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

Optimal. Leaf size=264 \[ -\frac {\, _2F_1\left (1,m;1+m;\frac {1}{2} (1+i \tan (e+f x))\right ) (a+i a \tan (e+f x))^m}{2 (i c+d)^3 f m}-\frac {d \left (2 i c d (3-m) m+c^2 \left (6-5 m+m^2\right )-d^2 \left (2-m+m^2\right )\right ) \, _2F_1\left (1,m;1+m;-\frac {d (1+i \tan (e+f x))}{i c-d}\right ) (a+i a \tan (e+f x))^m}{2 \left (c^2+d^2\right )^3 f m}-\frac {d (a+i a \tan (e+f x))^m}{2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^2}-\frac {d (c (4-m)+i d m) (a+i a \tan (e+f x))^m}{2 \left (c^2+d^2\right )^2 f (c+d \tan (e+f x))} \]

[Out]

-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))

________________________________________________________________________________________

Rubi [A]
time = 0.63, antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3642, 3679, 3681, 3562, 70, 3680} \begin {gather*} -\frac {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 \, _2F_1\left (1,m;m+1;-\frac {d (i \tan (e+f x)+1)}{i c-d}\right )}{2 f m \left (c^2+d^2\right )^3}-\frac {d (c (4-m)+i d m) (a+i a \tan (e+f x))^m}{2 f \left (c^2+d^2\right )^2 (c+d \tan (e+f x))}-\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}-\frac {(a+i a \tan (e+f x))^m \, _2F_1\left (1,m;m+1;\frac {1}{2} (i \tan (e+f x)+1)\right )}{2 f m (d+i c)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 70

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] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 3562

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[-b/d, Subst[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 3642

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[d*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 + d^2))), x] - Dist[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)*T
an[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 3679

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*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] - Dist[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] /; FreeQ[{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 3680

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[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 3681

Int[(((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*tan[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(A*b + a*B)/(b*c + a*d), Int[(a + b*Tan[e + f*x])^m, x], x] - Dist[(B*c
- A*d)/(b*c + a*d), Int[(a + b*Tan[e + f*x])^m*((a - b*Tan[e + f*x])/(c + d*Tan[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]

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^3} \, dx &=-\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 {\int \frac {(a+i a \tan (e+f x))^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 \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))}+\frac {\int \frac {(a+i a \tan (e+f x))^m \left (a^2 \left (2 c^2+i c d (5-m) m-d^2 \left (2-m+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}{2 a^2 \left (c^2+d^2\right )^2}\\ &=-\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))}+\frac {\int (a+i a \tan (e+f x))^m \, dx}{(c-i d)^3}-\frac {\left (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 )\right ) \int \frac {(a-i a \tan (e+f x)) (a+i a \tan (e+f x))^m}{c+d \tan (e+f x)} \, dx}{2 a (c+i d)^2 (i c+d)^3}\\ &=-\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))}-\frac {a \text {Subst}\left (\int \frac {(a+x)^{-1+m}}{a-x} \, dx,x,i a \tan (e+f x)\right )}{(i c+d)^3 f}-\frac {\left (a 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 )\right ) \text {Subst}\left (\int \frac {(a+i a x)^{-1+m}}{c+d x} \, dx,x,\tan (e+f x)\right )}{2 (c+i d)^2 (i c+d)^3 f}\\ &=-\frac {\, _2F_1\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 ) \, _2F_1\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))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [F]
time = 47.35, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]
time = 3.99, size = 0, normalized size = 0.00 \[\int \frac {\left (a +i a \tan \left (f x +e \right )\right )^{m}}{\left (c +d \tan \left (f x +e \right )\right )^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________