3.163 \(\int \sec (e+f x) (a+a \sec (e+f x))^m (c-c \sec (e+f x))^{-2-m} \, dx\)

Optimal. Leaf size=104 \[ \frac {\tan (e+f x) (a \sec (e+f x)+a)^{m+1} (c-c \sec (e+f x))^{-m-2}}{a f \left (4 m^2+8 m+3\right )}-\frac {\tan (e+f x) (a \sec (e+f x)+a)^m (c-c \sec (e+f x))^{-m-2}}{f (2 m+1)} \]

[Out]

-(a+a*sec(f*x+e))^m*(c-c*sec(f*x+e))^(-2-m)*tan(f*x+e)/f/(1+2*m)+(a+a*sec(f*x+e))^(1+m)*(c-c*sec(f*x+e))^(-2-m
)*tan(f*x+e)/a/f/(4*m^2+8*m+3)

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Rubi [A]  time = 0.22, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {3951, 3950} \[ \frac {\tan (e+f x) (a \sec (e+f x)+a)^{m+1} (c-c \sec (e+f x))^{-m-2}}{a f \left (4 m^2+8 m+3\right )}-\frac {\tan (e+f x) (a \sec (e+f x)+a)^m (c-c \sec (e+f x))^{-m-2}}{f (2 m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a + a*Sec[e + f*x])^m*(c - c*Sec[e + f*x])^(-2 - m)*Tan[e + f*x])/(f*(1 + 2*m))) + ((a + a*Sec[e + f*x])^(
1 + m)*(c - c*Sec[e + f*x])^(-2 - m)*Tan[e + f*x])/(a*f*(3 + 8*m + 4*m^2))

Rule 3950

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

Rule 3951

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

Rubi steps

\begin {align*} \int \sec (e+f x) (a+a \sec (e+f x))^m (c-c \sec (e+f x))^{-2-m} \, dx &=-\frac {(a+a \sec (e+f x))^m (c-c \sec (e+f x))^{-2-m} \tan (e+f x)}{f (1+2 m)}-\frac {\int \sec (e+f x) (a+a \sec (e+f x))^{1+m} (c-c \sec (e+f x))^{-2-m} \, dx}{a (1+2 m)}\\ &=-\frac {(a+a \sec (e+f x))^m (c-c \sec (e+f x))^{-2-m} \tan (e+f x)}{f (1+2 m)}+\frac {(a+a \sec (e+f x))^{1+m} (c-c \sec (e+f x))^{-2-m} \tan (e+f x)}{a f \left (3+8 m+4 m^2\right )}\\ \end {align*}

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Mathematica [C]  time = 3.05, size = 250, normalized size = 2.40 \[ \frac {i 2^{m+3} \left (1+e^{i (e+f x)}\right ) \left (-i e^{-\frac {1}{2} i (e+f x)} \left (-1+e^{i (e+f x)}\right )\right )^{-2 m} \left (\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^{-m} \left (\frac {\left (1+e^{i (e+f x)}\right )^2}{1+e^{2 i (e+f x)}}\right )^m \left ((m+1) e^{2 i (e+f x)}-e^{i (e+f x)}+m+1\right ) \sin ^{2 (m+2)}\left (\frac {1}{2} (e+f x)\right ) \sec ^{m+2}(e+f x) (\sec (e+f x)+1)^{-m} (a (\sec (e+f x)+1))^m (c-c \sec (e+f x))^{-m-2}}{f (2 m+1) (2 m+3) \left (-1+e^{i (e+f x)}\right )^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(I*2^(3 + m)*(1 + E^(I*(e + f*x)))*((1 + E^(I*(e + f*x)))^2/(1 + E^((2*I)*(e + f*x))))^m*(1 - E^(I*(e + f*x))
+ m + E^((2*I)*(e + f*x))*(1 + m))*Sec[e + f*x]^(2 + m)*(a*(1 + Sec[e + f*x]))^m*(c - c*Sec[e + f*x])^(-2 - m)
*Sin[(e + f*x)/2]^(2*(2 + m)))/((-1 + E^(I*(e + f*x)))^3*(((-I)*(-1 + E^(I*(e + f*x))))/E^((I/2)*(e + f*x)))^(
2*m)*(E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x))))^m*f*(1 + 2*m)*(3 + 2*m)*(1 + Sec[e + f*x])^m)

