3.73 \(\int \sec (e+f x) (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=85 \[ -\frac {8 c^2 \tan (e+f x) (a \sec (e+f x)+a)^2}{35 f \sqrt {c-c \sec (e+f x)}}-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a)^2 \sqrt {c-c \sec (e+f x)}}{7 f} \]

[Out]

-8/35*c^2*(a+a*sec(f*x+e))^2*tan(f*x+e)/f/(c-c*sec(f*x+e))^(1/2)-2/7*c*(a+a*sec(f*x+e))^2*(c-c*sec(f*x+e))^(1/
2)*tan(f*x+e)/f

________________________________________________________________________________________

Rubi [A]  time = 0.21, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {3955, 3953} \[ -\frac {8 c^2 \tan (e+f x) (a \sec (e+f x)+a)^2}{35 f \sqrt {c-c \sec (e+f x)}}-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a)^2 \sqrt {c-c \sec (e+f x)}}{7 f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]*(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^(3/2),x]

[Out]

(-8*c^2*(a + a*Sec[e + f*x])^2*Tan[e + f*x])/(35*f*Sqrt[c - c*Sec[e + f*x]]) - (2*c*(a + a*Sec[e + f*x])^2*Sqr
t[c - c*Sec[e + f*x]]*Tan[e + f*x])/(7*f)

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)]

Rule 3955

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

Rubi steps

\begin {align*} \int \sec (e+f x) (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2} \, dx &=-\frac {2 c (a+a \sec (e+f x))^2 \sqrt {c-c \sec (e+f x)} \tan (e+f x)}{7 f}+\frac {1}{7} (4 c) \int \sec (e+f x) (a+a \sec (e+f x))^2 \sqrt {c-c \sec (e+f x)} \, dx\\ &=-\frac {8 c^2 (a+a \sec (e+f x))^2 \tan (e+f x)}{35 f \sqrt {c-c \sec (e+f x)}}-\frac {2 c (a+a \sec (e+f x))^2 \sqrt {c-c \sec (e+f x)} \tan (e+f x)}{7 f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.78, size = 66, normalized size = 0.78 \[ \frac {8 a^2 c \cos ^4\left (\frac {1}{2} (e+f x)\right ) (9 \cos (e+f x)-5) \cot \left (\frac {1}{2} (e+f x)\right ) \sec ^3(e+f x) \sqrt {c-c \sec (e+f x)}}{35 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]*(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^(3/2),x]

[Out]

(8*a^2*c*Cos[(e + f*x)/2]^4*(-5 + 9*Cos[e + f*x])*Cot[(e + f*x)/2]*Sec[e + f*x]^3*Sqrt[c - c*Sec[e + f*x]])/(3
5*f)

________________________________________________________________________________________

fricas [A]  time = 0.44, size = 105, normalized size = 1.24 \[ \frac {2 \, {\left (9 \, a^{2} c \cos \left (f x + e\right )^{4} + 22 \, a^{2} c \cos \left (f x + e\right )^{3} + 12 \, a^{2} c \cos \left (f x + e\right )^{2} - 6 \, a^{2} c \cos \left (f x + e\right ) - 5 \, a^{2} c\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{35 \, f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^2*(c-c*sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

2/35*(9*a^2*c*cos(f*x + e)^4 + 22*a^2*c*cos(f*x + e)^3 + 12*a^2*c*cos(f*x + e)^2 - 6*a^2*c*cos(f*x + e) - 5*a^
2*c)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))/(f*cos(f*x + e)^3*sin(f*x + e))

________________________________________________________________________________________

giac [A]  time = 3.40, size = 60, normalized size = 0.71 \[ -\frac {16 \, \sqrt {2} {\left (7 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )} c^{4} + 5 \, c^{5}\right )} a^{2}}{35 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{\frac {7}{2}} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^2*(c-c*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

