3.777 \(\int (d \sec (e+f x))^n (a+b \sec (e+f x))^3 \, dx\)

Optimal. Leaf size=251 \[ -\frac{a d \left (a^2 (n+1)+3 b^2 n\right ) \sin (e+f x) (d \sec (e+f x))^{n-1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1-n}{2},\frac{3-n}{2},\cos ^2(e+f x)\right )}{f \left (1-n^2\right ) \sqrt{\sin ^2(e+f x)}}+\frac{b \left (3 a^2 (n+2)+b^2 (n+1)\right ) \sin (e+f x) (d \sec (e+f x))^n \text{Hypergeometric2F1}\left (\frac{1}{2},-\frac{n}{2},\frac{2-n}{2},\cos ^2(e+f x)\right )}{f n (n+2) \sqrt{\sin ^2(e+f x)}}+\frac{a b^2 (2 n+5) \tan (e+f x) (d \sec (e+f x))^n}{f (n+1) (n+2)}+\frac{b^2 \tan (e+f x) (a+b \sec (e+f x)) (d \sec (e+f x))^n}{f (n+2)} \]

[Out]

-((a*d*(3*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])) + (b*(b^2*(1 + n) + 3*a^2*(2 + n))*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*(2 + n)*Sqrt[Sin[e + f*x]^2]) + (
a*b^2*(5 + 2*n)*(d*Sec[e + f*x])^n*Tan[e + f*x])/(f*(1 + n)*(2 + n)) + (b^2*(d*Sec[e + f*x])^n*(a + b*Sec[e +
f*x])*Tan[e + f*x])/(f*(2 + n))

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Rubi [A]  time = 0.349022, antiderivative size = 251, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3842, 4047, 3772, 2643, 4046} \[ -\frac{a d \left (a^2 (n+1)+3 b^2 n\right ) \sin (e+f x) (d \sec (e+f x))^{n-1} \, _2F_1\left (\frac{1}{2},\frac{1-n}{2};\frac{3-n}{2};\cos ^2(e+f x)\right )}{f \left (1-n^2\right ) \sqrt{\sin ^2(e+f x)}}+\frac{b \left (3 a^2 (n+2)+b^2 (n+1)\right ) \sin (e+f x) (d \sec (e+f x))^n \, _2F_1\left (\frac{1}{2},-\frac{n}{2};\frac{2-n}{2};\cos ^2(e+f x)\right )}{f n (n+2) \sqrt{\sin ^2(e+f x)}}+\frac{a b^2 (2 n+5) \tan (e+f x) (d \sec (e+f x))^n}{f (n+1) (n+2)}+\frac{b^2 \tan (e+f x) (a+b \sec (e+f x)) (d \sec (e+f x))^n}{f (n+2)} \]

Antiderivative was successfully verified.

[In]

Int[(d*Sec[e + f*x])^n*(a + b*Sec[e + f*x])^3,x]

[Out]

-((a*d*(3*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])) + (b*(b^2*(1 + n) + 3*a^2*(2 + n))*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*(2 + n)*Sqrt[Sin[e + f*x]^2]) + (
a*b^2*(5 + 2*n)*(d*Sec[e + f*x])^n*Tan[e + f*x])/(f*(1 + n)*(2 + n)) + (b^2*(d*Sec[e + f*x])^n*(a + b*Sec[e +
f*x])*Tan[e + f*x])/(f*(2 + n))

Rule 3842

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(b^2*
Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n)/(f*(m + n - 1)), x] + Dist[1/(d*(m + n - 1)), In
t[(a + b*Csc[e + f*x])^(m - 3)*(d*Csc[e + f*x])^n*Simp[a^3*d*(m + n - 1) + a*b^2*d*n + b*(b^2*d*(m + n - 2) +
3*a^2*d*(m + n - 1))*Csc[e + f*x] + a*b^2*d*(3*m + 2*n - 4)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f
, n}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] &&  !Integ
erQ[m])

