3.162 \(\int \sec (e+f x) (a+a \sec (e+f x))^m (c-c \sec (e+f x))^{-3-m} \, dx\)

Optimal. Leaf size=169 \[ -\frac {2 \tan (e+f x) (a \sec (e+f x)+a)^{m+2} (c-c \sec (e+f x))^{-m-3}}{a^2 f (2 m+1) \left (4 m^2+16 m+15\right )}+\frac {2 \tan (e+f x) (a \sec (e+f x)+a)^{m+1} (c-c \sec (e+f x))^{-m-3}}{a f \left (4 m^2+8 m+3\right )}-\frac {\tan (e+f x) (a \sec (e+f x)+a)^m (c-c \sec (e+f x))^{-m-3}}{f (2 m+1)} \]

[Out]

-(a+a*sec(f*x+e))^m*(c-c*sec(f*x+e))^(-3-m)*tan(f*x+e)/f/(1+2*m)+2*(a+a*sec(f*x+e))^(1+m)*(c-c*sec(f*x+e))^(-3
-m)*tan(f*x+e)/a/f/(4*m^2+8*m+3)-2*(a+a*sec(f*x+e))^(2+m)*(c-c*sec(f*x+e))^(-3-m)*tan(f*x+e)/a^2/f/(1+2*m)/(4*
m^2+16*m+15)

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Rubi [A]  time = 0.37, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {3951, 3950} \[ -\frac {2 \tan (e+f x) (a \sec (e+f x)+a)^{m+2} (c-c \sec (e+f x))^{-m-3}}{a^2 f (2 m+1) \left (4 m^2+16 m+15\right )}+\frac {2 \tan (e+f x) (a \sec (e+f x)+a)^{m+1} (c-c \sec (e+f x))^{-m-3}}{a f \left (4 m^2+8 m+3\right )}-\frac {\tan (e+f x) (a \sec (e+f x)+a)^m (c-c \sec (e+f x))^{-m-3}}{f (2 m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a + a*Sec[e + f*x])^m*(c - c*Sec[e + f*x])^(-3 - m)*Tan[e + f*x])/(f*(1 + 2*m))) + (2*(a + a*Sec[e + f*x])
^(1 + m)*(c - c*Sec[e + f*x])^(-3 - m)*Tan[e + f*x])/(a*f*(3 + 8*m + 4*m^2)) - (2*(a + a*Sec[e + f*x])^(2 + m)
*(c - c*Sec[e + f*x])^(-3 - m)*Tan[e + f*x])/(a^2*f*(1 + 2*m)*(15 + 16*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))^{-3-m} \, dx &=-\frac {(a+a \sec (e+f x))^m (c-c \sec (e+f x))^{-3-m} \tan (e+f x)}{f (1+2 m)}-\frac {2 \int \sec (e+f x) (a+a \sec (e+f x))^{1+m} (c-c \sec (e+f x))^{-3-m} \, dx}{a (1+2 m)}\\ &=-\frac {(a+a \sec (e+f x))^m (c-c \sec (e+f x))^{-3-m} \tan (e+f x)}{f (1+2 m)}+\frac {2 (a+a \sec (e+f x))^{1+m} (c-c \sec (e+f x))^{-3-m} \tan (e+f x)}{a f \left (3+8 m+4 m^2\right )}+\frac {2 \int \sec (e+f x) (a+a \sec (e+f x))^{2+m} (c-c \sec (e+f x))^{-3-m} \, dx}{a^2 \left (3+8 m+4 m^2\right )}\\ &=-\frac {(a+a \sec (e+f x))^m (c-c \sec (e+f x))^{-3-m} \tan (e+f x)}{f (1+2 m)}+\frac {2 (a+a \sec (e+f x))^{1+m} (c-c \sec (e+f x))^{-3-m} \tan (e+f x)}{a f \left (3+8 m+4 m^2\right )}-\frac {2 (a+a \sec (e+f x))^{2+m} (c-c \sec (e+f x))^{-3-m} \tan (e+f x)}{a^2 f (5+2 m) \left (3+8 m+4 m^2\right )}\\ \end {align*}

