3.204 \(\int \tan ^m(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=290 \[ \frac {8 a^4 (A-i B) \tan ^{m+1}(c+d x) \, _2F_1(1,m+1;m+2;i \tan (c+d x))}{d (m+1)}-\frac {2 \left (A (m+4)^2-i B \left (m^2+8 m+19\right )\right ) \left (a^4+i a^4 \tan (c+d x)\right ) \tan ^{m+1}(c+d x)}{d (m+2) (m+3) (m+4)}-\frac {2 a^4 \left (A \left (2 m^3+19 m^2+60 m+64\right )-i B \left (2 m^3+19 m^2+60 m+67\right )\right ) \tan ^{m+1}(c+d x)}{d (m+1) (m+2) (m+3) (m+4)}-\frac {(A (m+4)-i B (m+7)) \left (a^2+i a^2 \tan (c+d x)\right )^2 \tan ^{m+1}(c+d x)}{d (m+3) (m+4)}+\frac {i a B (a+i a \tan (c+d x))^3 \tan ^{m+1}(c+d x)}{d (m+4)} \]

[Out]

-2*a^4*(A*(2*m^3+19*m^2+60*m+64)-I*B*(2*m^3+19*m^2+60*m+67))*tan(d*x+c)^(1+m)/d/(3+m)/(4+m)/(m^2+3*m+2)+8*a^4*
(A-I*B)*hypergeom([1, 1+m],[2+m],I*tan(d*x+c))*tan(d*x+c)^(1+m)/d/(1+m)+I*a*B*tan(d*x+c)^(1+m)*(a+I*a*tan(d*x+
c))^3/d/(4+m)-(A*(4+m)-I*B*(7+m))*tan(d*x+c)^(1+m)*(a^2+I*a^2*tan(d*x+c))^2/d/(3+m)/(4+m)-2*(A*(4+m)^2-I*B*(m^
2+8*m+19))*tan(d*x+c)^(1+m)*(a^4+I*a^4*tan(d*x+c))/d/(4+m)/(m^2+5*m+6)

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Rubi [A]  time = 1.07, antiderivative size = 290, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {3594, 3592, 3537, 12, 64} \[ \frac {8 a^4 (A-i B) \tan ^{m+1}(c+d x) \, _2F_1(1,m+1;m+2;i \tan (c+d x))}{d (m+1)}-\frac {2 a^4 \left (A \left (2 m^3+19 m^2+60 m+64\right )-i B \left (2 m^3+19 m^2+60 m+67\right )\right ) \tan ^{m+1}(c+d x)}{d (m+1) (m+2) (m+3) (m+4)}-\frac {2 \left (A (m+4)^2-i B \left (m^2+8 m+19\right )\right ) \left (a^4+i a^4 \tan (c+d x)\right ) \tan ^{m+1}(c+d x)}{d (m+2) (m+3) (m+4)}-\frac {(A (m+4)-i B (m+7)) \left (a^2+i a^2 \tan (c+d x)\right )^2 \tan ^{m+1}(c+d x)}{d (m+3) (m+4)}+\frac {i a B (a+i a \tan (c+d x))^3 \tan ^{m+1}(c+d x)}{d (m+4)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^4*(A*(64 + 60*m + 19*m^2 + 2*m^3) - I*B*(67 + 60*m + 19*m^2 + 2*m^3))*Tan[c + d*x]^(1 + m))/(d*(1 + m)*(
2 + m)*(3 + m)*(4 + m)) + (8*a^4*(A - I*B)*Hypergeometric2F1[1, 1 + m, 2 + m, I*Tan[c + d*x]]*Tan[c + d*x]^(1
+ m))/(d*(1 + m)) + (I*a*B*Tan[c + d*x]^(1 + m)*(a + I*a*Tan[c + d*x])^3)/(d*(4 + m)) - ((A*(4 + m) - I*B*(7 +
m))*Tan[c + d*x]^(1 + m)*(a^2 + I*a^2*Tan[c + d*x])^2)/(d*(3 + m)*(4 + m)) - (2*(A*(4 + m)^2 - I*B*(19 + 8*m
+ m^2))*Tan[c + d*x]^(1 + m)*(a^4 + I*a^4*Tan[c + d*x]))/(d*(2 + m)*(3 + m)*(4 + m))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-(d/(b*c)), 0])))

