\(\int \sqrt {d \sec (e+f x)} (a+a \sec (e+f x))^m \, dx\) [348]
Optimal result
Integrand size = 25, antiderivative size = 96 \[
\int \sqrt {d \sec (e+f x)} (a+a \sec (e+f x))^m \, dx=-\frac {2 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\sec (e+f x),-\sec (e+f x)\right ) \sqrt {d \sec (e+f x)} (1+\sec (e+f x))^{-\frac {1}{2}-m} (a+a \sec (e+f x))^m \tan (e+f x)}{f \sqrt {1-\sec (e+f x)}}
\]
[Out]
-2*AppellF1(1/2,1/2-m,1/2,3/2,-sec(f*x+e),sec(f*x+e))*(1+sec(f*x+e))^(-1/2-m)*(a+a*sec(f*x+e))^m*(d*sec(f*x+e)
)^(1/2)*tan(f*x+e)/f/(1-sec(f*x+e))^(1/2)
Rubi [A] (verified)
Time = 0.14 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of
steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {3913, 3912, 138}
\[
\int \sqrt {d \sec (e+f x)} (a+a \sec (e+f x))^m \, dx=-\frac {2 \tan (e+f x) \sqrt {d \sec (e+f x)} (\sec (e+f x)+1)^{-m-\frac {1}{2}} (a \sec (e+f x)+a)^m \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\sec (e+f x),-\sec (e+f x)\right )}{f \sqrt {1-\sec (e+f x)}}
\]
[In]
Int[Sqrt[d*Sec[e + f*x]]*(a + a*Sec[e + f*x])^m,x]
[Out]
(-2*AppellF1[1/2, 1/2, 1/2 - m, 3/2, Sec[e + f*x], -Sec[e + f*x]]*Sqrt[d*Sec[e + f*x]]*(1 + Sec[e + f*x])^(-1/
2 - m)*(a + a*Sec[e + f*x])^m*Tan[e + f*x])/(f*Sqrt[1 - Sec[e + f*x]])
Rule 138
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[c^n*e^p*((b*x)^(m +
1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
Rule 3912
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 3913
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[a^Int
Part[m]*((a + b*Csc[e + f*x])^FracPart[m]/(1 + (b/a)*Csc[e + f*x])^FracPart[m]), Int[(1 + (b/a)*Csc[e + f*x])^
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]
Rubi steps \begin{align*}
\text {integral}& = \left ((1+\sec (e+f x))^{-m} (a+a \sec (e+f x))^m\right ) \int \sqrt {d \sec (e+f x)} (1+\sec (e+f x))^m \, dx \\ & = -\frac {\left (d (1+\sec (e+f x))^{-\frac {1}{2}-m} (a+a \sec (e+f x))^m \tan (e+f x)\right ) \text {Subst}\left (\int \frac {(1+x)^{-\frac {1}{2}+m}}{\sqrt {1-x} \sqrt {d x}} \, dx,x,\sec (e+f x)\right )}{f \sqrt {1-\sec (e+f x)}} \\ & = -\frac {2 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\sec (e+f x),-\sec (e+f x)\right ) \sqrt {d \sec (e+f x)} (1+\sec (e+f x))^{-\frac {1}{2}-m} (a+a \sec (e+f x))^m \tan (e+f x)}{f \sqrt {1-\sec (e+f x)}} \\
\end{align*}
Mathematica [B] (warning: unable to verify)
Leaf count is larger than twice the leaf count of optimal. \(2225\) vs. \(2(96)=192\).
Time = 16.98 (sec) , antiderivative size = 2225, normalized size of antiderivative = 23.18
\[
\int \sqrt {d \sec (e+f x)} (a+a \sec (e+f x))^m \, dx=\text {Result too large to show}
\]
[In]
Integrate[Sqrt[d*Sec[e + f*x]]*(a + a*Sec[e + f*x])^m,x]
[Out]
(2^(1 + m)*AppellF1[1/2, 1/2 + m, 1/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sqrt[d*Sec[e + f*x]]*(Cos
[(e + f*x)/2]^2*Sec[e + f*x])^(1/2 + m)*(a*(1 + Sec[e + f*x]))^m*Tan[(e + f*x)/2])/(f*Sqrt[Sec[(e + f*x)/2]^2]
*(AppellF1[1/2, 1/2 + m, 1/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - ((AppellF1[3/2, 1/2 + m, 3/2, 5/
2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - (1 + 2*m)*AppellF1[3/2, 3/2 + m, 1/2, 5/2, Tan[(e + f*x)/2]^2, -
Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)/3)*((2^m*AppellF1[1/2, 1/2 + m, 1/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e
+ f*x)/2]^2]*Sqrt[Sec[(e + f*x)/2]^2]*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^(1/2 + m))/(AppellF1[1/2, 1/2 + m, 1/
2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - ((AppellF1[3/2, 1/2 + m, 3/2, 5/2, Tan[(e + f*x)/2]^2, -Tan
[(e + f*x)/2]^2] - (1 + 2*m)*AppellF1[3/2, 3/2 + m, 1/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e
+ f*x)/2]^2)/3) - (2^m*AppellF1[1/2, 1/2 + m, 1/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Cos[(e + f*
x)/2]^2*Sec[e + f*x])^(1/2 + m)*Tan[(e + f*x)/2]^2)/(Sqrt[Sec[(e + f*x)/2]^2]*(AppellF1[1/2, 1/2 + m, 1/2, 3/2
, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - ((AppellF1[3/2, 1/2 + m, 3/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e +
f*x)/2]^2] - (1 + 2*m)*AppellF1[3/2, 3/2 + m, 1/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x
)/2]^2)/3)) + (2^(1 + m)*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^(1/2 + m)*Tan[(e + f*x)/2]*(-1/6*(AppellF1[3/2, 1/2
