3.347 \(\int \frac{(d \tan (e+f x))^n}{a+b \sec (e+f x)} \, dx\)

Optimal. Leaf size=266 \[ \frac{d \left (-\tan ^2(e+f x)\right )^{\frac{1-n}{2}+\frac{n-1}{2}} (d \tan (e+f x))^{n-1} \left (-\frac{b (1-\sec (e+f x))}{a+b \sec (e+f x)}\right )^{\frac{1-n}{2}} \left (\frac{b (\sec (e+f x)+1)}{a+b \sec (e+f x)}\right )^{\frac{1-n}{2}} F_1\left (1-n;\frac{1-n}{2},\frac{1-n}{2};2-n;\frac{a+b}{a+b \sec (e+f x)},\frac{a-b}{a+b \sec (e+f x)}\right )}{a f (1-n)}-\frac{d \left (-\tan ^2(e+f x)\right )^{\frac{1-n}{2}+\frac{n+1}{2}} (d \tan (e+f x))^{n-1} \text{Hypergeometric2F1}\left (1,\frac{n+1}{2},\frac{n+3}{2},-\tan ^2(e+f x)\right )}{a f (n+1)} \]

[Out]

(d*AppellF1[1 - n, (1 - n)/2, (1 - n)/2, 2 - n, (a + b)/(a + b*Sec[e + f*x]), (a - b)/(a + b*Sec[e + f*x])]*(-
((b*(1 - Sec[e + f*x]))/(a + b*Sec[e + f*x])))^((1 - n)/2)*((b*(1 + Sec[e + f*x]))/(a + b*Sec[e + f*x]))^((1 -
n)/2)*(d*Tan[e + f*x])^(-1 + n)*(-Tan[e + f*x]^2)^((1 - n)/2 + (-1 + n)/2))/(a*f*(1 - n)) - (d*Hypergeometric
2F1[1, (1 + n)/2, (3 + n)/2, -Tan[e + f*x]^2]*(d*Tan[e + f*x])^(-1 + n)*(-Tan[e + f*x]^2)^((1 - n)/2 + (1 + n)
/2))/(a*f*(1 + n))

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Rubi [F]  time = 0.0498594, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int \frac{(d \tan (e+f x))^n}{a+b \sec (e+f x)} \, dx \]

Verification is Not applicable to the result.

[In]

Int[(d*Tan[e + f*x])^n/(a + b*Sec[e + f*x]),x]

[Out]

Defer[Int][(d*Tan[e + f*x])^n/(a + b*Sec[e + f*x]), x]

Rubi steps

\begin{align*} \int \frac{(d \tan (e+f x))^n}{a+b \sec (e+f x)} \, dx &=\int \frac{(d \tan (e+f x))^n}{a+b \sec (e+f x)} \, dx\\ \end{align*}

Mathematica [B]  time = 4.5795, size = 786, normalized size = 2.95 \[ \frac{2 \tan \left (\frac{1}{2} (e+f x)\right ) (d \tan (e+f x))^n \left ((a+b) F_1\left (\frac{n+1}{2};n,1;\frac{n+3}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )-b F_1\left (\frac{n+1}{2};n,1;\frac{n+3}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),\frac{(a-b) \tan ^2\left (\frac{1}{2} (e+f x)\right )}{a+b}\right )\right )}{f (a+b \sec (e+f x)) \left (\sec ^2\left (\frac{1}{2} (e+f x)\right ) \left ((a+b) F_1\left (\frac{n+1}{2};n,1;\frac{n+3}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )-b F_1\left (\frac{n+1}{2};n,1;\frac{n+3}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),\frac{(a-b) \tan ^2\left (\frac{1}{2} (e+f x)\right )}{a+b}\right )\right )-\frac{2 (n+1) \tan ^2\left (\frac{1}{2} (e+f x)\right ) \sec ^2\left (\frac{1}{2} (e+f x)\right ) \left ((a+b)^2 \left (F_1\left (\frac{n+3}{2};n,2;\frac{n+5}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )-n F_1\left (\frac{n+3}{2};n+1,1;\frac{n+5}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )\right )+b n (a+b) F_1\left (\frac{n+3}{2};n+1,1;\frac{n+5}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),\frac{(a-b) \tan ^2\left (\frac{1}{2} (e+f x)\right )}{a+b}\right )+b (a-b) F_1\left (\frac{n+3}{2};n,2;\frac{n+5}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),\frac{(a-b) \tan ^2\left (\frac{1}{2} (e+f x)\right )}{a+b}\right )\right )}{(n+3) (a+b)}+2 n \tan \left (\frac{1}{2} (e+f x)\right ) \csc (e+f x) \sec (e+f x) \left ((a+b) F_1\left (\frac{n+1}{2};n,1;\frac{n+3}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )-b F_1\left (\frac{n+1}{2};n,1;\frac{n+3}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),\frac{(a-b) \tan ^2\left (\frac{1}{2} (e+f x)\right )}{a+b}\right )\right )-16 n \sin ^5\left (\frac{1}{2} (e+f x)\right ) \cos \left (\frac{1}{2} (e+f x)\right ) \csc ^3(e+f x) \sec (e+f x) \left ((a+b) F_1\left (\frac{n+1}{2};n,1;\frac{n+3}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )-b F_1\left (\frac{n+1}{2};n,1;\frac{n+3}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),\frac{(a-b) \tan ^2\left (\frac{1}{2} (e+f x)\right )}{a+b}\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(d*Tan[e + f*x])^n/(a + b*Sec[e + f*x]),x]

