3.220 \(\int \tan ^2(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=164 \[ \frac {(B+i A) (a+i a \tan (c+d x))^n \, _2F_1\left (1,n;n+1;\frac {1}{2} (i \tan (c+d x)+1)\right )}{2 d n}-\frac {(B n+i A (n+2)) (a+i a \tan (c+d x))^{n+1}}{a d (n+1) (n+2)}+\frac {B \tan ^2(c+d x) (a+i a \tan (c+d x))^n}{d (n+2)}-\frac {2 B (a+i a \tan (c+d x))^n}{d n (n+2)} \]

[Out]

-2*B*(a+I*a*tan(d*x+c))^n/d/n/(2+n)+1/2*(I*A+B)*hypergeom([1, n],[1+n],1/2+1/2*I*tan(d*x+c))*(a+I*a*tan(d*x+c)
)^n/d/n+B*tan(d*x+c)^2*(a+I*a*tan(d*x+c))^n/d/(2+n)-(B*n+I*A*(2+n))*(a+I*a*tan(d*x+c))^(1+n)/a/d/(1+n)/(2+n)

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Rubi [A]  time = 0.31, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {3597, 3592, 3527, 3481, 68} \[ \frac {(B+i A) (a+i a \tan (c+d x))^n \, _2F_1\left (1,n;n+1;\frac {1}{2} (i \tan (c+d x)+1)\right )}{2 d n}-\frac {(B n+i A (n+2)) (a+i a \tan (c+d x))^{n+1}}{a d (n+1) (n+2)}+\frac {B \tan ^2(c+d x) (a+i a \tan (c+d x))^n}{d (n+2)}-\frac {2 B (a+i a \tan (c+d x))^n}{d n (n+2)} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^2*(a + I*a*Tan[c + d*x])^n*(A + B*Tan[c + d*x]),x]

[Out]

(-2*B*(a + I*a*Tan[c + d*x])^n)/(d*n*(2 + n)) + ((I*A + B)*Hypergeometric2F1[1, n, 1 + n, (1 + I*Tan[c + d*x])
/2]*(a + I*a*Tan[c + d*x])^n)/(2*d*n) + (B*Tan[c + d*x]^2*(a + I*a*Tan[c + d*x])^n)/(d*(2 + n)) - ((B*n + I*A*
(2 + n))*(a + I*a*Tan[c + d*x])^(1 + n))/(a*d*(1 + n)*(2 + n))

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 3481

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 3527

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d*
(a + b*Tan[e + f*x])^m)/(f*m), x] + Dist[(b*c + a*d)/b, Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a, b, c,
d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] &&  !LtQ[m, 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 3597

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*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n)/(f*(m + n)), x] +
Dist[1/(a*(m + n)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Simp[a*A*c*(m + n) - B*(b*c*m + a*
d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*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] && GtQ[n, 0]

Rubi steps

\begin {align*} \int \tan ^2(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx &=\frac {B \tan ^2(c+d x) (a+i a \tan (c+d x))^n}{d (2+n)}+\frac {\int \tan (c+d x) (a+i a \tan (c+d x))^n (-2 a B-a (i B n-A (2+n)) \tan (c+d x)) \, dx}{a (2+n)}\\ &=\frac {B \tan ^2(c+d x) (a+i a \tan (c+d x))^n}{d (2+n)}-\frac {(B n+i A (2+n)) (a+i a \tan (c+d x))^{1+n}}{a d (1+n) (2+n)}+\frac {\int (a+i a \tan (c+d x))^n (a (i B n-A (2+n))-2 a B \tan (c+d x)) \, dx}{a (2+n)}\\ &=-\frac {2 B (a+i a \tan (c+d x))^n}{d n (2+n)}+\frac {B \tan ^2(c+d x) (a+i a \tan (c+d x))^n}{d (2+n)}-\frac {(B n+i A (2+n)) (a+i a \tan (c+d x))^{1+n}}{a d (1+n) (2+n)}+(-A+i B) \int (a+i a \tan (c+d x))^n \, dx\\ &=-\frac {2 B (a+i a \tan (c+d x))^n}{d n (2+n)}+\frac {B \tan ^2(c+d x) (a+i a \tan (c+d x))^n}{d (2+n)}-\frac {(B n+i A (2+n)) (a+i a \tan (c+d x))^{1+n}}{a d (1+n) (2+n)}+\frac {(a (i A+B)) \operatorname {Subst}\left (\int \frac {(a+x)^{-1+n}}{a-x} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac {2 B (a+i a \tan (c+d x))^n}{d n (2+n)}+\frac {(i A+B) \, _2F_1\left (1,n;1+n;\frac {1}{2} (1+i \tan (c+d x))\right ) (a+i a \tan (c+d x))^n}{2 d n}+\frac {B \tan ^2(c+d x) (a+i a \tan (c+d x))^n}{d (2+n)}-\frac {(B n+i A (2+n)) (a+i a \tan (c+d x))^{1+n}}{a d (1+n) (2+n)}\\ \end {align*}

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Mathematica [F]  time = 22.69, size = 0, normalized size = 0.00 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Tan[c + d*x]^2*(a + I*a*Tan[c + d*x])^n*(A + B*Tan[c + d*x]),x]

[Out]

Integrate[Tan[c + d*x]^2*(a + I*a*Tan[c + d*x])^n*(A + B*Tan[c + d*x]), x]

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fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left ({\left (A - i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - {\left (A - 3 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (A + 3 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + A + i \, B\right )} \left (\frac {2 \, a e^{\left (2 i \, d x + 2 i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{n}}{e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

integral(-((A - I*B)*e^(6*I*d*x + 6*I*c) - (A - 3*I*B)*e^(4*I*d*x + 4*I*c) - (A + 3*I*B)*e^(2*I*d*x + 2*I*c) +
A + I*B)*(2*a*e^(2*I*d*x + 2*I*c)/(e^(2*I*d*x + 2*I*c) + 1))^n/(e^(6*I*d*x + 6*I*c) + 3*e^(4*I*d*x + 4*I*c) +
3*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 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^n*tan(d*x + c)^2, x)

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maple [F]  time = 5.48, size = 0, normalized size = 0.00 \[ \int \left (\tan ^{2}\left (d x +c \right )\right ) \left (a +i a \tan \left (d x +c \right )\right )^{n} \left (A +B \tan \left (d x +c \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c)),x)

[Out]

int(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c)),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^n*tan(d*x + c)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {tan}\left (c+d\,x\right )}^2\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^2*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^n,x)

[Out]

int(tan(c + d*x)^2*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^n, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**2*(a+I*a*tan(d*x+c))**n*(A+B*tan(d*x+c)),x)

[Out]

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

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