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

Optimal. Leaf size=272 \[ -\frac{\left (\sqrt{-b^2} \left (\frac{a^2}{b^2}-n+1\right )-a n\right ) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a-\sqrt{-b^2}}\right )}{4 b d (n+1) \left (\frac{a^2}{b^2}+1\right ) \left (a-\sqrt{-b^2}\right )}+\frac{b \left (\sqrt{-b^2} \left (\frac{a^2}{b^2}-n+1\right )+a n\right ) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a+\sqrt{-b^2}}\right )}{4 d (n+1) \left (a^2+b^2\right ) \left (a+\sqrt{-b^2}\right )}+\frac{\cos ^2(c+d x) (a \tan (c+d x)+b) (a+b \tan (c+d x))^{n+1}}{2 d \left (a^2+b^2\right )} \]

[Out]

-((Sqrt[-b^2]*(1 + a^2/b^2 - n) - a*n)*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a - Sqrt[-b^2]
)]*(a + b*Tan[c + d*x])^(1 + n))/(4*(1 + a^2/b^2)*b*(a - Sqrt[-b^2])*d*(1 + n)) + (b*(Sqrt[-b^2]*(1 + a^2/b^2
- n) + a*n)*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a + Sqrt[-b^2])]*(a + b*Tan[c + d*x])^(1
+ n))/(4*(a^2 + b^2)*(a + Sqrt[-b^2])*d*(1 + n)) + (Cos[c + d*x]^2*(b + a*Tan[c + d*x])*(a + b*Tan[c + d*x])^(
1 + n))/(2*(a^2 + b^2)*d)

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Rubi [A]  time = 0.457396, antiderivative size = 272, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3506, 741, 831, 68} \[ -\frac{\left (\sqrt{-b^2} \left (\frac{a^2}{b^2}-n+1\right )-a n\right ) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a-\sqrt{-b^2}}\right )}{4 b d (n+1) \left (\frac{a^2}{b^2}+1\right ) \left (a-\sqrt{-b^2}\right )}+\frac{b \left (\sqrt{-b^2} \left (\frac{a^2}{b^2}-n+1\right )+a n\right ) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a+\sqrt{-b^2}}\right )}{4 d (n+1) \left (a^2+b^2\right ) \left (a+\sqrt{-b^2}\right )}+\frac{\cos ^2(c+d x) (a \tan (c+d x)+b) (a+b \tan (c+d x))^{n+1}}{2 d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[-b^2]*(1 + a^2/b^2 - n) - a*n)*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a - Sqrt[-b^2]
)]*(a + b*Tan[c + d*x])^(1 + n))/(4*(1 + a^2/b^2)*b*(a - Sqrt[-b^2])*d*(1 + n)) + (b*(Sqrt[-b^2]*(1 + a^2/b^2
- n) + a*n)*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a + Sqrt[-b^2])]*(a + b*Tan[c + d*x])^(1
+ n))/(4*(a^2 + b^2)*(a + Sqrt[-b^2])*d*(1 + n)) + (Cos[c + d*x]^2*(b + a*Tan[c + d*x])*(a + b*Tan[c + d*x])^(
1 + n))/(2*(a^2 + b^2)*d)

Rule 3506

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

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 831

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

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]

