A \left (8-6 m+m^2\right )+B \left (-2-2 m+m^2\right )\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\tan ^2(c+d x)\right ) \tan (c+d x)}{2+m}+\frac {6 (A+i B)}{(-i+\tan (c+d x))^4}+\frac {2 (-i A (-5+m)+B (-1+m))}{(-i+\tan (c+d x))^3}-\frac {A \left (13-7 m+m^2\right )+i B \left (-1-3 m+m^2\right )}{(-i+\tan (c+d x))^2}+\frac {(-2+m) \left (B \left (2+2 m-m^2\right )+i A \left (8-6 m+m^2\right )\right )}{-i+\tan (c+d x)}\right )}{48 a^4 d} \]

[In]

Integrate[(Tan[c + d*x]^m*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(Tan[c + d*x]^(1 + m)*(((3 - 4*m + m^2)*(I*B*(-1 + m^2) + A*(1 - 4*m + m^2))*Hypergeometric2F1[1, (1 + m)/2, (
3 + m)/2, -Tan[c + d*x]^2])/(1 + m) + ((-2 + m)*m*((-I)*A*(8 - 6*m + m^2) + B*(-2 - 2*m + m^2))*Hypergeometric
2F1[1, (2 + m)/2, (4 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x])/(2 + m) + (6*(A + I*B))/(-I + Tan[c + d*x])^4 + (2
*((-I)*A*(-5 + m) + B*(-1 + m)))/(-I + Tan[c + d*x])^3 - (A*(13 - 7*m + m^2) + I*B*(-1 - 3*m + m^2))/(-I + Tan
[c + d*x])^2 + ((-2 + m)*(B*(2 + 2*m - m^2) + I*A*(8 - 6*m + m^2)))/(-I + Tan[c + d*x])))/(48*a^4*d)

Maple [F]

\[\int \frac {\left (\tan ^{m}\left (d x +c \right )\right ) \left (A +B \tan \left (d x +c \right )\right )}{\left (a +i a \tan \left (d x +c \right )\right )^{4}}d x\]

[In]

int(tan(d*x+c)^m*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^4,x)

[Out]

int(tan(d*x+c)^m*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^4,x)

Fricas [F]

\[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{m}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}} \,d x } \]

[In]

integrate(tan(d*x+c)^m*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

integral(1/16*((A - I*B)*e^(8*I*d*x + 8*I*c) + 2*(2*A - I*B)*e^(6*I*d*x + 6*I*c) + 6*A*e^(4*I*d*x + 4*I*c) + 2
*(2*A + I*B)*e^(2*I*d*x + 2*I*c) + A + I*B)*((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^m*e^(-8*I
*d*x - 8*I*c)/a^4, x)

Sympy [F]

\[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=\frac {\int \frac {A \tan ^{m}{\left (c + d x \right )}}{\tan ^{4}{\left (c + d x \right )} - 4 i \tan ^{3}{\left (c + d x \right )} - 6 \tan ^{2}{\left (c + d x \right )} + 4 i \tan {\left (c + d x \right )} + 1}\, dx + \int \frac {B \tan {\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}}{\tan ^{4}{\left (c + d x \right )} - 4 i \tan ^{3}{\left (c + d x \right )} - 6 \tan ^{2}{\left (c + d x \right )} + 4 i \tan {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]

[In]

integrate(tan(d*x+c)**m*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**4,x)

[Out]

(Integral(A*tan(c + d*x)**m/(tan(c + d*x)**4 - 4*I*tan(c + d*x)**3 - 6*tan(c + d*x)**2 + 4*I*tan(c + d*x) + 1)
, x) + Integral(B*tan(c + d*x)*tan(c + d*x)**m/(tan(c + d*x)**4 - 4*I*tan(c + d*x)**3 - 6*tan(c + d*x)**2 + 4*
I*tan(c + d*x) + 1), x))/a**4

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(tan(d*x+c)^m*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{m}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}} \,d x } \]

[In]

integrate(tan(d*x+c)^m*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*tan(d*x + c)^m/(I*a*tan(d*x + c) + a)^4, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^m\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4} \,d x \]

[In]

int((tan(c + d*x)^m*(A + B*tan(c + d*x)))/(a + a*tan(c + d*x)*1i)^4,x)

[Out]

int((tan(c + d*x)^m*(A + B*tan(c + d*x)))/(a + a*tan(c + d*x)*1i)^4, x)