\(\int (a+a \sec (c+d x))^n \tan ^2(c+d x) \, dx\) [226]

Optimal result

Integrand size = 21, antiderivative size = 106 \[ \int (a+a \sec (c+d x))^n \tan ^2(c+d x) \, dx=\frac {2^{3+n} \operatorname {AppellF1}\left (\frac {3}{2},2+n,1,\frac {5}{2},-\frac {a-a \sec (c+d x)}{a+a \sec (c+d x)},\frac {a-a \sec (c+d x)}{a+a \sec (c+d x)}\right ) \left (\frac {1}{1+\sec (c+d x)}\right )^{3+n} (a+a \sec (c+d x))^n \tan ^3(c+d x)}{3 d} \]

[Out]

1/3*2^(3+n)*AppellF1(3/2,2+n,1,5/2,(-a+a*sec(d*x+c))/(a+a*sec(d*x+c)),(a-a*sec(d*x+c))/(a+a*sec(d*x+c)))*(1/(1
+sec(d*x+c)))^(3+n)*(a+a*sec(d*x+c))^n*tan(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {3974} \[ \int (a+a \sec (c+d x))^n \tan ^2(c+d x) \, dx=\frac {2^{n+3} \tan ^3(c+d x) \left (\frac {1}{\sec (c+d x)+1}\right )^{n+3} (a \sec (c+d x)+a)^n \operatorname {AppellF1}\left (\frac {3}{2},n+2,1,\frac {5}{2},-\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a},\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a}\right )}{3 d} \]

[In]

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

[Out]

(2^(3 + n)*AppellF1[3/2, 2 + n, 1, 5/2, -((a - a*Sec[c + d*x])/(a + a*Sec[c + d*x])), (a - a*Sec[c + d*x])/(a
+ a*Sec[c + d*x])]*((1 + Sec[c + d*x])^(-1))^(3 + n)*(a + a*Sec[c + d*x])^n*Tan[c + d*x]^3)/(3*d)

Rule 3974

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[(-2^(m
+ n + 1))*(e*Cot[c + d*x])^(m + 1)*((a + b*Csc[c + d*x])^n/(d*e*(m + 1)))*(a/(a + b*Csc[c + d*x]))^(m + n + 1
)*AppellF1[(m + 1)/2, m + n, 1, (m + 3)/2, -(a - b*Csc[c + d*x])/(a + b*Csc[c + d*x]), (a - b*Csc[c + d*x])/(a
+ b*Csc[c + d*x])], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[n]

Rubi steps \begin{align*} \text {integral}& = \frac {2^{3+n} \operatorname {AppellF1}\left (\frac {3}{2},2+n,1,\frac {5}{2},-\frac {a-a \sec (c+d x)}{a+a \sec (c+d x)},\frac {a-a \sec (c+d x)}{a+a \sec (c+d x)}\right ) \left (\frac {1}{1+\sec (c+d x)}\right )^{3+n} (a+a \sec (c+d x))^n \tan ^3(c+d x)}{3 d} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(910\) vs. \(2(106)=212\).

Time = 6.33 (sec) , antiderivative size = 910, normalized size of antiderivative = 8.58 \[ \int (a+a \sec (c+d x))^n \tan ^2(c+d x) \, dx=\frac {(a (1+\sec (c+d x)))^n \left (-\frac {4 \operatorname {Hypergeometric2F1}\left (-1-n,n,-n,\frac {1}{2} \left (1-\tan \left (\frac {1}{2} (c+d x)\right )\right )\right ) \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^n (1+\sec (c+d x))^{-n} \left (1+\tan \left (\frac {1}{2} (c+d x)\right )\right )^n}{(1+n) \left (-1+\tan \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {\operatorname {Hypergeometric2F1}\left (1-n,2+n,2-n,\frac {1}{2} \left (1-\tan \left (\frac {1}{2} (c+d x)\right )\right )\right ) \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^n (1+\sec (c+d x))^{-n} \left (-1+\tan \left (\frac {1}{2} (c+d x)\right )\right ) \left (1+\tan \left (\frac {1}{2} (c+d x)\right )\right )^n}{-1+n}-\frac {120 \operatorname {AppellF1}\left (\frac {1}{2},n,1,\frac {3}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \cos (c+d x) \sin (c+d x) \left (3 \operatorname {AppellF1}\left (\frac {1}{2},n,1,\frac {3}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-2 \left (\operatorname {AppellF1}\left (\frac {3}{2},n,2,\frac {5}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-n \operatorname {AppellF1}\left (\frac {3}{2},1+n,1,\frac {5}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{45 \operatorname {AppellF1}\left (\frac {1}{2},n,1,\frac {3}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^2 \cos ^2\left (\frac {1}{2} (c+d x)\right ) (1+2 n-2 n \cos (c+d x)+\cos (2 (c+d x)))+6 \operatorname {AppellF1}\left (\frac {1}{2},n,1,\frac {3}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sin ^2\left (\frac {1}{2} (c+d x)\right ) \left (-5 \operatorname {AppellF1}\left (\frac {3}{2},n,2,\frac {5}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+2 n-2 (2+n) \cos (c+d x)+\cos (2 (c+d x)))+5 n \operatorname {AppellF1}\left (\frac {3}{2},1+n,1,\frac {5}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+2 n-2 (2+n) \cos (c+d x)+\cos (2 (c+d x)))-48 \left (2 \operatorname {AppellF1}\left (\frac {5}{2},n,3,\frac {7}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-2 n \operatorname {AppellF1}\left (\frac {5}{2},1+n,2,\frac {7}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+n (1+n) \operatorname {AppellF1}\left (\frac {5}{2},2+n,1,\frac {7}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) \cot (c+d x) \csc (c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )\right )+40 \left (\operatorname {AppellF1}\left (\frac {3}{2},n,2,\frac {5}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-n \operatorname {AppellF1}\left (\frac {3}{2},1+n,1,\frac {5}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right )^2 \cos (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )}{4 d} \]