+ m, 3/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2]) + ((1/2 + m)*App
ellF1[3/2, 3/2 + m, 1/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/3)
)/(Sqrt[Sec[(e + f*x)/2]^2]*(AppellF1[1/2, 1/2 + m, 1/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - ((App
ellF1[3/2, 1/2 + m, 3/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - (1 + 2*m)*AppellF1[3/2, 3/2 + m, 1/2,
5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)/3)) - (2^(1 + m)*AppellF1[1/2, 1/2 + m, 1/
2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^(1/2 + m)*Tan[(e + f*x)/2]*
(-1/6*(AppellF1[3/2, 1/2 + m, 3/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f
*x)/2]) + ((1/2 + m)*AppellF1[3/2, 3/2 + m, 1/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2
]^2*Tan[(e + f*x)/2])/3 - ((AppellF1[3/2, 1/2 + m, 3/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - (1 + 2
*m)*AppellF1[3/2, 3/2 + m, 1/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Sec[(e + f*x)/2]^2*Tan[(e + f*x
)/2])/3 - (Tan[(e + f*x)/2]^2*((-9*AppellF1[5/2, 1/2 + m, 5/2, 7/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*S
ec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/10 + (3*(1/2 + m)*AppellF1[5/2, 3/2 + m, 3/2, 7/2, Tan[(e + f*x)/2]^2, -Ta
n[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/5 - (1 + 2*m)*((-3*AppellF1[5/2, 3/2 + m, 3/2, 7/2, Tan
[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/10 + (3*(3/2 + m)*AppellF1[5/2, 5/2
+ m, 1/2, 7/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/5)))/3))/(Sqrt[S
ec[(e + f*x)/2]^2]*(AppellF1[1/2, 1/2 + m, 1/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - ((AppellF1[3/2
, 1/2 + m, 3/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - (1 + 2*m)*AppellF1[3/2, 3/2 + m, 1/2, 5/2, Tan
[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)/3)^2) + (2^(1 + m)*(1/2 + m)*AppellF1[1/2, 1/2 + m,
1/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^(-1/2 + m)*Tan[(e + f*x)
/2]*(-(Cos[(e + f*x)/2]*Sec[e + f*x]*Sin[(e + f*x)/2]) + Cos[(e + f*x)/2]^2*Sec[e + f*x]*Tan[e + f*x]))/(Sqrt[
Sec[(e + f*x)/2]^2]*(AppellF1[1/2, 1/2 + m, 1/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - ((AppellF1[3/
2, 1/2 + m, 3/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - (1 + 2*m)*AppellF1[3/2, 3/2 + m, 1/2, 5/2, Ta
n[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)/3))))
Maple [F]
\[\int \sqrt {d \sec \left (f x +e \right )}\, \left (a +a \sec \left (f x +e \right )\right )^{m}d x\]
[In]
int((d*sec(f*x+e))^(1/2)*(a+a*sec(f*x+e))^m,x)
[Out]
int((d*sec(f*x+e))^(1/2)*(a+a*sec(f*x+e))^m,x)
Fricas [F]
\[
\int \sqrt {d \sec (e+f x)} (a+a \sec (e+f x))^m \, dx=\int { \sqrt {d \sec \left (f x + e\right )} {\left (a \sec \left (f x + e\right ) + a\right )}^{m} \,d x }
\]
[In]
integrate((d*sec(f*x+e))^(1/2)*(a+a*sec(f*x+e))^m,x, algorithm="fricas")
[Out]
integral(sqrt(d*sec(f*x + e))*(a*sec(f*x + e) + a)^m, x)
Sympy [F]
\[
\int \sqrt {d \sec (e+f x)} (a+a \sec (e+f x))^m \, dx=\int \left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{m} \sqrt {d \sec {\left (e + f x \right )}}\, dx
\]
[In]
integrate((d*sec(f*x+e))**(1/2)*(a+a*sec(f*x+e))**m,x)
[Out]
Integral((a*(sec(e + f*x) + 1))**m*sqrt(d*sec(e + f*x)), x)
Maxima [F]
\[
\int \sqrt {d \sec (e+f x)} (a+a \sec (e+f x))^m \, dx=\int { \sqrt {d \sec \left (f x + e\right )} {\left (a \sec \left (f x + e\right ) + a\right )}^{m} \,d x }
\]
[In]
integrate((d*sec(f*x+e))^(1/2)*(a+a*sec(f*x+e))^m,x, algorithm="maxima")
[Out]
integrate(sqrt(d*sec(f*x + e))*(a*sec(f*x + e) + a)^m, x)
Giac [F]
\[
\int \sqrt {d \sec (e+f x)} (a+a \sec (e+f x))^m \, dx=\int { \sqrt {d \sec \left (f x + e\right )} {\left (a \sec \left (f x + e\right ) + a\right )}^{m} \,d x }
\]
[In]
integrate((d*sec(f*x+e))^(1/2)*(a+a*sec(f*x+e))^m,x, algorithm="giac")
[Out]
integrate(sqrt(d*sec(f*x + e))*(a*sec(f*x + e) + a)^m, x)
Mupad [F(-1)]
Timed out. \[
\int \sqrt {d \sec (e+f x)} (a+a \sec (e+f x))^m \, dx=\int {\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^m\,\sqrt {\frac {d}{\cos \left (e+f\,x\right )}} \,d x
\]
[In]
int((a + a/cos(e + f*x))^m*(d/cos(e + f*x))^(1/2),x)
[Out]
int((a + a/cos(e + f*x))^m*(d/cos(e + f*x))^(1/2), x)