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

Optimal. Leaf size=50 \[ \frac{(d \tan (a+b x))^{n+1} \, _2F_1\left (2,\frac{n+1}{2};\frac{n+3}{2};-\tan ^2(a+b x)\right )}{b d (n+1)} \]

[Out]

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

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Rubi [A]  time = 0.0466357, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2607, 364} \[ \frac{(d \tan (a+b x))^{n+1} \, _2F_1\left (2,\frac{n+1}{2};\frac{n+3}{2};-\tan ^2(a+b x)\right )}{b d (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2607

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

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \cos ^2(a+b x) (d \tan (a+b x))^n \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(d x)^n}{\left (1+x^2\right )^2} \, dx,x,\tan (a+b x)\right )}{b}\\ &=\frac{\, _2F_1\left (2,\frac{1+n}{2};\frac{3+n}{2};-\tan ^2(a+b x)\right ) (d \tan (a+b x))^{1+n}}{b d (1+n)}\\ \end{align*}

Mathematica [C]  time = 4.37625, size = 939, normalized size = 18.78 \[ \frac{2 \left (F_1\left (\frac{n+1}{2};n,1;\frac{n+3}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )-4 F_1\left (\frac{n+1}{2};n,2;\frac{n+3}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )+4 F_1\left (\frac{n+1}{2};n,3;\frac{n+3}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )\right ) \cos ^2(a+b x) \tan \left (\frac{1}{2} (a+b x)\right ) (d \tan (a+b x))^n}{b \left (\frac{2 (n+1) \left (-F_1\left (\frac{n+3}{2};n,2;\frac{n+5}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )+8 F_1\left (\frac{n+3}{2};n,3;\frac{n+5}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )-12 F_1\left (\frac{n+3}{2};n,4;\frac{n+5}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )+n F_1\left (\frac{n+3}{2};n+1,1;\frac{n+5}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )-4 n F_1\left (\frac{n+3}{2};n+1,2;\frac{n+5}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )+4 n F_1\left (\frac{n+3}{2};n+1,3;\frac{n+5}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )\right ) \tan ^2\left (\frac{1}{2} (a+b x)\right ) \sec ^2\left (\frac{1}{2} (a+b x)\right )}{n+3}+\left (F_1\left (\frac{n+1}{2};n,1;\frac{n+3}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )-4 F_1\left (\frac{n+1}{2};n,2;\frac{n+3}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )+4 F_1\left (\frac{n+1}{2};n,3;\frac{n+3}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )\right ) \sec ^2\left (\frac{1}{2} (a+b x)\right )+n \left (F_1\left (\frac{n+1}{2};n,1;\frac{n+3}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )-4 F_1\left (\frac{n+1}{2};n,2;\frac{n+3}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )+4 F_1\left (\frac{n+1}{2};n,3;\frac{n+3}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )\right ) \sec (a+b x) \sec ^2\left (\frac{1}{2} (a+b x)\right )-2 n \left (F_1\left (\frac{n+1}{2};n,1;\frac{n+3}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )-4 F_1\left (\frac{n+1}{2};n,2;\frac{n+3}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )+4 F_1\left (\frac{n+1}{2};n,3;\frac{n+3}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )\right ) \sec (a+b x) \tan ^2\left (\frac{1}{2} (a+b x)\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*(AppellF1[(1 + n)/2, n, 1, (3 + n)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] - 4*AppellF1[(1 + n)/2, n, 2
, (3 + n)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] + 4*AppellF1[(1 + n)/2, n, 3, (3 + n)/2, Tan[(a + b*x)/2
]^2, -Tan[(a + b*x)/2]^2])*Cos[a + b*x]^2*Tan[(a + b*x)/2]*(d*Tan[a + b*x])^n)/(b*((AppellF1[(1 + n)/2, n, 1,
(3 + n)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] - 4*AppellF1[(1 + n)/2, n, 2, (3 + n)/2, Tan[(a + b*x)/2]^
2, -Tan[(a + b*x)/2]^2] + 4*AppellF1[(1 + n)/2, n, 3, (3 + n)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2])*Sec
[(a + b*x)/2]^2 + n*(AppellF1[(1 + n)/2, n, 1, (3 + n)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] - 4*AppellF
1[(1 + n)/2, n, 2, (3 + n)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] + 4*AppellF1[(1 + n)/2, n, 3, (3 + n)/2
, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2])*Sec[(a + b*x)/2]^2*Sec[a + b*x] + (2*(1 + n)*(-AppellF1[(3 + n)/2,
n, 2, (5 + n)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] + 8*AppellF1[(3 + n)/2, n, 3, (5 + n)/2, Tan[(a + b
*x)/2]^2, -Tan[(a + b*x)/2]^2] - 12*AppellF1[(3 + n)/2, n, 4, (5 + n)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]
^2] + n*AppellF1[(3 + n)/2, 1 + n, 1, (5 + n)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] - 4*n*AppellF1[(3 +
n)/2, 1 + n, 2, (5 + n)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] + 4*n*AppellF1[(3 + n)/2, 1 + n, 3, (5 + n
)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2])*Sec[(a + b*x)/2]^2*Tan[(a + b*x)/2]^2)/(3 + n) - 2*n*(AppellF1[
(1 + n)/2, n, 1, (3 + n)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] - 4*AppellF1[(1 + n)/2, n, 2, (3 + n)/2,
Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] + 4*AppellF1[(1 + n)/2, n, 3, (3 + n)/2, Tan[(a + b*x)/2]^2, -Tan[(a
+ b*x)/2]^2])*Sec[a + b*x]*Tan[(a + b*x)/2]^2))

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Maple [F]  time = 0.651, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( bx+a \right ) \right ) ^{2} \left ( d\tan \left ( bx+a \right ) \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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