3.8.10 \(\int \tan (c+d x) (a+b \tan (c+d x))^n \, dx\) [710]

Optimal. Leaf size=127 \[ -\frac {\, _2F_1\left (1,1+n;2+n;\frac {a+b \tan (c+d x)}{a-i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 (a-i b) d (1+n)}-\frac {\, _2F_1\left (1,1+n;2+n;\frac {a+b \tan (c+d x)}{a+i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 (a+i b) d (1+n)} \]

[Out]

-1/2*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)-1/2*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)

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Rubi [A]
time = 0.09, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3620, 3618, 70} \begin {gather*} -\frac {(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+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)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3618

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c*(
d/f), Subst[Int[(a + (b/d)*x)^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 3620

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]

Rubi steps

\begin {align*} \int \tan (c+d x) (a+b \tan (c+d x))^n \, dx &=\frac {1}{2} i \int (1-i \tan (c+d x)) (a+b \tan (c+d x))^n \, dx-\frac {1}{2} i \int (1+i \tan (c+d x)) (a+b \tan (c+d x))^n \, dx\\ &=\frac {\text {Subst}\left (\int \frac {(a-i b x)^n}{-1+x} \, dx,x,i \tan (c+d x)\right )}{2 d}+\frac {\text {Subst}\left (\int \frac {(a+i b x)^n}{-1+x} \, dx,x,-i \tan (c+d x)\right )}{2 d}\\ &=-\frac {\, _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 {\, _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)}\\ \end {align*}

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Mathematica [A]
time = 0.19, size = 117, normalized size = 0.92 \begin {gather*} -\frac {\left ((a+i b) \, _2F_1\left (1,1+n;2+n;\frac {a+b \tan (c+d x)}{a-i b}\right )+(a-i b) \, _2F_1\left (1,1+n;2+n;\frac {a+b \tan (c+d x)}{a+i b}\right )\right ) (a+b \tan (c+d x))^{1+n}}{2 (a-i b) (a+i b) d (1+n)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(((a + I*b)*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a - I*b)] + (a - I*b)*Hypergeometric
2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a + I*b)])*(a + b*Tan[c + d*x])^(1 + n))/((a - I*b)*(a + I*b)*d*(1
+ n))

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Maple [F]
time = 0.23, size = 0, normalized size = 0.00 \[\int \tan \left (d x +c \right ) \left (a +b \tan \left (d x +c \right )\right )^{n}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \tan {\left (c + d x \right )}\right )^{n} \tan {\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \mathrm {tan}\left (c+d\,x\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^n \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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