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

Optimal. Leaf size=266 \[ \frac {d \left (-\tan ^2(e+f x)\right )^{\frac {1-n}{2}+\frac {n-1}{2}} (d \tan (e+f x))^{n-1} \left (-\frac {b (1-\sec (e+f x))}{a+b \sec (e+f x)}\right )^{\frac {1-n}{2}} \left (\frac {b (\sec (e+f x)+1)}{a+b \sec (e+f x)}\right )^{\frac {1-n}{2}} F_1\left (1-n;\frac {1-n}{2},\frac {1-n}{2};2-n;\frac {a+b}{a+b \sec (e+f x)},\frac {a-b}{a+b \sec (e+f x)}\right )}{a f (1-n)}-\frac {d \left (-\tan ^2(e+f x)\right )^{\frac {1-n}{2}+\frac {n+1}{2}} (d \tan (e+f x))^{n-1} \, _2F_1\left (1,\frac {n+1}{2};\frac {n+3}{2};-\tan ^2(e+f x)\right )}{a f (n+1)} \]

[Out]

d*AppellF1(1-n,1/2-1/2*n,1/2-1/2*n,2-n,(a-b)/(a+b*sec(f*x+e)),(a+b)/(a+b*sec(f*x+e)))*(-b*(1-sec(f*x+e))/(a+b*
sec(f*x+e)))^(1/2-1/2*n)*(b*(1+sec(f*x+e))/(a+b*sec(f*x+e)))^(1/2-1/2*n)*(d*tan(f*x+e))^(-1+n)/a/f/(1-n)+d*hyp
ergeom([1, 1/2+1/2*n],[3/2+1/2*n],-tan(f*x+e)^2)*(d*tan(f*x+e))^(-1+n)*tan(f*x+e)^2/a/f/(1+n)

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Rubi [F]  time = 0.05, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, 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*}

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Mathematica [B]  time = 4.74, 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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fricas [F]  time = 0.93, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (d \tan \left (f x + e\right )\right )^{n}}{b \sec \left (f x + e\right ) + a}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \tan \left (f x + e\right )\right )^{n}}{b \sec \left (f x + e\right ) + a}\,{d x} \]

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)

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maple [F]  time = 2.48, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \tan \left (f x +e \right )\right )^{n}}{a +b \sec \left (f x +e \right )}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \tan \left (f x + e\right )\right )^{n}}{b \sec \left (f x + e\right ) + a}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\cos \left (e+f\,x\right )\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^n}{b+a\,\cos \left (e+f\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \tan {\left (e + f x \right )}\right )^{n}}{a + b \sec {\left (e + f x \right )}}\, dx \]

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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