3.4.67 \(\int \cos ^2(a+b x) (d \tan (a+b x))^n \, dx\) [367]

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 50, normalized size of antiderivative = 1.00, 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 = {2687, 371} \begin {gather*} \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)} \end {gather*}

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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 2687

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])

Rubi steps

\begin {align*} \int \cos ^2(a+b x) (d \tan (a+b x))^n \, dx &=\frac {\text {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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 4.55, size = 939, normalized size = 18.78 \begin {gather*} \frac {2 \left (F_1\left (\frac {1+n}{2};n,1;\frac {3+n}{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 {1+n}{2};n,2;\frac {3+n}{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 {1+n}{2};n,3;\frac {3+n}{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 (\left (F_1\left (\frac {1+n}{2};n,1;\frac {3+n}{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 {1+n}{2};n,2;\frac {3+n}{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 {1+n}{2};n,3;\frac {3+n}{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 {1+n}{2};n,1;\frac {3+n}{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 {1+n}{2};n,2;\frac {3+n}{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 {1+n}{2};n,3;\frac {3+n}{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 ) \sec (a+b x)+\frac {2 (1+n) \left (-F_1\left (\frac {3+n}{2};n,2;\frac {5+n}{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 {3+n}{2};n,3;\frac {5+n}{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 {3+n}{2};n,4;\frac {5+n}{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 {3+n}{2};1+n,1;\frac {5+n}{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 {3+n}{2};1+n,2;\frac {5+n}{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 {3+n}{2};1+n,3;\frac {5+n}{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 ) \tan ^2\left (\frac {1}{2} (a+b x)\right )}{3+n}-2 n \left (F_1\left (\frac {1+n}{2};n,1;\frac {3+n}{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 {1+n}{2};n,2;\frac {3+n}{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 {1+n}{2};n,3;\frac {3+n}{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 )} \end {gather*}

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 = 0.31, size = 0, normalized size = 0.00 \[\int \left (\cos ^{2}\left (b x +a \right )\right ) \left (d \tan \left (b x +a \right )\right )^{n}\, dx\]

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 = 0.00, 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(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 = 0.00, 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(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (d \tan {\left (a + b x \right )}\right )^{n} \cos ^{2}{\left (a + b x \right )}\, dx \end {gather*}

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 = 0.00, 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(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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int {\cos \left (a+b\,x\right )}^2\,{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^n \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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