3.778 \(\int (d \sec (e+f x))^n (a+b \sec (e+f x))^2 \, dx\)

Optimal. Leaf size=181 \[ -\frac {d \left (a^2 (n+1)+b^2 n\right ) \sin (e+f x) (d \sec (e+f x))^{n-1} \, _2F_1\left (\frac {1}{2},\frac {1-n}{2};\frac {3-n}{2};\cos ^2(e+f x)\right )}{f \left (1-n^2\right ) \sqrt {\sin ^2(e+f x)}}+\frac {2 a b \sin (e+f x) (d \sec (e+f x))^n \, _2F_1\left (\frac {1}{2},-\frac {n}{2};\frac {2-n}{2};\cos ^2(e+f x)\right )}{f n \sqrt {\sin ^2(e+f x)}}+\frac {b^2 \tan (e+f x) (d \sec (e+f x))^n}{f (n+1)} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3788, 3772, 2643, 4046} \[ -\frac {d \left (a^2 (n+1)+b^2 n\right ) \sin (e+f x) (d \sec (e+f x))^{n-1} \, _2F_1\left (\frac {1}{2},\frac {1-n}{2};\frac {3-n}{2};\cos ^2(e+f x)\right )}{f \left (1-n^2\right ) \sqrt {\sin ^2(e+f x)}}+\frac {2 a b \sin (e+f x) (d \sec (e+f x))^n \, _2F_1\left (\frac {1}{2},-\frac {n}{2};\frac {2-n}{2};\cos ^2(e+f x)\right )}{f n \sqrt {\sin ^2(e+f x)}}+\frac {b^2 \tan (e+f x) (d \sec (e+f x))^n}{f (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[(d*Sec[e + f*x])^n*(a + b*Sec[e + f*x])^2,x]

[Out]

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

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3772

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x])^(n - 1)*((Sin[c + d*x]/b)^(n - 1)
*Int[1/(Sin[c + d*x]/b)^n, x]), x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 3788

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[(2*a*b)/
d, Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b
, d, e, f, n}, x]

Rule 4046

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> -Simp[(C*Cot[
e + f*x]*(b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x]
/; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 0.36, size = 171, normalized size = 0.94 \[ \frac {\sqrt {-\tan ^2(e+f x)} \csc (e+f x) \sec (e+f x) (d \sec (e+f x))^n \left (a^2 \left (n^2+3 n+2\right ) \cos ^2(e+f x) \, _2F_1\left (\frac {1}{2},\frac {n}{2};\frac {n+2}{2};\sec ^2(e+f x)\right )+b n \left (2 a (n+2) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {n+1}{2};\frac {n+3}{2};\sec ^2(e+f x)\right )+b (n+1) \, _2F_1\left (\frac {1}{2},\frac {n+2}{2};\frac {n+4}{2};\sec ^2(e+f x)\right )\right )\right )}{f n (n+1) (n+2)} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Sec[e + f*x])^n*(a + b*Sec[e + f*x])^2,x]

[Out]

(Csc[e + f*x]*(a^2*(2 + 3*n + n^2)*Cos[e + f*x]^2*Hypergeometric2F1[1/2, n/2, (2 + n)/2, Sec[e + f*x]^2] + b*n
*(2*a*(2 + n)*Cos[e + f*x]*Hypergeometric2F1[1/2, (1 + n)/2, (3 + n)/2, Sec[e + f*x]^2] + b*(1 + n)*Hypergeome
tric2F1[1/2, (2 + n)/2, (4 + n)/2, Sec[e + f*x]^2]))*Sec[e + f*x]*(d*Sec[e + f*x])^n*Sqrt[-Tan[e + f*x]^2])/(f
*n*(1 + n)*(2 + n))

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{2} \sec \left (f x + e\right )^{2} + 2 \, a b \sec \left (f x + e\right ) + a^{2}\right )} \left (d \sec \left (f x + e\right )\right )^{n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^n*(a+b*sec(f*x+e))^2,x, algorithm="fricas")

[Out]

integral((b^2*sec(f*x + e)^2 + 2*a*b*sec(f*x + e) + a^2)*(d*sec(f*x + e))^n, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (f x + e\right ) + a\right )}^{2} \left (d \sec \left (f x + e\right )\right )^{n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^n*(a+b*sec(f*x+e))^2,x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e) + a)^2*(d*sec(f*x + e))^n, x)

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maple [F]  time = 8.19, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x +e \right )\right )^{n} \left (a +b \sec \left (f x +e \right )\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^n*(a+b*sec(f*x+e))^2,x)

[Out]

int((d*sec(f*x+e))^n*(a+b*sec(f*x+e))^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (f x + e\right ) + a\right )}^{2} \left (d \sec \left (f x + e\right )\right )^{n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^n*(a+b*sec(f*x+e))^2,x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e) + a)^2*(d*sec(f*x + e))^n, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+\frac {b}{\cos \left (e+f\,x\right )}\right )}^2\,{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(e + f*x))^2*(d/cos(e + f*x))^n,x)

[Out]

int((a + b/cos(e + f*x))^2*(d/cos(e + f*x))^n, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \sec {\left (e + f x \right )}\right )^{n} \left (a + b \sec {\left (e + f x \right )}\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))**n*(a+b*sec(f*x+e))**2,x)

[Out]

Integral((d*sec(e + f*x))**n*(a + b*sec(e + f*x))**2, x)

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