3.158 \(\int \sec (e+f x) (a+a \sec (e+f x))^m \sqrt {c-c \sec (e+f x)} \, dx\)

Optimal. Leaf size=46 \[ -\frac {2 c \tan (e+f x) (a \sec (e+f x)+a)^m}{f (2 m+1) \sqrt {c-c \sec (e+f x)}} \]

[Out]

-2*c*(a+a*sec(f*x+e))^m*tan(f*x+e)/f/(1+2*m)/(c-c*sec(f*x+e))^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {3953} \[ -\frac {2 c \tan (e+f x) (a \sec (e+f x)+a)^m}{f (2 m+1) \sqrt {c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]*(a + a*Sec[e + f*x])^m*Sqrt[c - c*Sec[e + f*x]],x]

[Out]

(-2*c*(a + a*Sec[e + f*x])^m*Tan[e + f*x])/(f*(1 + 2*m)*Sqrt[c - c*Sec[e + f*x]])

Rule 3953

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

Rubi steps

\begin {align*} \int \sec (e+f x) (a+a \sec (e+f x))^m \sqrt {c-c \sec (e+f x)} \, dx &=-\frac {2 c (a+a \sec (e+f x))^m \tan (e+f x)}{f (1+2 m) \sqrt {c-c \sec (e+f x)}}\\ \end {align*}

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Mathematica [C]  time = 18.90, size = 163, normalized size = 3.54 \[ \frac {\sqrt {2} e^{-\frac {1}{2} i (e+f x)} \left (1+e^{i (e+f x)}\right ) \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \left (\frac {\left (1+e^{i (e+f x)}\right )^2}{1+e^{2 i (e+f x)}}\right )^m \csc \left (\frac {1}{2} (e+f x)\right ) \sqrt {c-c \sec (e+f x)} (\sec (e+f x)+1)^{-m} (a (\sec (e+f x)+1))^m}{(2 f m+f) \sqrt {\sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]*(a + a*Sec[e + f*x])^m*Sqrt[c - c*Sec[e + f*x]],x]

[Out]

(Sqrt[2]*(1 + E^(I*(e + f*x)))*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)))]*((1 + E^(I*(e + f*x)))^2/(1 + E
^((2*I)*(e + f*x))))^m*Csc[(e + f*x)/2]*(a*(1 + Sec[e + f*x]))^m*Sqrt[c - c*Sec[e + f*x]])/(E^((I/2)*(e + f*x)
)*(f + 2*f*m)*Sqrt[Sec[e + f*x]]*(1 + Sec[e + f*x])^m)

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fricas [A]  time = 0.47, size = 70, normalized size = 1.52 \[ \frac {2 \, \left (\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}\right )^{m} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} {\left (\cos \left (f x + e\right ) + 1\right )}}{{\left (2 \, f m + f\right )} \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^m*(c-c*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

2*((a*cos(f*x + e) + a)/cos(f*x + e))^m*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))*(cos(f*x + e) + 1)/((2*f*m + f
)*sin(f*x + e))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {-c \sec \left (f x + e\right ) + c} {\left (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^m*(c-c*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(-c*sec(f*x + e) + c)*(a*sec(f*x + e) + a)^m*sec(f*x + e), x)

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maple [F]  time = 1.86, size = 0, normalized size = 0.00 \[ \int \sec \left (f x +e \right ) \left (a +a \sec \left (f x +e \right )\right )^{m} \sqrt {c -c \sec \left (f x +e \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(a+a*sec(f*x+e))^m*(c-c*sec(f*x+e))^(1/2),x)

[Out]

int(sec(f*x+e)*(a+a*sec(f*x+e))^m*(c-c*sec(f*x+e))^(1/2),x)

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maxima [B]  time = 0.46, size = 114, normalized size = 2.48 \[ \frac {2^{m + \frac {3}{2}} \left (-a\right )^{m} \sqrt {c} e^{\left (-m \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right ) - m \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )\right )}}{f {\left (2 \, m + 1\right )} \sqrt {\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1} \sqrt {\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^m*(c-c*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

2^(m + 3/2)*(-a)^m*sqrt(c)*e^(-m*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1) - m*log(sin(f*x + e)/(cos(f*x + e) +
1) - 1))/(f*(2*m + 1)*sqrt(sin(f*x + e)/(cos(f*x + e) + 1) + 1)*sqrt(sin(f*x + e)/(cos(f*x + e) + 1) - 1))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^m\,\sqrt {c-\frac {c}{\cos \left (e+f\,x\right )}}}{\cos \left (e+f\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + a/cos(e + f*x))^m*(c - c/cos(e + f*x))^(1/2))/cos(e + f*x),x)

[Out]

int(((a + a/cos(e + f*x))^m*(c - c/cos(e + f*x))^(1/2))/cos(e + f*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{m} \sqrt {- c \left (\sec {\left (e + f x \right )} - 1\right )} \sec {\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))**m*(c-c*sec(f*x+e))**(1/2),x)

[Out]

Integral((a*(sec(e + f*x) + 1))**m*sqrt(-c*(sec(e + f*x) - 1))*sec(e + f*x), x)

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