-16/35*sqrt(2)*(7*(c*tan(1/2*f*x + 1/2*e)^2 - c)*c^4 + 5*c^5)*a^2/((c*tan(1/2*f*x + 1/2*e)^2 - c)^(7/2)*f)

________________________________________________________________________________________

maple [A]  time = 1.58, size = 65, normalized size = 0.76 \[ -\frac {2 a^{2} \left (9 \cos \left (f x +e \right )-5\right ) \left (\sin ^{5}\left (f x +e \right )\right ) \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}}}{35 f \left (-1+\cos \left (f x +e \right )\right )^{4} \cos \left (f x +e \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(a+a*sec(f*x+e))^2*(c-c*sec(f*x+e))^(3/2),x)

[Out]

-2/35*a^2/f*(9*cos(f*x+e)-5)*sin(f*x+e)^5*(c*(-1+cos(f*x+e))/cos(f*x+e))^(3/2)/(-1+cos(f*x+e))^4/cos(f*x+e)^2

________________________________________________________________________________________

maxima [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))^2*(c-c*sec(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

mupad [B]  time = 6.14, size = 384, normalized size = 4.52 \[ \frac {\sqrt {c-\frac {c}{\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {a^2\,c\,2{}\mathrm {i}}{f}+\frac {a^2\,c\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,18{}\mathrm {i}}{35\,f}\right )}{{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-1}-\frac {\sqrt {c-\frac {c}{\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {a^2\,c\,16{}\mathrm {i}}{7\,f}-\frac {a^2\,c\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,16{}\mathrm {i}}{7\,f}\right )}{\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-1\right )\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^3}-\frac {\sqrt {c-\frac {c}{\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {a^2\,c\,4{}\mathrm {i}}{f}-\frac {a^2\,c\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,44{}\mathrm {i}}{35\,f}\right )}{\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-1\right )\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}+\frac {\sqrt {c-\frac {c}{\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {a^2\,c\,24{}\mathrm {i}}{5\,f}-\frac {a^2\,c\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,72{}\mathrm {i}}{35\,f}\right )}{\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-1\right )\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + a/cos(e + f*x))^2*(c - c/cos(e + f*x))^(3/2))/cos(e + f*x),x)

[Out]

((c - c/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2))^(1/2)*((a^2*c*2i)/f + (a^2*c*exp(e*1i + f*x*1i)*18i)/
(35*f)))/(exp(e*1i + f*x*1i) - 1) - ((c - c/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2))^(1/2)*((a^2*c*16i
)/(7*f) - (a^2*c*exp(e*1i + f*x*1i)*16i)/(7*f)))/((exp(e*1i + f*x*1i) - 1)*(exp(e*2i + f*x*2i) + 1)^3) - ((c -
c/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2))^(1/2)*((a^2*c*4i)/f - (a^2*c*exp(e*1i + f*x*1i)*44i)/(35*f
)))/((exp(e*1i + f*x*1i) - 1)*(exp(e*2i + f*x*2i) + 1)) + ((c - c/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)
/2))^(1/2)*((a^2*c*24i)/(5*f) - (a^2*c*exp(e*1i + f*x*1i)*72i)/(35*f)))/((exp(e*1i + f*x*1i) - 1)*(exp(e*2i +
f*x*2i) + 1)^2)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \left (\int c \sqrt {- c \sec {\left (e + f x \right )} + c} \sec {\left (e + f x \right )}\, dx + \int c \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{2}{\left (e + f x \right )}\, dx + \int \left (- c \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{3}{\left (e + f x \right )}\right )\, dx + \int \left (- c \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{4}{\left (e + f x \right )}\right )\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))**2*(c-c*sec(f*x+e))**(3/2),x)

[Out]

a**2*(Integral(c*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x), x) + Integral(c*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x
)**2, x) + Integral(-c*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x)**3, x) + Integral(-c*sqrt(-c*sec(e + f*x) + c)*s
ec(e + f*x)**4, x))

________________________________________________________________________________________