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

Optimal. Leaf size=72 \[ \frac{\cos (a+b x) \cos ^2(a+b x)^{n/2} (d \tan (a+b x))^{n+1} \, _2F_1\left (\frac{n}{2},\frac{n+1}{2};\frac{n+3}{2};\sin ^2(a+b x)\right )}{b d (n+1)} \]

[Out]

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

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Rubi [A]  time = 0.03143, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.059, Rules used = {2617} \[ \frac{\cos (a+b x) \cos ^2(a+b x)^{n/2} (d \tan (a+b x))^{n+1} \, _2F_1\left (\frac{n}{2},\frac{n+1}{2};\frac{n+3}{2};\sin ^2(a+b x)\right )}{b d (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2617

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*Sec[e +
f*x])^m*(b*Tan[e + f*x])^(n + 1)*(Cos[e + f*x]^2)^((m + n + 1)/2)*Hypergeometric2F1[(n + 1)/2, (m + n + 1)/2,
(n + 3)/2, Sin[e + f*x]^2])/(b*f*(n + 1)), x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !IntegerQ[(n - 1)/2] &&  !In
tegerQ[m/2]

Rubi steps

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

Mathematica [C]  time = 2.3333, size = 452, normalized size = 6.28 \[ -\frac{2 \sin \left (\frac{1}{2} (a+b x)\right ) \cos \left (\frac{1}{2} (a+b x)\right ) \cos (a+b x) \left (F_1\left (\frac{n+1}{2};n,1;\frac{n+3}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )-2 F_1\left (\frac{n+1}{2};n,2;\frac{n+3}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )\right ) (d \tan (a+b x))^n}{b (n+1) \left (\frac{\sec ^2\left (\frac{1}{2} (a+b x)\right ) \left ((n+3) (\cos (a+b x)+1) F_1\left (\frac{n+1}{2};n,2;\frac{n+3}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )-(\cos (a+b x)-1) \left (F_1\left (\frac{n+3}{2};n,2;\frac{n+5}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )-4 F_1\left (\frac{n+3}{2};n,3;\frac{n+5}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )-n F_1\left (\frac{n+3}{2};n+1,1;\frac{n+5}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )+2 n F_1\left (\frac{n+3}{2};n+1,2;\frac{n+5}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )\right )\right )}{n+3}-F_1\left (\frac{n+1}{2};n,1;\frac{n+3}{2};\tan ^2\left (\frac{1}{2} (a+b x)\right ),-\tan ^2\left (\frac{1}{2} (a+b x)\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

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Maple [F]  time = 0.767, size = 0, normalized size = 0. \begin{align*} \int \cos \left ( bx+a \right ) \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)*(d*tan(b*x+a))^n,x)

[Out]

int(cos(b*x+a)*(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 )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*tan(b*x + a))^n*cos(b*x + a), 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 ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((d*tan(b*x + a))^n*cos(b*x + a), 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{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((d*tan(a + b*x))**n*cos(a + b*x), 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 )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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