[In]

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

[Out]

((a*(1 + Sec[c + d*x]))^n*((-4*Hypergeometric2F1[-1 - n, n, -n, (1 - Tan[(c + d*x)/2])/2]*(Cos[(c + d*x)/2]^2*
Sec[c + d*x])^n*(1 + Tan[(c + d*x)/2])^n)/((1 + n)*(1 + Sec[c + d*x])^n*(-1 + Tan[(c + d*x)/2])) - (Hypergeome
tric2F1[1 - n, 2 + n, 2 - n, (1 - Tan[(c + d*x)/2])/2]*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^n*(-1 + Tan[(c + d*x)
/2])*(1 + Tan[(c + d*x)/2])^n)/((-1 + n)*(1 + Sec[c + d*x])^n) - (120*AppellF1[1/2, n, 1, 3/2, Tan[(c + d*x)/2
]^2, -Tan[(c + d*x)/2]^2]*Cos[(c + d*x)/2]^2*Cos[c + d*x]*Sin[c + d*x]*(3*AppellF1[1/2, n, 1, 3/2, Tan[(c + d*
x)/2]^2, -Tan[(c + d*x)/2]^2] - 2*(AppellF1[3/2, n, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - n*Appel
lF1[3/2, 1 + n, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2))/(45*AppellF1[1/2, n, 1,
3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]^2*Cos[(c + d*x)/2]^2*(1 + 2*n - 2*n*Cos[c + d*x] + Cos[2*(c + d
*x)]) + 6*AppellF1[1/2, n, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sin[(c + d*x)/2]^2*(-5*AppellF1[3/
2, n, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(1 + 2*n - 2*(2 + n)*Cos[c + d*x] + Cos[2*(c + d*x)]) +
5*n*AppellF1[3/2, 1 + n, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(1 + 2*n - 2*(2 + n)*Cos[c + d*x] +
Cos[2*(c + d*x)]) - 48*(2*AppellF1[5/2, n, 3, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 2*n*AppellF1[5/
2, 1 + n, 2, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + n*(1 + n)*AppellF1[5/2, 2 + n, 1, 7/2, Tan[(c + d
*x)/2]^2, -Tan[(c + d*x)/2]^2])*Cot[c + d*x]*Csc[c + d*x]*Sin[(c + d*x)/2]^4) + 40*(AppellF1[3/2, n, 2, 5/2, T
an[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - n*AppellF1[3/2, 1 + n, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]
^2])^2*Cos[c + d*x]*Sin[(c + d*x)/2]^2*Tan[(c + d*x)/2]^2)))/(4*d)

Maple [F]

\[\int \left (a +a \sec \left (d x +c \right )\right )^{n} \tan \left (d x +c \right )^{2}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int (a+a \sec (c+d x))^n \tan ^2(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{2} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int (a+a \sec (c+d x))^n \tan ^2(c+d x) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{n} \tan ^{2}{\left (c + d x \right )}\, dx \]

[In]

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

[Out]

Integral((a*(sec(c + d*x) + 1))**n*tan(c + d*x)**2, x)

Maxima [F]

\[ \int (a+a \sec (c+d x))^n \tan ^2(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{2} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int (a+a \sec (c+d x))^n \tan ^2(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{2} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int (a+a \sec (c+d x))^n \tan ^2(c+d x) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^2\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n \,d x \]

[In]

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

[Out]

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