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fricas [A]  time = 0.46, size = 93, normalized size = 0.89 \[ -\frac {{\left (2 \, {\left (m + 1\right )} \cos \left (f x + e\right ) - 1\right )} \left (\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}\right )^{m} \left (\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}\right )^{-m - 2} \sin \left (f x + e\right )}{{\left (4 \, f m^{2} + 8 \, f m + 3 \, f\right )} \cos \left (f x + e\right )^{2}} \]

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))^(-2-m),x, algorithm="fricas")

[Out]

-(2*(m + 1)*cos(f*x + e) - 1)*((a*cos(f*x + e) + a)/cos(f*x + e))^m*((c*cos(f*x + e) - c)/cos(f*x + e))^(-m -
2)*sin(f*x + e)/((4*f*m^2 + 8*f*m + 3*f)*cos(f*x + e)^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sec \left (f x + e\right ) + a\right )}^{m} {\left (-c \sec \left (f x + e\right ) + c\right )}^{-m - 2} \sec \left (f x + e\right )\,{d x} \]

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))^(-2-m),x, algorithm="giac")

[Out]

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

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maple [F]  time = 3.81, size = 0, normalized size = 0.00 \[ \int \sec \left (f x +e \right ) \left (a +a \sec \left (f x +e \right )\right )^{m} \left (c -c \sec \left (f x +e \right )\right )^{-2-m}\, dx \]

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))^(-2-m),x)

[Out]

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

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maxima [A]  time = 0.83, size = 107, normalized size = 1.03 \[ -\frac {{\left (\left (-a\right )^{m} {\left (2 \, m + 1\right )} - \frac {\left (-a\right )^{m} {\left (2 \, m + 3\right )} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} c^{-m - 2} {\left (\cos \left (f x + e\right ) + 1\right )}^{3}}{2 \, {\left (4 \, m^{2} + 8 \, m + 3\right )} f \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )^{2 \, m} \sin \left (f x + e\right )^{3}} \]

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))^(-2-m),x, algorithm="maxima")

[Out]

-1/2*((-a)^m*(2*m + 1) - (-a)^m*(2*m + 3)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2)*c^(-m - 2)*(cos(f*x + e) + 1)^3
/((4*m^2 + 8*m + 3)*f*(sin(f*x + e)/(cos(f*x + e) + 1))^(2*m)*sin(f*x + e)^3)

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mupad [B]  time = 8.00, size = 145, normalized size = 1.39 \[ \frac {\sin \left (e+f\,x\right )\,{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^m\,1{}\mathrm {i}}{f\,{\cos \left (e+f\,x\right )}^2\,{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{m+2}\,\left (m^2\,4{}\mathrm {i}+m\,8{}\mathrm {i}+3{}\mathrm {i}\right )}-\frac {\sin \left (2\,e+2\,f\,x\right )\,\left (2\,m+2\right )\,{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^m\,1{}\mathrm {i}}{2\,f\,{\cos \left (e+f\,x\right )}^2\,{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{m+2}\,\left (m^2\,4{}\mathrm {i}+m\,8{}\mathrm {i}+3{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sin(e + f*x)*(a + a/cos(e + f*x))^m*1i)/(f*cos(e + f*x)^2*(c - c/cos(e + f*x))^(m + 2)*(m*8i + m^2*4i + 3i))
- (sin(2*e + 2*f*x)*(2*m + 2)*(a + a/cos(e + f*x))^m*1i)/(2*f*cos(e + f*x)^2*(c - c/cos(e + f*x))^(m + 2)*(m*8
i + m^2*4i + 3i))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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))**(-2-m),x)

[Out]

Timed out

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