3.205 \(\int \tan ^m(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=205 \[ \frac {4 a^3 (A-i B) \tan ^{m+1}(c+d x) \, _2F_1(1,m+1;m+2;i \tan (c+d x))}{d (m+1)}-\frac {a^3 \left (A \left (2 m^2+11 m+15\right )-i B \left (2 m^2+11 m+17\right )\right ) \tan ^{m+1}(c+d x)}{d (m+1) (m+2) (m+3)}-\frac {(A (m+3)-i B (m+5)) \left (a^3+i a^3 \tan (c+d x)\right ) \tan ^{m+1}(c+d x)}{d (m+2) (m+3)}+\frac {i a B (a+i a \tan (c+d x))^2 \tan ^{m+1}(c+d x)}{d (m+3)} \]

[Out]

-a^3*(A*(2*m^2+11*m+15)-I*B*(2*m^2+11*m+17))*tan(d*x+c)^(1+m)/d/(3+m)/(m^2+3*m+2)+4*a^3*(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))^2/d/(3+m)-(A*(3+m)
-I*B*(5+m))*tan(d*x+c)^(1+m)*(a^3+I*a^3*tan(d*x+c))/d/(2+m)/(3+m)

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Rubi [A]  time = 0.64, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ \frac {4 a^3 (A-i B) \tan ^{m+1}(c+d x) \, _2F_1(1,m+1;m+2;i \tan (c+d x))}{d (m+1)}-\frac {a^3 \left (A \left (2 m^2+11 m+15\right )-i B \left (2 m^2+11 m+17\right )\right ) \tan ^{m+1}(c+d x)}{d (m+1) (m+2) (m+3)}-\frac {(A (m+3)-i B (m+5)) \left (a^3+i a^3 \tan (c+d x)\right ) \tan ^{m+1}(c+d x)}{d (m+2) (m+3)}+\frac {i a B (a+i a \tan (c+d x))^2 \tan ^{m+1}(c+d x)}{d (m+3)} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^m*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

-((a^3*(A*(15 + 11*m + 2*m^2) - I*B*(17 + 11*m + 2*m^2))*Tan[c + d*x]^(1 + m))/(d*(1 + m)*(2 + m)*(3 + m))) +
(4*a^3*(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])^2)/(d*(3 + m)) - ((A*(3 + m) - I*B*(5 + m))*Tan[c + d*x]^(1 + m)*
(a^3 + I*a^3*Tan[c + d*x]))/(d*(2 + m)*(3 + 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))^3 (A+B \tan (c+d x)) \, dx &=\frac {i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^2}{d (3+m)}+\frac {\int \tan ^m(c+d x) (a+i a \tan (c+d x))^2 (-a (i B (1+m)-A (3+m))+a (i A (3+m)+B (5+m)) \tan (c+d x)) \, dx}{3+m}\\ &=\frac {i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^2}{d (3+m)}-\frac {(A (3+m)-i B (5+m)) \tan ^{1+m}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{d (2+m) (3+m)}+\frac {\int \tan ^m(c+d x) (a+i a \tan (c+d x)) \left (-a^2 \left (i B \left (7+9 m+2 m^2\right )-A \left (9+9 m+2 m^2\right )\right )+a^2 \left (i A \left (15+11 m+2 m^2\right )+B \left (17+11 m+2 m^2\right )\right ) \tan (c+d x)\right ) \, dx}{6+5 m+m^2}\\ &=-\frac {a^3 \left (A \left (15+11 m+2 m^2\right )-i B \left (17+11 m+2 m^2\right )\right ) \tan ^{1+m}(c+d x)}{d (1+m) \left (6+5 m+m^2\right )}+\frac {i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^2}{d (3+m)}-\frac {(A (3+m)-i B (5+m)) \tan ^{1+m}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{d (2+m) (3+m)}+\frac {\int \tan ^m(c+d x) \left (4 a^3 (A-i B) (2+m) (3+m)+4 a^3 (i A+B) (2+m) (3+m) \tan (c+d x)\right ) \, dx}{6+5 m+m^2}\\ &=-\frac {a^3 \left (A \left (15+11 m+2 m^2\right )-i B \left (17+11 m+2 m^2\right )\right ) \tan ^{1+m}(c+d x)}{d (1+m) \left (6+5 m+m^2\right )}+\frac {i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^2}{d (3+m)}-\frac {(A (3+m)-i B (5+m)) \tan ^{1+m}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{d (2+m) (3+m)}+\frac {\left (16 i a^6 (A-i B)^2 (2+m) (3+m)\right ) \operatorname {Subst}\left (\int \frac {4^{-m} \left (\frac {x}{a^3 (i A+B) (2+m) (3+m)}\right )^m}{16 a^6 (i A+B)^2 (2+m)^2 (3+m)^2+4 a^3 (A-i B) (2+m) (3+m) x} \, dx,x,4 a^3 (i A+B) (2+m) (3+m) \tan (c+d x)\right )}{d}\\ &=-\frac {a^3 \left (A \left (15+11 m+2 m^2\right )-i B \left (17+11 m+2 m^2\right )\right ) \tan ^{1+m}(c+d x)}{d (1+m) \left (6+5 m+m^2\right )}+\frac {i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^2}{d (3+m)}-\frac {(A (3+m)-i B (5+m)) \tan ^{1+m}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{d (2+m) (3+m)}+\frac {\left (i 4^{2-m} a^6 (A-i B)^2 (2+m) (3+m)\right ) \operatorname {Subst}\left (\int \frac {\left (\frac {x}{a^3 (i A+B) (2+m) (3+m)}\right )^m}{16 a^6 (i A+B)^2 (2+m)^2 (3+m)^2+4 a^3 (A-i B) (2+m) (3+m) x} \, dx,x,4 a^3 (i A+B) (2+m) (3+m) \tan (c+d x)\right )}{d}\\ &=-\frac {a^3 \left (A \left (15+11 m+2 m^2\right )-i B \left (17+11 m+2 m^2\right )\right ) \tan ^{1+m}(c+d x)}{d (1+m) \left (6+5 m+m^2\right )}+\frac {4 a^3 (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))^2}{d (3+m)}-\frac {(A (3+m)-i B (5+m)) \tan ^{1+m}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{d (2+m) (3+m)}\\ \end {align*}

