3.72 \(\int \sec (e+f x) (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=128 \[ -\frac {64 c^3 \tan (e+f x) (a \sec (e+f x)+a)^2}{315 f \sqrt {c-c \sec (e+f x)}}-\frac {16 c^2 \tan (e+f x) (a \sec (e+f x)+a)^2 \sqrt {c-c \sec (e+f x)}}{63 f}-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a)^2 (c-c \sec (e+f x))^{3/2}}{9 f} \]

[Out]

-2/9*c*(a+a*sec(f*x+e))^2*(c-c*sec(f*x+e))^(3/2)*tan(f*x+e)/f-64/315*c^3*(a+a*sec(f*x+e))^2*tan(f*x+e)/f/(c-c*
sec(f*x+e))^(1/2)-16/63*c^2*(a+a*sec(f*x+e))^2*(c-c*sec(f*x+e))^(1/2)*tan(f*x+e)/f

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Rubi [A]  time = 0.33, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {3955, 3953} \[ -\frac {64 c^3 \tan (e+f x) (a \sec (e+f x)+a)^2}{315 f \sqrt {c-c \sec (e+f x)}}-\frac {16 c^2 \tan (e+f x) (a \sec (e+f x)+a)^2 \sqrt {c-c \sec (e+f x)}}{63 f}-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a)^2 (c-c \sec (e+f x))^{3/2}}{9 f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]*(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^(5/2),x]

[Out]

(-64*c^3*(a + a*Sec[e + f*x])^2*Tan[e + f*x])/(315*f*Sqrt[c - c*Sec[e + f*x]]) - (16*c^2*(a + a*Sec[e + f*x])^
2*Sqrt[c - c*Sec[e + f*x]]*Tan[e + f*x])/(63*f) - (2*c*(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^(3/2)*Tan[e
+ f*x])/(9*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))^{5/2} \, dx &=-\frac {2 c (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{9 f}+\frac {1}{9} (8 c) \int \sec (e+f x) (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2} \, dx\\ &=-\frac {16 c^2 (a+a \sec (e+f x))^2 \sqrt {c-c \sec (e+f x)} \tan (e+f x)}{63 f}-\frac {2 c (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{9 f}+\frac {1}{63} \left (32 c^2\right ) \int \sec (e+f x) (a+a \sec (e+f x))^2 \sqrt {c-c \sec (e+f x)} \, dx\\ &=-\frac {64 c^3 (a+a \sec (e+f x))^2 \tan (e+f x)}{315 f \sqrt {c-c \sec (e+f x)}}-\frac {16 c^2 (a+a \sec (e+f x))^2 \sqrt {c-c \sec (e+f x)} \tan (e+f x)}{63 f}-\frac {2 c (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{9 f}\\ \end {align*}

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Mathematica [A]  time = 1.24, size = 78, normalized size = 0.61 \[ \frac {4 a^2 c^2 \cos ^4\left (\frac {1}{2} (e+f x)\right ) (-220 \cos (e+f x)+107 \cos (2 (e+f x))+177) \cot \left (\frac {1}{2} (e+f x)\right ) \sec ^4(e+f x) \sqrt {c-c \sec (e+f x)}}{315 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]*(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^(5/2),x]

[Out]

(4*a^2*c^2*Cos[(e + f*x)/2]^4*(177 - 220*Cos[e + f*x] + 107*Cos[2*(e + f*x)])*Cot[(e + f*x)/2]*Sec[e + f*x]^4*
Sqrt[c - c*Sec[e + f*x]])/(315*f)

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fricas [A]  time = 0.44, size = 131, normalized size = 1.02 \[ \frac {2 \, {\left (107 \, a^{2} c^{2} \cos \left (f x + e\right )^{5} + 211 \, a^{2} c^{2} \cos \left (f x + e\right )^{4} + 26 \, a^{2} c^{2} \cos \left (f x + e\right )^{3} - 118 \, a^{2} c^{2} \cos \left (f x + e\right )^{2} - 5 \, a^{2} c^{2} \cos \left (f x + e\right ) + 35 \, a^{2} c^{2}\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{315 \, f \cos \left (f x + e\right )^{4} \sin \left (f x + e\right )} \]

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))^(5/2),x, algorithm="fricas")

[Out]

2/315*(107*a^2*c^2*cos(f*x + e)^5 + 211*a^2*c^2*cos(f*x + e)^4 + 26*a^2*c^2*cos(f*x + e)^3 - 118*a^2*c^2*cos(f
*x + e)^2 - 5*a^2*c^2*cos(f*x + e) + 35*a^2*c^2)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))/(f*cos(f*x + e)^4*sin
(f*x + e))