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

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 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 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))^3 \, dx &=\frac{b^2 (d \sec (e+f x))^n (a+b \sec (e+f x)) \tan (e+f x)}{f (2+n)}+\frac{\int (d \sec (e+f x))^n \left (a d \left (b^2 n+a^2 (2+n)\right )+b d \left (b^2 (1+n)+3 a^2 (2+n)\right ) \sec (e+f x)+a b^2 d (5+2 n) \sec ^2(e+f x)\right ) \, dx}{d (2+n)}\\ &=\frac{b^2 (d \sec (e+f x))^n (a+b \sec (e+f x)) \tan (e+f x)}{f (2+n)}+\frac{\int (d \sec (e+f x))^n \left (a d \left (b^2 n+a^2 (2+n)\right )+a b^2 d (5+2 n) \sec ^2(e+f x)\right ) \, dx}{d (2+n)}+\frac{\left (b \left (b^2 (1+n)+3 a^2 (2+n)\right )\right ) \int (d \sec (e+f x))^{1+n} \, dx}{d (2+n)}\\ &=\frac{a b^2 (5+2 n) (d \sec (e+f x))^n \tan (e+f x)}{f (1+n) (2+n)}+\frac{b^2 (d \sec (e+f x))^n (a+b \sec (e+f x)) \tan (e+f x)}{f (2+n)}+\left (a \left (a^2+\frac{3 b^2 n}{1+n}\right )\right ) \int (d \sec (e+f x))^n \, dx+\frac{\left (b \left (b^2 (1+n)+3 a^2 (2+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 )^{-1-n} \, dx}{d (2+n)}\\ &=\frac{b \left (b^2 (1+n)+3 a^2 (2+n)\right ) \, _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 (2+n) \sqrt{\sin ^2(e+f x)}}+\frac{a b^2 (5+2 n) (d \sec (e+f x))^n \tan (e+f x)}{f (1+n) (2+n)}+\frac{b^2 (d \sec (e+f x))^n (a+b \sec (e+f x)) \tan (e+f x)}{f (2+n)}+\left (a \left (a^2+\frac{3 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{a \left (a^2+\frac{3 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{b \left (b^2 (1+n)+3 a^2 (2+n)\right ) \, _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 (2+n) \sqrt{\sin ^2(e+f x)}}+\frac{a b^2 (5+2 n) (d \sec (e+f x))^n \tan (e+f x)}{f (1+n) (2+n)}+\frac{b^2 (d \sec (e+f x))^n (a+b \sec (e+f x)) \tan (e+f x)}{f (2+n)}\\ \end{align*}

Mathematica [A]  time = 0.832583, size = 231, normalized size = 0.92 \[ -\frac{\left (-\tan ^2(e+f x)\right )^{3/2} \csc ^3(e+f x) (d \sec (e+f x))^n \left (b n \left (3 a^2 \left (n^2+5 n+6\right ) \cos ^2(e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{n+1}{2},\frac{n+3}{2},\sec ^2(e+f x)\right )+b (n+1) \left (3 a (n+3) \cos (e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{n+2}{2},\frac{n+4}{2},\sec ^2(e+f x)\right )+b (n+2) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{n+3}{2},\frac{n+5}{2},\sec ^2(e+f x)\right )\right )\right )+a^3 \left (n^3+6 n^2+11 n+6\right ) \cos ^3(e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{n}{2},\frac{n+2}{2},\sec ^2(e+f x)\right )\right )}{f n (n+1) (n+2) (n+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Sec[e + f*x])^n*(a + b*Sec[e + f*x])^3,x]

[Out]

-((Csc[e + f*x]^3*(a^3*(6 + 11*n + 6*n^2 + n^3)*Cos[e + f*x]^3*Hypergeometric2F1[1/2, n/2, (2 + n)/2, Sec[e +
f*x]^2] + b*n*(3*a^2*(6 + 5*n + n^2)*Cos[e + f*x]^2*Hypergeometric2F1[1/2, (1 + n)/2, (3 + n)/2, Sec[e + f*x]^
2] + b*(1 + n)*(3*a*(3 + n)*Cos[e + f*x]*Hypergeometric2F1[1/2, (2 + n)/2, (4 + n)/2, Sec[e + f*x]^2] + b*(2 +
n)*Hypergeometric2F1[1/2, (3 + n)/2, (5 + n)/2, Sec[e + f*x]^2])))*(d*Sec[e + f*x])^n*(-Tan[e + f*x]^2)^(3/2)
)/(f*n*(1 + n)*(2 + n)*(3 + n)))

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Maple [F]  time = 3.113, size = 0, normalized size = 0. \begin{align*} \int \left ( d\sec \left ( fx+e \right ) \right ) ^{n} \left ( a+b\sec \left ( fx+e \right ) \right ) ^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^n*(a+b*sec(f*x+e))^3,x)

[Out]

int((d*sec(f*x+e))^n*(a+b*sec(f*x+e))^3,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (f x + e\right ) + a\right )}^{3} \left (d \sec \left (f x + e\right )\right )^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^n*(a+b*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e) + a)^3*(d*sec(f*x + e))^n, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b^{3} \sec \left (f x + e\right )^{3} + 3 \, a b^{2} \sec \left (f x + e\right )^{2} + 3 \, a^{2} b \sec \left (f x + e\right ) + a^{3}\right )} \left (d \sec \left (f x + e\right )\right )^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^n*(a+b*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

integral((b^3*sec(f*x + e)^3 + 3*a*b^2*sec(f*x + e)^2 + 3*a^2*b*sec(f*x + e) + a^3)*(d*sec(f*x + e))^n, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \sec{\left (e + f x \right )}\right )^{n} \left (a + b \sec{\left (e + f x \right )}\right )^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))**n*(a+b*sec(f*x+e))**3,x)

[Out]

Integral((d*sec(e + f*x))**n*(a + b*sec(e + f*x))**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (f x + e\right ) + a\right )}^{3} \left (d \sec \left (f x + e\right )\right )^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^n*(a+b*sec(f*x+e))^3,x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e) + a)^3*(d*sec(f*x + e))^n, x)