[Out]

(2*((a + b)*AppellF1[(1 + n)/2, n, 1, (3 + n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - b*AppellF1[(1 + n)
/2, n, 1, (3 + n)/2, Tan[(e + f*x)/2]^2, ((a - b)*Tan[(e + f*x)/2]^2)/(a + b)])*Tan[(e + f*x)/2]*(d*Tan[e + f*
x])^n)/(f*(a + b*Sec[e + f*x])*(((a + b)*AppellF1[(1 + n)/2, n, 1, (3 + n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*
x)/2]^2] - b*AppellF1[(1 + n)/2, n, 1, (3 + n)/2, Tan[(e + f*x)/2]^2, ((a - b)*Tan[(e + f*x)/2]^2)/(a + b)])*S
ec[(e + f*x)/2]^2 - 16*n*((a + b)*AppellF1[(1 + n)/2, n, 1, (3 + n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2
] - b*AppellF1[(1 + n)/2, n, 1, (3 + n)/2, Tan[(e + f*x)/2]^2, ((a - b)*Tan[(e + f*x)/2]^2)/(a + b)])*Cos[(e +
f*x)/2]*Csc[e + f*x]^3*Sec[e + f*x]*Sin[(e + f*x)/2]^5 + 2*n*((a + b)*AppellF1[(1 + n)/2, n, 1, (3 + n)/2, Ta
n[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - b*AppellF1[(1 + n)/2, n, 1, (3 + n)/2, Tan[(e + f*x)/2]^2, ((a - b)*T
an[(e + f*x)/2]^2)/(a + b)])*Csc[e + f*x]*Sec[e + f*x]*Tan[(e + f*x)/2] - (2*(1 + n)*((a - b)*b*AppellF1[(3 +
n)/2, n, 2, (5 + n)/2, Tan[(e + f*x)/2]^2, ((a - b)*Tan[(e + f*x)/2]^2)/(a + b)] + (a + b)^2*(AppellF1[(3 + n)
/2, n, 2, (5 + n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - n*AppellF1[(3 + n)/2, 1 + n, 1, (5 + n)/2, Tan
[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]) + b*(a + b)*n*AppellF1[(3 + n)/2, 1 + n, 1, (5 + n)/2, Tan[(e + f*x)/2]
^2, ((a - b)*Tan[(e + f*x)/2]^2)/(a + b)])*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2]^2)/((a + b)*(3 + n))))

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Maple [F]  time = 0.668, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( d\tan \left ( fx+e \right ) \right ) ^{n}}{a+b\sec \left ( fx+e \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(f*x+e))^n/(a+b*sec(f*x+e)),x)

[Out]

int((d*tan(f*x+e))^n/(a+b*sec(f*x+e)),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^n/(a+b*sec(f*x+e)),x, algorithm="maxima")

[Out]

integrate((d*tan(f*x + e))^n/(b*sec(f*x + e) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^n/(a+b*sec(f*x+e)),x, algorithm="fricas")

[Out]

integral((d*tan(f*x + e))^n/(b*sec(f*x + e) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))**n/(a+b*sec(f*x+e)),x)

[Out]

Integral((d*tan(e + f*x))**n/(a + b*sec(e + f*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^n/(a+b*sec(f*x+e)),x, algorithm="giac")

[Out]

integrate((d*tan(f*x + e))^n/(b*sec(f*x + e) + a), x)