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giac [A]  time = 2.74, size = 88, normalized size = 0.69 \[ -\frac {32 \, \sqrt {2} {\left (63 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{2} c^{3} + 90 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )} c^{4} + 35 \, c^{5}\right )} a^{2} c^{2}}{315 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{\frac {9}{2}} f} \]

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))^(5/2),x, algorithm="giac")

[Out]

-32/315*sqrt(2)*(63*(c*tan(1/2*f*x + 1/2*e)^2 - c)^2*c^3 + 90*(c*tan(1/2*f*x + 1/2*e)^2 - c)*c^4 + 35*c^5)*a^2
*c^2/((c*tan(1/2*f*x + 1/2*e)^2 - c)^(9/2)*f)

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maple [A]  time = 1.64, size = 75, normalized size = 0.59 \[ -\frac {2 a^{2} \left (107 \left (\cos ^{2}\left (f x +e \right )\right )-110 \cos \left (f x +e \right )+35\right ) \left (\sin ^{5}\left (f x +e \right )\right ) \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}}}{315 f \left (-1+\cos \left (f x +e \right )\right )^{5} \cos \left (f x +e \right )^{2}} \]

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))^(5/2),x)

[Out]

-2/315*a^2/f*(107*cos(f*x+e)^2-110*cos(f*x+e)+35)*sin(f*x+e)^5*(c*(-1+cos(f*x+e))/cos(f*x+e))^(5/2)/(-1+cos(f*
x+e))^5/cos(f*x+e)^2

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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))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [B]  time = 7.99, size = 503, normalized size = 3.93 \[ \frac {\left (\frac {a^2\,c^2\,2{}\mathrm {i}}{f}+\frac {a^2\,c^2\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,214{}\mathrm {i}}{315\,f}\right )\,\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}}}}{{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-1}+\frac {\left (\frac {a^2\,c^2\,32{}\mathrm {i}}{9\,f}+\frac {a^2\,c^2\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,32{}\mathrm {i}}{9\,f}\right )\,\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 ({\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 )}^4}-\frac {\left (\frac {a^2\,c^2\,64{}\mathrm {i}}{7\,f}+\frac {a^2\,c^2\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,320{}\mathrm {i}}{63\,f}\right )\,\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 ({\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 {\left (\frac {a^2\,c^2\,48{}\mathrm {i}}{5\,f}+\frac {a^2\,c^2\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,368{}\mathrm {i}}{105\,f}\right )\,\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 ({\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}-\frac {\left (\frac {a^2\,c^2\,16{}\mathrm {i}}{3\,f}+\frac {a^2\,c^2\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,208{}\mathrm {i}}{315\,f}\right )\,\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 ({\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 )} \]

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))^(5/2))/cos(e + f*x),x)

[Out]

(((a^2*c^2*2i)/f + (a^2*c^2*exp(e*1i + f*x*1i)*214i)/(315*f))*(c - c/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*
1i)/2))^(1/2))/(exp(e*1i + f*x*1i) - 1) + (((a^2*c^2*32i)/(9*f) + (a^2*c^2*exp(e*1i + f*x*1i)*32i)/(9*f))*(c -
c/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2))^(1/2))/((exp(e*1i + f*x*1i) - 1)*(exp(e*2i + f*x*2i) + 1)^
4) - (((a^2*c^2*64i)/(7*f) + (a^2*c^2*exp(e*1i + f*x*1i)*320i)/(63*f))*(c - c/(exp(- e*1i - f*x*1i)/2 + exp(e*
1i + f*x*1i)/2))^(1/2))/((exp(e*1i + f*x*1i) - 1)*(exp(e*2i + f*x*2i) + 1)^3) + (((a^2*c^2*48i)/(5*f) + (a^2*c
^2*exp(e*1i + f*x*1i)*368i)/(105*f))*(c - c/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2))^(1/2))/((exp(e*1i
+ f*x*1i) - 1)*(exp(e*2i + f*x*2i) + 1)^2) - (((a^2*c^2*16i)/(3*f) + (a^2*c^2*exp(e*1i + f*x*1i)*208i)/(315*f
))*(c - c/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2))^(1/2))/((exp(e*1i + f*x*1i) - 1)*(exp(e*2i + f*x*2i
) + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \left (\int c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec {\left (e + f x \right )}\, dx + \int \left (- 2 c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{3}{\left (e + f x \right )}\right )\, dx + \int c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{5}{\left (e + f x \right )}\, dx\right ) \]

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))**(5/2),x)

[Out]

a**2*(Integral(c**2*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x), x) + Integral(-2*c**2*sqrt(-c*sec(e + f*x) + c)*se
c(e + f*x)**3, x) + Integral(c**2*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x)**5, x))

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