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Mathematica [B]  time = 12.64, size = 1305, normalized size = 6.37 \[ -\frac {4 i (A-i B) e^{-3 i c} \left (-1+e^{2 i (c+d x)}\right )^m \left (-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}\right )^m \cos ^4(c+d x) \left (-\frac {\left (1+e^{2 i c}\right ) \left (-1+e^{2 i (c+d x)}\right ) \, _2F_1\left (1,m+1;m+2;\frac {1-e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right ) \left (1+e^{2 i (c+d x)}\right )^{-m-1}}{m+1}-\frac {\, _2F_1\left (1,m;m+1;\frac {1-e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right ) \left (1+e^{2 i (c+d x)}\right )^{-m}}{m}+\frac {2^{-m} \, _2F_1\left (m,m;m+1;\frac {1}{2} \left (1-e^{2 i (c+d x)}\right )\right )}{m}\right ) (i \tan (c+d x) a+a)^3 (A+B \tan (c+d x)) \left (\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{-m}}{d \left (1+e^{2 i c}\right ) (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac {\cos ^4(c+d x) \left (\frac {\sec (c+d x) \left (\frac {1}{2} \cos (3 c)-\frac {1}{2} i \sin (3 c)\right ) (-i A \cos (c-d x)-3 B \cos (c-d x)+i A \cos (c+d x)+3 B \cos (c+d x)+3 A \sin (c-d x)-5 i B \sin (c-d x)-3 A \sin (c+d x)+5 i B \sin (c+d x)) \sec ^2(c)}{m+1}+\frac {(3 A \cos (c)-5 i B \cos (c)+i A \sin (c)+3 B \sin (c)) (i \sin (3 c)-\cos (3 c)) \tan (c) \sec (c)}{m+1}\right ) \tan ^m(c+d x) (i \tan (c+d x) a+a)^3 (A+B \tan (c+d x))}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac {\cos ^4(c+d x) \left (-\frac {i B \sec (c) (\cos (3 c)-i \sin (3 c)) \sin (d x) \sec ^3(c+d x)}{m+3}-\frac {i B (\cos (3 c)-i \sin (3 c)) \tan (c) \sec ^2(c+d x)}{m+3}-\frac {i B \sec (c) (2 \cos (3 c)-2 i \sin (3 c)) \sin (d x) \sec (c+d x)}{(m+1) (m+3)}-\frac {i (2 B \cos (3 c)-2 i B \sin (3 c)) \tan (c)}{(m+1) (m+3)}\right ) \tan ^m(c+d x) (i \tan (c+d x) a+a)^3 (A+B \tan (c+d x))}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac {\cos ^4(c+d x) \left (\frac {(A-3 i B) (-2 m+\cos (2 c)-3) \left (-\frac {1}{2} i \cos (3 c)-\frac {1}{2} \sin (3 c)\right ) \sec ^2(c)}{(m+1) (m+2)}+\frac {(i A \cos (c-d x)+3 B \cos (c-d x)-i A \cos (c+d x)-3 B \cos (c+d x)) \sec (c+d x) \left (\frac {1}{2} \cos (3 c)-\frac {1}{2} i \sin (3 c)\right ) \sec ^2(c)}{m+1}+\frac {(A-3 i B) \sec ^2(c+d x) (-i \cos (3 c)-\sin (3 c))}{m+2}\right ) \tan ^m(c+d x) (i \tan (c+d x) a+a)^3 (A+B \tan (c+d x))}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac {2^{2-m} (i A+B) e^{-i c} \left (-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}\right )^m \cos ^4(c+d x) \left (2^m \, _2F_1\left (1,m;m+1;-\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )-\left (1+e^{2 i (c+d x)}\right )^m \, _2F_1\left (m,m;m+1;\frac {1}{2} \left (1-e^{2 i (c+d x)}\right )\right )\right ) (i \tan (c+d x) a+a)^3 (A+B \tan (c+d x))}{d \left (1+e^{2 i c}\right ) m (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Tan[c + d*x]^m*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

