3.342 \(\int \sec ^3(e+f x) (a+a \sec (e+f x))^m \, dx\)

Optimal. Leaf size=155 \[ \frac{2^{m+\frac{1}{2}} \left (m^2+m+1\right ) \tan (e+f x) (\sec (e+f x)+1)^{-m-\frac{1}{2}} (a \sec (e+f x)+a)^m \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{2}-m,\frac{3}{2},\frac{1}{2} (1-\sec (e+f x))\right )}{f (m+1) (m+2)}-\frac{\tan (e+f x) (a \sec (e+f x)+a)^m}{f \left (m^2+3 m+2\right )}+\frac{\tan (e+f x) (a \sec (e+f x)+a)^{m+1}}{a f (m+2)} \]

[Out]

-(((a + a*Sec[e + f*x])^m*Tan[e + f*x])/(f*(2 + 3*m + m^2))) + (2^(1/2 + m)*(1 + m + m^2)*Hypergeometric2F1[1/
2, 1/2 - m, 3/2, (1 - Sec[e + f*x])/2]*(1 + Sec[e + f*x])^(-1/2 - m)*(a + a*Sec[e + f*x])^m*Tan[e + f*x])/(f*(
1 + m)*(2 + m)) + ((a + a*Sec[e + f*x])^(1 + m)*Tan[e + f*x])/(a*f*(2 + m))

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Rubi [A]  time = 0.194901, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3800, 4001, 3828, 3827, 69} \[ \frac{2^{m+\frac{1}{2}} \left (m^2+m+1\right ) \tan (e+f x) (\sec (e+f x)+1)^{-m-\frac{1}{2}} (a \sec (e+f x)+a)^m \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{1}{2} (1-\sec (e+f x))\right )}{f (m+1) (m+2)}-\frac{\tan (e+f x) (a \sec (e+f x)+a)^m}{f \left (m^2+3 m+2\right )}+\frac{\tan (e+f x) (a \sec (e+f x)+a)^{m+1}}{a f (m+2)} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^3*(a + a*Sec[e + f*x])^m,x]

[Out]

-(((a + a*Sec[e + f*x])^m*Tan[e + f*x])/(f*(2 + 3*m + m^2))) + (2^(1/2 + m)*(1 + m + m^2)*Hypergeometric2F1[1/
2, 1/2 - m, 3/2, (1 - Sec[e + f*x])/2]*(1 + Sec[e + f*x])^(-1/2 - m)*(a + a*Sec[e + f*x])^m*Tan[e + f*x])/(f*(
1 + m)*(2 + m)) + ((a + a*Sec[e + f*x])^(1 + m)*Tan[e + f*x])/(a*f*(2 + m))

Rule 3800

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(Cot[e + f*x]*(a
+ b*Csc[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(b
*(m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]

Rule 4001

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

Rule 3828

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[(a^In
tPart[m]*(a + b*Csc[e + f*x])^FracPart[m])/(1 + (b*Csc[e + f*x])/a)^FracPart[m], Int[(1 + (b*Csc[e + f*x])/a)^
m*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m] &&  !GtQ
[a, 0]

Rule 3827

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[(a^2*
d*Cot[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]]), Subst[Int[((d*x)^(n - 1)*(a + b*x)^(m -
1/2))/Sqrt[a - b*x], x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !In
tegerQ[m] && GtQ[a, 0]

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

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

Mathematica [A]  time = 0.621114, size = 123, normalized size = 0.79 \[ \frac{\tan (e+f x) (\sec (e+f x)+1)^{-m-\frac{1}{2}} (a (\sec (e+f x)+1))^m \left (2^{m+\frac{3}{2}} \left (m^2+m+1\right ) \text{Hypergeometric2F1}\left (\frac{1}{2},-m-\frac{1}{2},\frac{3}{2},\frac{1}{2} (1-\sec (e+f x))\right )+((2 m+1) \sec (e+f x)+m-1) (\sec (e+f x)+1)^{m+\frac{1}{2}}\right )}{f (m+2) (2 m+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^3*(a + a*Sec[e + f*x])^m,x]

[Out]

((1 + Sec[e + f*x])^(-1/2 - m)*(a*(1 + Sec[e + f*x]))^m*(2^(3/2 + m)*(1 + m + m^2)*Hypergeometric2F1[1/2, -1/2
- m, 3/2, (1 - Sec[e + f*x])/2] + (1 + Sec[e + f*x])^(1/2 + m)*(-1 + m + (1 + 2*m)*Sec[e + f*x]))*Tan[e + f*x
])/(f*(2 + m)*(1 + 2*m))

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Maple [F]  time = 0.279, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( fx+e \right ) \right ) ^{3} \left ( a+a\sec \left ( fx+e \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^3*(a+a*sec(f*x+e))^m,x)

[Out]

int(sec(f*x+e)^3*(a+a*sec(f*x+e))^m,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sec(f*x + e) + a)^m*sec(f*x + e)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*sec(f*x + e) + a)^m*sec(f*x + e)^3, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \left (\sec{\left (e + f x \right )} + 1\right )\right )^{m} \sec ^{3}{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**3*(a+a*sec(f*x+e))**m,x)

[Out]

Integral((a*(sec(e + f*x) + 1))**m*sec(e + f*x)**3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sec(f*x + e) + a)^m*sec(f*x + e)^3, x)