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Mathematica [C]  time = 9.08, size = 321, normalized size = 1.90 \[ -\frac {i 2^{m+3} \left (1+e^{i (e+f x)}\right ) \left (-i e^{-\frac {1}{2} i (e+f x)} \left (-1+e^{i (e+f x)}\right )\right )^{-2 m} \left (\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^{-m} \left (\frac {\left (1+e^{i (e+f x)}\right )^2}{1+e^{2 i (e+f x)}}\right )^m \left (\left (4 m^2+12 m+7\right ) e^{4 i (e+f x)}+\left (8 m^2+24 m+22\right ) e^{2 i (e+f x)}-4 (2 m+3) e^{i (e+f x)}-4 (2 m+3) e^{3 i (e+f x)}+4 m^2+12 m+7\right ) \sin ^{-2 (-m-3)}\left (\frac {e}{2}+\frac {f x}{2}\right ) \sec ^{m+3}(e+f x) (\sec (e+f x)+1)^{-m} (a (\sec (e+f x)+1))^m (c-c \sec (e+f x))^{-m-3}}{f (2 m+1) (2 m+3) (2 m+5) \left (-1+e^{i (e+f x)}\right )^5} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[e + f*x]*(a + a*Sec[e + f*x])^m*(c - c*Sec[e + f*x])^(-3 - 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*(7 + 12*m + 4*m^2
- 4*E^(I*(e + f*x))*(3 + 2*m) - 4*E^((3*I)*(e + f*x))*(3 + 2*m) + E^((4*I)*(e + f*x))*(7 + 12*m + 4*m^2) + E^(
(2*I)*(e + f*x))*(22 + 24*m + 8*m^2))*Sec[e + f*x]^(3 + m)*(a*(1 + Sec[e + f*x]))^m*(c - c*Sec[e + f*x])^(-3 -
m))/((-1 + E^(I*(e + f*x)))^5*(((-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)*(5 + 2*m)*(1 + Sec[e + f*x])^m*Sin[e/2 + (f*x)/2]^(2*(-3 - m)))

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fricas [A]  time = 0.53, size = 120, normalized size = 0.71 \[ -\frac {{\left ({\left (4 \, m^{2} + 12 \, m + 7\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (2 \, m + 3\right )} \cos \left (f x + e\right ) + 2\right )} \left (\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}\right )^{m} \left (\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}\right )^{-m - 3} \sin \left (f x + e\right )}{{\left (8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m + 15 \, f\right )} \cos \left (f x + e\right )^{3}} \]

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))^(-3-m),x, algorithm="fricas")

[Out]

-((4*m^2 + 12*m + 7)*cos(f*x + e)^2 - 2*(2*m + 3)*cos(f*x + e) + 2)*((a*cos(f*x + e) + a)/cos(f*x + e))^m*((c*
cos(f*x + e) - c)/cos(f*x + e))^(-m - 3)*sin(f*x + e)/((8*f*m^3 + 36*f*m^2 + 46*f*m + 15*f)*cos(f*x + e)^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sec \left (f x + e\right ) + a\right )}^{m} {\left (-c \sec \left (f x + e\right ) + c\right )}^{-m - 3} \sec \left (f x + e\right )\,{d x} \]

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))^(-3-m),x, algorithm="giac")

[Out]

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

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maple [F]  time = 3.88, size = 0, normalized size = 0.00 \[ \int \sec \left (f x +e \right ) \left (a +a \sec \left (f x +e \right )\right )^{m} \left (c -c \sec \left (f x +e \right )\right )^{-3-m}\, dx \]

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))^(-3-m),x)

[Out]