Rule 3537

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

Rule 3592

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

Rule 3594

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

Rubi steps

\begin {align*} \int \tan ^m(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx &=\frac {i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^3}{d (4+m)}+\frac {\int \tan ^m(c+d x) (a+i a \tan (c+d x))^3 (-a (i B (1+m)-A (4+m))+a (i A (4+m)+B (7+m)) \tan (c+d x)) \, dx}{4+m}\\ &=\frac {i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^3}{d (4+m)}-\frac {(A (4+m)-i B (7+m)) \tan ^{1+m}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d (3+m) (4+m)}+\frac {\int \tan ^m(c+d x) (a+i a \tan (c+d x))^2 \left (-2 a^2 \left (i B \left (5+6 m+m^2\right )-A \left (8+6 m+m^2\right )\right )+2 a^2 \left (i A (4+m)^2+B \left (19+8 m+m^2\right )\right ) \tan (c+d x)\right ) \, dx}{12+7 m+m^2}\\ &=\frac {i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^3}{d (4+m)}-\frac {(A (4+m)-i B (7+m)) \tan ^{1+m}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d (3+m) (4+m)}-\frac {2 \left (A (4+m)^2-i B \left (19+8 m+m^2\right )\right ) \tan ^{1+m}(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{d (2+m) \left (12+7 m+m^2\right )}+\frac {\int \tan ^m(c+d x) (a+i a \tan (c+d x)) \left (-2 a^3 \left (i B \left (29+44 m+17 m^2+2 m^3\right )-A \left (32+44 m+17 m^2+2 m^3\right )\right )+2 a^3 \left (i A \left (64+60 m+19 m^2+2 m^3\right )+B \left (67+60 m+19 m^2+2 m^3\right )\right ) \tan (c+d x)\right ) \, dx}{24+26 m+9 m^2+m^3}\\ &=-\frac {2 a^4 \left (A \left (64+60 m+19 m^2+2 m^3\right )-i B \left (67+60 m+19 m^2+2 m^3\right )\right ) \tan ^{1+m}(c+d x)}{d (1+m) \left (24+26 m+9 m^2+m^3\right )}+\frac {i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^3}{d (4+m)}-\frac {(A (4+m)-i B (7+m)) \tan ^{1+m}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d (3+m) (4+m)}-\frac {2 \left (A (4+m)^2-i B \left (19+8 m+m^2\right )\right ) \tan ^{1+m}(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{d (2+m) \left (12+7 m+m^2\right )}+\frac {\int \tan ^m(c+d x) \left (8 a^4 (A-i B) (2+m) (3+m) (4+m)+8 a^4 (i A+B) (2+m) (3+m) (4+m) \tan (c+d x)\right ) \, dx}{24+26 m+9 m^2+m^3}\\ &=-\frac {2 a^4 \left (A \left (64+60 m+19 m^2+2 m^3\right )-i B \left (67+60 m+19 m^2+2 m^3\right )\right ) \tan ^{1+m}(c+d x)}{d (1+m) \left (24+26 m+9 m^2+m^3\right )}+\frac {i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^3}{d (4+m)}-\frac {(A (4+m)-i B (7+m)) \tan ^{1+m}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d (3+m) (4+m)}-\frac {2 \left (A (4+m)^2-i B \left (19+8 m+m^2\right )\right ) \tan ^{1+m}(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{d (2+m) \left (12+7 m+m^2\right )}+\frac {\left (64 i a^8 (A-i B)^2 (2+m) (3+m) (4+m)\right ) \operatorname {Subst}\left (\int \frac {8^{-m} \left (\frac {x}{a^4 (i A+B) (2+m) (3+m) (4+m)}\right )^m}{64 a^8 (i A+B)^2 (2+m)^2 (3+m)^2 (4+m)^2+8 a^4 (A-i B) (2+m) (3+m) (4+m) x} \, dx,x,8 a^4 (i A+B) (2+m) (3+m) (4+m) \tan (c+d x)\right )}{d}\\ &=-\frac {2 a^4 \left (A \left (64+60 m+19 m^2+2 m^3\right )-i B \left (67+60 m+19 m^2+2 m^3\right )\right ) \tan ^{1+m}(c+d x)}{d (1+m) \left (24+26 m+9 m^2+m^3\right )}+\frac {i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^3}{d (4+m)}-\frac {(A (4+m)-i B (7+m)) \tan ^{1+m}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d (3+m) (4+m)}-\frac {2 \left (A (4+m)^2-i B \left (19+8 m+m^2\right )\right ) \tan ^{1+m}(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{d (2+m) \left (12+7 m+m^2\right )}+\frac {\left (i 8^{2-m} a^8 (A-i B)^2 (2+m) (3+m) (4+m)\right ) \operatorname {Subst}\left (\int \frac {\left (\frac {x}{a^4 (i A+B) (2+m) (3+m) (4+m)}\right )^m}{64 a^8 (i A+B)^2 (2+m)^2 (3+m)^2 (4+m)^2+8 a^4 (A-i B) (2+m) (3+m) (4+m) x} \, dx,x,8 a^4 (i A+B) (2+m) (3+m) (4+m) \tan (c+d x)\right )}{d}\\ &=-\frac {2 a^4 \left (A \left (64+60 m+19 m^2+2 m^3\right )-i B \left (67+60 m+19 m^2+2 m^3\right )\right ) \tan ^{1+m}(c+d x)}{d (1+m) \left (24+26 m+9 m^2+m^3\right )}+\frac {8 a^4 (A-i B) \, _2F_1(1,1+m;2+m;i \tan (c+d x)) \tan ^{1+m}(c+d x)}{d (1+m)}+\frac {i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^3}{d (4+m)}-\frac {(A (4+m)-i B (7+m)) \tan ^{1+m}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d (3+m) (4+m)}-\frac {2 \left (A (4+m)^2-i B \left (19+8 m+m^2\right )\right ) \tan ^{1+m}(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{d (2+m) \left (12+7 m+m^2\right )}\\ \end {align*}