(2^(2 - m)*(I*A + B)*(((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x))))^m*Cos[c + d*x]^4*(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])^3*(A + B*Tan[c + d*x]))
/(d*E^(I*c)*(1 + E^((2*I)*c))*m*(Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[c + d*x])) - ((4*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]^4*
(-(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)*Hypergeometric2F1
[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])^3*(A + B*Tan[c + d*x]))/(d*E^((3*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])^3*(A*Cos[c + d*x]
+ B*Sin[c + d*x])) + (Cos[c + d*x]^4*(((A - (3*I)*B)*Sec[c + d*x]^2*((-I)*Cos[3*c] - Sin[3*c]))/(2 + m) + ((A
- (3*I)*B)*(-3 - 2*m + Cos[2*c])*Sec[c]^2*((-1/2*I)*Cos[3*c] - Sin[3*c]/2))/((1 + m)*(2 + m)) + ((I*A*Cos[c -
d*x] + 3*B*Cos[c - d*x] - I*A*Cos[c + d*x] - 3*B*Cos[c + d*x])*Sec[c]^2*Sec[c + d*x]*(Cos[3*c]/2 - (I/2)*Sin[3
*c]))/(1 + m))*Tan[c + d*x]^m*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/(d*(Cos[d*x] + I*Sin[d*x])^3*(A*C
os[c + d*x] + B*Sin[c + d*x])) + (Cos[c + d*x]^4*((Sec[c]^2*Sec[c + d*x]*(Cos[3*c]/2 - (I/2)*Sin[3*c])*((-I)*A
*Cos[c - d*x] - 3*B*Cos[c - d*x] + I*A*Cos[c + d*x] + 3*B*Cos[c + d*x] + 3*A*Sin[c - d*x] - (5*I)*B*Sin[c - d*
x] - 3*A*Sin[c + d*x] + (5*I)*B*Sin[c + d*x]))/(1 + m) + (Sec[c]*(3*A*Cos[c] - (5*I)*B*Cos[c] + I*A*Sin[c] + 3
*B*Sin[c])*(-Cos[3*c] + I*Sin[3*c])*Tan[c])/(1 + m))*Tan[c + d*x]^m*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*
x]))/(d*(Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[c + d*x])) + (Cos[c + d*x]^4*(((-I)*B*Sec[c]*Sec[c +
d*x]^3*(Cos[3*c] - I*Sin[3*c])*Sin[d*x])/(3 + m) - (I*B*Sec[c]*Sec[c + d*x]*(2*Cos[3*c] - (2*I)*Sin[3*c])*Sin
[d*x])/((1 + m)*(3 + m)) - (I*B*Sec[c + d*x]^2*(Cos[3*c] - I*Sin[3*c])*Tan[c])/(3 + m) - (I*(2*B*Cos[3*c] - (2
*I)*B*Sin[3*c])*Tan[c])/((1 + m)*(3 + m)))*Tan[c + d*x]^m*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/(d*(C
os[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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fricas [F]  time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {8 \, {\left ({\left (A - i \, B\right )} a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} + {\left (A + i \, B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{m}}{e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

integral(8*((A - I*B)*a^3*e^(8*I*d*x + 8*I*c) + (A + I*B)*a^3*e^(6*I*d*x + 6*I*c))*((-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) + 4*e^(6*I*d*x + 6*I*c) + 6*e^(4*I*d*x + 4*I*c) + 4*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 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} \tan \left (d x + c\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^3*tan(d*x + c)^m, x)

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maple [F]  time = 1.28, size = 0, normalized size = 0.00 \[ \int \left (\tan ^{m}\left (d x +c \right )\right ) \left (a +i a \tan \left (d x +c \right )\right )^{3} \left (A +B \tan \left (d x +c \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x)

[Out]

int(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} \tan \left (d x + c\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^3*tan(d*x + c)^m, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\mathrm {tan}\left (c+d\,x\right )}^m\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3 \,d x \]

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)^3,x)

[Out]

int(tan(c + d*x)^m*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**m*(a+I*a*tan(d*x+c))**3*(A+B*tan(d*x+c)),x)

[Out]

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

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