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

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maxima [A]  time = 0.66, size = 156, normalized size = 0.92 \[ \frac {{\left ({\left (4 \, m^{2} + 8 \, m + 3\right )} \left (-a\right )^{m} - \frac {2 \, {\left (4 \, m^{2} + 12 \, m + 5\right )} \left (-a\right )^{m} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {{\left (4 \, m^{2} + 16 \, m + 15\right )} \left (-a\right )^{m} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )} c^{-m - 3} {\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{4 \, {\left (8 \, m^{3} + 36 \, m^{2} + 46 \, m + 15\right )} f \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )^{2 \, m} \sin \left (f x + e\right )^{5}} \]

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))^(-3-m),x, algorithm="maxima")

[Out]

1/4*((4*m^2 + 8*m + 3)*(-a)^m - 2*(4*m^2 + 12*m + 5)*(-a)^m*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + (4*m^2 + 16*
m + 15)*(-a)^m*sin(f*x + e)^4/(cos(f*x + e) + 1)^4)*c^(-m - 3)*(cos(f*x + e) + 1)^5/((8*m^3 + 36*m^2 + 46*m +
15)*f*(sin(f*x + e)/(cos(f*x + e) + 1))^(2*m)*sin(f*x + e)^5)

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mupad [B]  time = 10.65, size = 290, normalized size = 1.72 \[ -\frac {\left (\cos \left (3\,e+3\,f\,x\right )-\sin \left (3\,e+3\,f\,x\right )\,1{}\mathrm {i}\right )\,\left (\frac {\sin \left (e+f\,x\right )\,{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^m\,\left (\cos \left (3\,e+3\,f\,x\right )+\sin \left (3\,e+3\,f\,x\right )\,1{}\mathrm {i}\right )\,\left (4\,m^2+12\,m+15\right )\,2{}\mathrm {i}}{f\,\left (m^3\,8{}\mathrm {i}+m^2\,36{}\mathrm {i}+m\,46{}\mathrm {i}+15{}\mathrm {i}\right )}-\frac {\sin \left (2\,e+2\,f\,x\right )\,\left (8\,m+12\right )\,{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^m\,\left (\cos \left (3\,e+3\,f\,x\right )+\sin \left (3\,e+3\,f\,x\right )\,1{}\mathrm {i}\right )\,2{}\mathrm {i}}{f\,\left (m^3\,8{}\mathrm {i}+m^2\,36{}\mathrm {i}+m\,46{}\mathrm {i}+15{}\mathrm {i}\right )}+\frac {\sin \left (3\,e+3\,f\,x\right )\,{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^m\,\left (\cos \left (3\,e+3\,f\,x\right )+\sin \left (3\,e+3\,f\,x\right )\,1{}\mathrm {i}\right )\,\left (4\,m^2+12\,m+7\right )\,2{}\mathrm {i}}{f\,\left (m^3\,8{}\mathrm {i}+m^2\,36{}\mathrm {i}+m\,46{}\mathrm {i}+15{}\mathrm {i}\right )}\right )}{8\,{\cos \left (e+f\,x\right )}^3\,{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{m+3}} \]

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 + 3)),x)

[Out]

-((cos(3*e + 3*f*x) - sin(3*e + 3*f*x)*1i)*((sin(e + f*x)*(a + a/cos(e + f*x))^m*(cos(3*e + 3*f*x) + sin(3*e +
3*f*x)*1i)*(12*m + 4*m^2 + 15)*2i)/(f*(m*46i + m^2*36i + m^3*8i + 15i)) - (sin(2*e + 2*f*x)*(8*m + 12)*(a + a
/cos(e + f*x))^m*(cos(3*e + 3*f*x) + sin(3*e + 3*f*x)*1i)*2i)/(f*(m*46i + m^2*36i + m^3*8i + 15i)) + (sin(3*e
+ 3*f*x)*(a + a/cos(e + f*x))^m*(cos(3*e + 3*f*x) + sin(3*e + 3*f*x)*1i)*(12*m + 4*m^2 + 7)*2i)/(f*(m*46i + m^
2*36i + m^3*8i + 15i))))/(8*cos(e + f*x)^3*(c - c/cos(e + f*x))^(m + 3))

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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))**(-3-m),x)

[Out]

Timed out

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