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Mathematica [B]  time = 14.23, size = 1805, normalized size = 6.22 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2^(3 - m)*(I*A + B)*(((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x))))^m*Cos[c + d*x]^5*(2^m*Hyper
geometric2F1[1, m, 1 + m, -((-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x))))] - (1 + E^((2*I)*(c + d*x)))
^m*Hypergeometric2F1[m, m, 1 + m, (1 - E^((2*I)*(c + d*x)))/2])*(a + I*a*Tan[c + d*x])^4*(A + B*Tan[c + d*x]))
/(d*E^((2*I)*c)*(1 + E^((2*I)*c))*m*(Cos[d*x] + I*Sin[d*x])^4*(A*Cos[c + d*x] + B*Sin[c + d*x])) - ((8*I)*(A -
I*B)*(-1 + E^((2*I)*(c + d*x)))^m*(((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x))))^m*Cos[c + d*x
]^5*(-(Hypergeometric2F1[1, m, 1 + m, (1 - E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]/((1 + E^((2*I)*(c +
d*x)))^m*m)) - ((1 + E^((2*I)*c))*(-1 + E^((2*I)*(c + d*x)))*(1 + E^((2*I)*(c + d*x)))^(-1 - m)*Hypergeometri
c2F1[1, 1 + m, 2 + m, (1 - E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))])/(1 + m) + Hypergeometric2F1[m, m,
1 + m, (1 - E^((2*I)*(c + d*x)))/2]/(2^m*m))*(a + I*a*Tan[c + d*x])^4*(A + B*Tan[c + d*x]))/(d*E^((4*I)*c)*(1
+ E^((2*I)*c))*((-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x))))^m*(Cos[d*x] + I*Sin[d*x])^4*(A*Cos[c + d
*x] + B*Sin[c + d*x])) + (Cos[c + d*x]^5*(((A - (2*I)*B)*Sec[c + d*x]^2*((-4*I)*Cos[4*c] - 4*Sin[4*c]))/(2 + m
) + ((A - (2*I)*B)*(-3 - 2*m + Cos[2*c])*Sec[c]^2*((-2*I)*Cos[4*c] - 2*Sin[4*c]))/((1 + m)*(2 + m)) + ((I*A*Co
s[c - d*x] + 2*B*Cos[c - d*x] - I*A*Cos[c + d*x] - 2*B*Cos[c + d*x])*Sec[c]^2*Sec[c + d*x]*(2*Cos[4*c] - (2*I)
*Sin[4*c]))/(1 + m))*Tan[c + d*x]^m*(a + I*a*Tan[c + d*x])^4*(A + B*Tan[c + d*x]))/(d*(Cos[d*x] + I*Sin[d*x])^
4*(A*Cos[c + d*x] + B*Sin[c + d*x])) + (Cos[c + d*x]^5*(((-8*B - 9*B*m - 2*B*m^2 + 8*B*Cos[2*c] + 3*B*m*Cos[2*
c])*Sec[c]^2*Sec[c + d*x]^2*(Cos[4*c]/2 - (I/2)*Sin[4*c]))/((2 + m)*(12 + 7*m + m^2)) + ((-(B*Cos[c - d*x]) +
B*Cos[c + d*x])*Sec[c]^2*Sec[c + d*x]^3*(Cos[4*c]/2 - (I/2)*Sin[4*c]))/(3 + m) + ((-(B*Cos[c - d*x]) + B*Cos[c
+ d*x])*Sec[c]^2*Sec[c + d*x]*(Cos[4*c] - I*Sin[4*c]))/((1 + m)*(3 + m)) + ((-11 - 10*m - 2*m^2 + 5*Cos[2*c]
+ 2*m*Cos[2*c])*Sec[c]^2*(B*Cos[4*c] - I*B*Sin[4*c]))/((1 + m)*(2 + m)*(3 + m)*(4 + m)) + (Sec[c + d*x]^4*(B*C
os[4*c] - I*B*Sin[4*c]))/(4 + m))*Tan[c + d*x]^m*(a + I*a*Tan[c + d*x])^4*(A + B*Tan[c + d*x]))/(d*(Cos[d*x] +
I*Sin[d*x])^4*(A*Cos[c + d*x] + B*Sin[c + d*x])) + (Cos[c + d*x]^5*((Sec[c]^2*Sec[c + d*x]^2*(B - B*Cos[2*c]
+ A*Sin[2*c] - (4*I)*B*Sin[2*c])*(Cos[4*c]/2 - (I/2)*Sin[4*c]))/(3 + m) + (Sec[c]^2*Sec[c + d*x]^3*(Cos[4*c]/2
- (I/2)*Sin[4*c])*(B*Cos[c - d*x] - B*Cos[c + d*x] - A*Sin[c - d*x] + (4*I)*B*Sin[c - d*x] + A*Sin[c + d*x] -
(4*I)*B*Sin[c + d*x]))/(3 + m) + (Sec[c]^2*Sec[c + d*x]*(Cos[4*c] - I*Sin[4*c])*(B*Cos[c - d*x] - B*Cos[c + d
*x] - A*Sin[c - d*x] + (4*I)*B*Sin[c - d*x] + A*Sin[c + d*x] - (4*I)*B*Sin[c + d*x]))/((1 + m)*(3 + m)) + (Sec
[c]*(A*Cos[c] - (4*I)*B*Cos[c] + B*Sin[c])*(2*Cos[4*c] - (2*I)*Sin[4*c])*Tan[c])/((1 + m)*(3 + m)))*Tan[c + d*
x]^m*(a + I*a*Tan[c + d*x])^4*(A + B*Tan[c + d*x]))/(d*(Cos[d*x] + I*Sin[d*x])^4*(A*Cos[c + d*x] + B*Sin[c + d
