3.789 \(\int \sec ^2(e+f x) (a+b \sec (e+f x))^m \, dx\)

Optimal. Leaf size=220 \[ \frac{\sqrt{2} (a+b) \tan (e+f x) (a+b \sec (e+f x))^m \left (\frac{a+b \sec (e+f x)}{a+b}\right )^{-m} F_1\left (\frac{1}{2};\frac{1}{2},-m-1;\frac{3}{2};\frac{1}{2} (1-\sec (e+f x)),\frac{b (1-\sec (e+f x))}{a+b}\right )}{b f \sqrt{\sec (e+f x)+1}}-\frac{\sqrt{2} a \tan (e+f x) (a+b \sec (e+f x))^m \left (\frac{a+b \sec (e+f x)}{a+b}\right )^{-m} F_1\left (\frac{1}{2};\frac{1}{2},-m;\frac{3}{2};\frac{1}{2} (1-\sec (e+f x)),\frac{b (1-\sec (e+f x))}{a+b}\right )}{b f \sqrt{\sec (e+f x)+1}} \]

[Out]

(Sqrt[2]*(a + b)*AppellF1[1/2, 1/2, -1 - m, 3/2, (1 - Sec[e + f*x])/2, (b*(1 - Sec[e + f*x]))/(a + b)]*(a + b*
Sec[e + f*x])^m*Tan[e + f*x])/(b*f*Sqrt[1 + Sec[e + f*x]]*((a + b*Sec[e + f*x])/(a + b))^m) - (Sqrt[2]*a*Appel
lF1[1/2, 1/2, -m, 3/2, (1 - Sec[e + f*x])/2, (b*(1 - Sec[e + f*x]))/(a + b)]*(a + b*Sec[e + f*x])^m*Tan[e + f*
x])/(b*f*Sqrt[1 + Sec[e + f*x]]*((a + b*Sec[e + f*x])/(a + b))^m)

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Rubi [A]  time = 0.221205, antiderivative size = 220, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3838, 3834, 139, 138} \[ \frac{\sqrt{2} (a+b) \tan (e+f x) (a+b \sec (e+f x))^m \left (\frac{a+b \sec (e+f x)}{a+b}\right )^{-m} F_1\left (\frac{1}{2};\frac{1}{2},-m-1;\frac{3}{2};\frac{1}{2} (1-\sec (e+f x)),\frac{b (1-\sec (e+f x))}{a+b}\right )}{b f \sqrt{\sec (e+f x)+1}}-\frac{\sqrt{2} a \tan (e+f x) (a+b \sec (e+f x))^m \left (\frac{a+b \sec (e+f x)}{a+b}\right )^{-m} F_1\left (\frac{1}{2};\frac{1}{2},-m;\frac{3}{2};\frac{1}{2} (1-\sec (e+f x)),\frac{b (1-\sec (e+f x))}{a+b}\right )}{b f \sqrt{\sec (e+f x)+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*(a + b)*AppellF1[1/2, 1/2, -1 - m, 3/2, (1 - Sec[e + f*x])/2, (b*(1 - Sec[e + f*x]))/(a + b)]*(a + b*
Sec[e + f*x])^m*Tan[e + f*x])/(b*f*Sqrt[1 + Sec[e + f*x]]*((a + b*Sec[e + f*x])/(a + b))^m) - (Sqrt[2]*a*Appel
lF1[1/2, 1/2, -m, 3/2, (1 - Sec[e + f*x])/2, (b*(1 - Sec[e + f*x]))/(a + b)]*(a + b*Sec[e + f*x])^m*Tan[e + f*
x])/(b*f*Sqrt[1 + Sec[e + f*x]]*((a + b*Sec[e + f*x])/(a + b))^m)

Rule 3838

Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Dist[a/b, Int[Csc[e +
f*x]*(a + b*Csc[e + f*x])^m, x], x] + Dist[1/b, Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1), x], x] /; Free
Q[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0]

Rule 3834

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

Rule 139

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(e + f*x)^
FracPart[p]/((b/(b*e - a*f))^IntPart[p]*((b*(e + f*x))/(b*e - a*f))^FracPart[p]), Int[(a + b*x)^m*(c + d*x)^n*
((b*e)/(b*e - a*f) + (b*f*x)/(b*e - a*f))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !GtQ[b/(b*e - a*f), 0]

Rule 138

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b*(m + 1
)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p), x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

Rubi steps

\begin{align*} \int \sec ^2(e+f x) (a+b \sec (e+f x))^m \, dx &=\frac{\int \sec (e+f x) (a+b \sec (e+f x))^{1+m} \, dx}{b}-\frac{a \int \sec (e+f x) (a+b \sec (e+f x))^m \, dx}{b}\\ &=-\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{(a+b x)^{1+m}}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\sec (e+f x)\right )}{b f \sqrt{1-\sec (e+f x)} \sqrt{1+\sec (e+f x)}}+\frac{(a \tan (e+f x)) \operatorname{Subst}\left (\int \frac{(a+b x)^m}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\sec (e+f x)\right )}{b f \sqrt{1-\sec (e+f x)} \sqrt{1+\sec (e+f x)}}\\ &=\frac{\left (a (a+b \sec (e+f x))^m \left (-\frac{a+b \sec (e+f x)}{-a-b}\right )^{-m} \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{\left (-\frac{a}{-a-b}-\frac{b x}{-a-b}\right )^m}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\sec (e+f x)\right )}{b f \sqrt{1-\sec (e+f x)} \sqrt{1+\sec (e+f x)}}+\frac{\left ((-a-b) (a+b \sec (e+f x))^m \left (-\frac{a+b \sec (e+f x)}{-a-b}\right )^{-m} \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{\left (-\frac{a}{-a-b}-\frac{b x}{-a-b}\right )^{1+m}}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\sec (e+f x)\right )}{b f \sqrt{1-\sec (e+f x)} \sqrt{1+\sec (e+f x)}}\\ &=\frac{\sqrt{2} (a+b) F_1\left (\frac{1}{2};\frac{1}{2},-1-m;\frac{3}{2};\frac{1}{2} (1-\sec (e+f x)),\frac{b (1-\sec (e+f x))}{a+b}\right ) (a+b \sec (e+f x))^m \left (\frac{a+b \sec (e+f x)}{a+b}\right )^{-m} \tan (e+f x)}{b f \sqrt{1+\sec (e+f x)}}-\frac{\sqrt{2} a F_1\left (\frac{1}{2};\frac{1}{2},-m;\frac{3}{2};\frac{1}{2} (1-\sec (e+f x)),\frac{b (1-\sec (e+f x))}{a+b}\right ) (a+b \sec (e+f x))^m \left (\frac{a+b \sec (e+f x)}{a+b}\right )^{-m} \tan (e+f x)}{b f \sqrt{1+\sec (e+f x)}}\\ \end{align*}

Mathematica [B]  time = 22.7531, size = 5564, normalized size = 25.29 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [F]  time = 0.229, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( fx+e \right ) \right ) ^{2} \left ( a+b\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)^2*(a+b*sec(f*x+e))^m,x)

[Out]

int(sec(f*x+e)^2*(a+b*sec(f*x+e))^m,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 )}^{m} \sec \left (f x + e\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \sec{\left (e + f x \right )}\right )^{m} \sec ^{2}{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*sec(e + f*x))**m*sec(e + f*x)**2, 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 )}^{m} \sec \left (f x + e\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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