Rubi steps

\begin{align*} \int \cos ^2(c+d x) (a+b \tan (c+d x))^n \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(a+x)^n}{\left (1+\frac{x^2}{b^2}\right )^2} \, dx,x,b \tan (c+d x)\right )}{b d}\\ &=\frac{\cos ^2(c+d x) (b+a \tan (c+d x)) (a+b \tan (c+d x))^{1+n}}{2 \left (a^2+b^2\right ) d}-\frac{b \operatorname{Subst}\left (\int \frac{(a+x)^n \left (-1-\frac{a^2}{b^2}+n+\frac{a n x}{b^2}\right )}{1+\frac{x^2}{b^2}} \, dx,x,b \tan (c+d x)\right )}{2 \left (a^2+b^2\right ) d}\\ &=\frac{\cos ^2(c+d x) (b+a \tan (c+d x)) (a+b \tan (c+d x))^{1+n}}{2 \left (a^2+b^2\right ) d}-\frac{b \operatorname{Subst}\left (\int \left (\frac{\left (-a n+\sqrt{-b^2} \left (-1-\frac{a^2}{b^2}+n\right )\right ) (a+x)^n}{2 \left (\sqrt{-b^2}-x\right )}+\frac{\left (a n+\sqrt{-b^2} \left (-1-\frac{a^2}{b^2}+n\right )\right ) (a+x)^n}{2 \left (\sqrt{-b^2}+x\right )}\right ) \, dx,x,b \tan (c+d x)\right )}{2 \left (a^2+b^2\right ) d}\\ &=\frac{\cos ^2(c+d x) (b+a \tan (c+d x)) (a+b \tan (c+d x))^{1+n}}{2 \left (a^2+b^2\right ) d}+\frac{\left (b \left (\sqrt{-b^2} \left (1+\frac{a^2}{b^2}-n\right )-a n\right )\right ) \operatorname{Subst}\left (\int \frac{(a+x)^n}{\sqrt{-b^2}+x} \, dx,x,b \tan (c+d x)\right )}{4 \left (a^2+b^2\right ) d}+\frac{\left (b \left (\sqrt{-b^2} \left (1+\frac{a^2}{b^2}-n\right )+a n\right )\right ) \operatorname{Subst}\left (\int \frac{(a+x)^n}{\sqrt{-b^2}-x} \, dx,x,b \tan (c+d x)\right )}{4 \left (a^2+b^2\right ) d}\\ &=-\frac{b \left (\sqrt{-b^2} \left (1+\frac{a^2}{b^2}-n\right )-a n\right ) \, _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}}{4 \left (a^2+b^2\right ) \left (a-\sqrt{-b^2}\right ) d (1+n)}+\frac{b \left (\sqrt{-b^2} \left (1+\frac{a^2}{b^2}-n\right )+a n\right ) \, _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}}{4 \left (a^2+b^2\right ) \left (a+\sqrt{-b^2}\right ) d (1+n)}+\frac{\cos ^2(c+d x) (b+a \tan (c+d x)) (a+b \tan (c+d x))^{1+n}}{2 \left (a^2+b^2\right ) d}\\ \end{align*}

Mathematica [A]  time = 1.24, size = 225, normalized size = 0.83 \[ \frac{(a+b \tan (c+d x))^{n+1} \left (-\frac{\left (\sqrt{-b^2} \left (a^2-b^2 (n-1)\right )-a b^2 n\right ) \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a-\sqrt{-b^2}}\right )}{(n+1) \left (a-\sqrt{-b^2}\right )}+\frac{\left (a^2 \sqrt{-b^2}+a b^2 n+\left (-b^2\right )^{3/2} (n-1)\right ) \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a+\sqrt{-b^2}}\right )}{(n+1) \left (a+\sqrt{-b^2}\right )}+2 b \cos ^2(c+d x) (a \tan (c+d x)+b)\right )}{4 b d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*Tan[c + d*x])^(1 + n)*(-(((Sqrt[-b^2]*(a^2 - b^2*(-1 + n)) - a*b^2*n)*Hypergeometric2F1[1, 1 + n, 2 +
n, (a + b*Tan[c + d*x])/(a - Sqrt[-b^2])])/((a - Sqrt[-b^2])*(1 + n))) + ((a^2*Sqrt[-b^2] + (-b^2)^(3/2)*(-1 +
n) + a*b^2*n)*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a + Sqrt[-b^2])])/((a + Sqrt[-b^2])*(1
+ n)) + 2*b*Cos[c + d*x]^2*(b + a*Tan[c + d*x])))/(4*b*(a^2 + b^2)*d)

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Maple [F]  time = 0.377, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( dx+c \right ) \right ) ^{2} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (d x + c\right ) + a\right )}^{n} \cos \left (d x + c\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \tan \left (d x + c\right ) + a\right )}^{n} \cos \left (d x + c\right )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (d x + c\right ) + a\right )}^{n} \cos \left (d x + c\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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