3.709 \(\int \tan ^2(c+d x) (a+b \tan (c+d x))^n \, dx\)

Optimal. Leaf size=193 \[ -\frac {b (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right )}{2 \sqrt {-b^2} d (n+1) \left (a-\sqrt {-b^2}\right )}+\frac {b (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right )}{2 \sqrt {-b^2} d (n+1) \left (a+\sqrt {-b^2}\right )}+\frac {(a+b \tan (c+d x))^{n+1}}{b d (n+1)} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3543, 3485, 712, 68} \[ -\frac {b (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right )}{2 \sqrt {-b^2} d (n+1) \left (a-\sqrt {-b^2}\right )}+\frac {b (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right )}{2 \sqrt {-b^2} d (n+1) \left (a+\sqrt {-b^2}\right )}+\frac {(a+b \tan (c+d x))^{n+1}}{b d (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a + b*Tan[c + d*x])^(1 + n)/(b*d*(1 + n)) - (b*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a - S
qrt[-b^2])]*(a + b*Tan[c + d*x])^(1 + n))/(2*Sqrt[-b^2]*(a - Sqrt[-b^2])*d*(1 + n)) + (b*Hypergeometric2F1[1,
1 + n, 2 + n, (a + b*Tan[c + d*x])/(a + Sqrt[-b^2])]*(a + b*Tan[c + d*x])^(1 + n))/(2*Sqrt[-b^2]*(a + Sqrt[-b^
2])*d*(1 + 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 712

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m, 1/(a + c*x^2
), x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[m]

Rule 3485

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[(a + x)^n/(b^2 + x^2), x], x
, b*Tan[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[a^2 + b^2, 0]

Rule 3543

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

Rubi steps

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

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Mathematica [C]  time = 0.20, size = 138, normalized size = 0.72 \[ \frac {(a+b \tan (c+d x))^{n+1} \left (i b (a+i b) \, _2F_1\left (1,n+1;n+2;\frac {a+b \tan (c+d x)}{a-i b}\right )+(a-i b) \left (-i b \, _2F_1\left (1,n+1;n+2;\frac {a+b \tan (c+d x)}{a+i b}\right )+2 a+2 i b\right )\right )}{2 b d (n+1) (b-i a) (b+i a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.93, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*tan(d*x + c) + a)^n*tan(d*x + c)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*tan(d*x + c) + a)^n*tan(d*x + c)^2, x)

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maple [F]  time = 0.71, size = 0, normalized size = 0.00 \[ \int \left (\tan ^{2}\left (d x +c \right )\right ) \left (a +b \tan \left (d x +c \right )\right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^2*(a+b*tan(d*x+c))^n,x)

[Out]

int(tan(d*x+c)^2*(a+b*tan(d*x+c))^n,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*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 \[ \int {\mathrm {tan}\left (c+d\,x\right )}^2\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^2*(a + b*tan(c + d*x))^n,x)

[Out]

int(tan(c + d*x)^2*(a + b*tan(c + d*x))^n, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**2*(a+b*tan(d*x+c))**n,x)

[Out]

Integral((a + b*tan(c + d*x))**n*tan(c + d*x)**2, x)

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