3.499 \(\int \cot (c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=190 \[ \frac {(B+i A) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {a+b \tan (c+d x)}{a-i b}\right )}{2 d (n+1) (b+i a)}+\frac {(A+i B) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {a+b \tan (c+d x)}{a+i b}\right )}{2 d (n+1) (a+i b)}-\frac {A (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {b \tan (c+d x)}{a}+1\right )}{a d (n+1)} \]

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {3613, 3539, 3537, 68, 3634, 65} \[ \frac {(B+i A) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {a+b \tan (c+d x)}{a-i b}\right )}{2 d (n+1) (b+i a)}+\frac {(A+i B) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {a+b \tan (c+d x)}{a+i b}\right )}{2 d (n+1) (a+i b)}-\frac {A (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {b \tan (c+d x)}{a}+1\right )}{a d (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 65

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

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 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 3539

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

Rule 3613

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

Rule 3634

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

Rubi steps

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

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Mathematica [A]  time = 0.42, size = 169, normalized size = 0.89 \[ \frac {(a+b \tan (c+d x))^{n+1} \left (a (a+i b) (A-i B) \, _2F_1\left (1,n+1;n+2;\frac {a+b \tan (c+d x)}{a-i b}\right )+(a-i b) \left (a (A+i B) \, _2F_1\left (1,n+1;n+2;\frac {a+b \tan (c+d x)}{a+i b}\right )-2 A (a+i b) \, _2F_1\left (1,n+1;n+2;\frac {b \tan (c+d x)}{a}+1\right )\right )\right )}{2 a d (n+1) (a-i b) (a+i b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B \cot \left (d x + c\right ) \tan \left (d x + c\right ) + A \cot \left (d x + c\right )\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*cot(d*x + c)*tan(d*x + c) + A*cot(d*x + c))*(b*tan(d*x + c) + a)^n, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*tan(d*x + c) + A)*(b*tan(d*x + c) + a)^n*cot(d*x + c), x)

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maple [F]  time = 1.57, size = 0, normalized size = 0.00 \[ \int \cot \left (d x +c \right ) \left (a +b \tan \left (d x +c \right )\right )^{n} \left (A +B \tan \left (d x +c \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)*(a+b*tan(d*x+c))^n*(A+B*tan(d*x+c)),x)

[Out]

int(cot(d*x+c)*(a+b*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 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \cot \left (d x + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*tan(d*x + c) + A)*(b*tan(d*x + c) + a)^n*cot(d*x + c), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)*(A + B*tan(c + d*x))*(a + b*tan(c + d*x))^n,x)

[Out]

int(cot(c + d*x)*(A + B*tan(c + d*x))*(a + b*tan(c + d*x))^n, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)*(a+b*tan(d*x+c))**n*(A+B*tan(d*x+c)),x)

[Out]

Integral((A + B*tan(c + d*x))*(a + b*tan(c + d*x))**n*cot(c + d*x), x)

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