*x])) + (Cos[c + d*x]^5*((Sec[c]^2*Sec[c + d*x]*(2*Cos[4*c] - (2*I)*Sin[4*c])*((-I)*A*Cos[c - d*x] - 2*B*Cos[c
- d*x] + I*A*Cos[c + d*x] + 2*B*Cos[c + d*x] + 2*A*Sin[c - d*x] - (3*I)*B*Sin[c - d*x] - 2*A*Sin[c + d*x] + (
3*I)*B*Sin[c + d*x]))/(1 + m) + (Sec[c]*(2*A*Cos[c] - (3*I)*B*Cos[c] + I*A*Sin[c] + 2*B*Sin[c])*(-4*Cos[4*c] +
(4*I)*Sin[4*c])*Tan[c])/(1 + m))*Tan[c + d*x]^m*(a + I*a*Tan[c + d*x])^4*(A + B*Tan[c + d*x]))/(d*(Cos[d*x] +
I*Sin[d*x])^4*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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fricas [F]  time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {16 \, {\left ({\left (A - i \, B\right )} a^{4} e^{\left (10 i \, d x + 10 i \, c\right )} + {\left (A + i \, B\right )} a^{4} e^{\left (8 i \, d x + 8 i \, c\right )}\right )} \left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{m}}{e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(16*((A - I*B)*a^4*e^(10*I*d*x + 10*I*c) + (A + I*B)*a^4*e^(8*I*d*x + 8*I*c))*((-I*e^(2*I*d*x + 2*I*c)
+ I)/(e^(2*I*d*x + 2*I*c) + 1))^m/(e^(10*I*d*x + 10*I*c) + 5*e^(8*I*d*x + 8*I*c) + 10*e^(6*I*d*x + 6*I*c) + 1
0*e^(4*I*d*x + 4*I*c) + 5*e^(2*I*d*x + 2*I*c) + 1), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} \tan \left (d x + c\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.34, size = 0, normalized size = 0.00 \[ \int \left (\tan ^{m}\left (d x +c \right )\right ) \left (a +i a \tan \left (d x +c \right )\right )^{4} \left (A +B \tan \left (d x +c \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} \tan \left (d x + c\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\mathrm {tan}\left (c+d\,x\right )}^m\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{4} \left (\int A \tan ^{m}{\left (c + d x \right )}\, dx + \int \left (- 6 A \tan ^{2}{\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\right )\, dx + \int A \tan ^{4}{\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\, dx + \int B \tan {\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\, dx + \int \left (- 6 B \tan ^{3}{\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\right )\, dx + \int B \tan ^{5}{\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\, dx + \int 4 i A \tan {\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\, dx + \int \left (- 4 i A \tan ^{3}{\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\right )\, dx + \int 4 i B \tan ^{2}{\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\, dx + \int \left (- 4 i B \tan ^{4}{\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\right )\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**4*(Integral(A*tan(c + d*x)**m, x) + Integral(-6*A*tan(c + d*x)**2*tan(c + d*x)**m, x) + Integral(A*tan(c +
d*x)**4*tan(c + d*x)**m, x) + Integral(B*tan(c + d*x)*tan(c + d*x)**m, x) + Integral(-6*B*tan(c + d*x)**3*tan(
c + d*x)**m, x) + Integral(B*tan(c + d*x)**5*tan(c + d*x)**m, x) + Integral(4*I*A*tan(c + d*x)*tan(c + d*x)**m
, x) + Integral(-4*I*A*tan(c + d*x)**3*tan(c + d*x)**m, x) + Integral(4*I*B*tan(c + d*x)**2*tan(c + d*x)**m, x
) + Integral(-4*I*B*tan(c + d*x)**4*tan(c + d*x)**m, x))

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