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

Optimal. Leaf size=122 \[ -\frac {64 c^3 \tan (e+f x) (a \sec (e+f x)+a)}{105 f \sqrt {c-c \sec (e+f x)}}-\frac {16 c^2 \tan (e+f x) (a \sec (e+f x)+a) \sqrt {c-c \sec (e+f x)}}{35 f}-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a) (c-c \sec (e+f x))^{3/2}}{7 f} \]

[Out]

-2/7*c*(a+a*sec(f*x+e))*(c-c*sec(f*x+e))^(3/2)*tan(f*x+e)/f-64/105*c^3*(a+a*sec(f*x+e))*tan(f*x+e)/f/(c-c*sec(
f*x+e))^(1/2)-16/35*c^2*(a+a*sec(f*x+e))*(c-c*sec(f*x+e))^(1/2)*tan(f*x+e)/f

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Rubi [A]  time = 0.20, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {3955, 3953} \[ -\frac {64 c^3 \tan (e+f x) (a \sec (e+f x)+a)}{105 f \sqrt {c-c \sec (e+f x)}}-\frac {16 c^2 \tan (e+f x) (a \sec (e+f x)+a) \sqrt {c-c \sec (e+f x)}}{35 f}-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a) (c-c \sec (e+f x))^{3/2}}{7 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-64*c^3*(a + a*Sec[e + f*x])*Tan[e + f*x])/(105*f*Sqrt[c - c*Sec[e + f*x]]) - (16*c^2*(a + a*Sec[e + f*x])*Sq
rt[c - c*Sec[e + f*x]]*Tan[e + f*x])/(35*f) - (2*c*(a + a*Sec[e + f*x])*(c - c*Sec[e + f*x])^(3/2)*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)) (c-c \sec (e+f x))^{5/2} \, dx &=-\frac {2 c (a+a \sec (e+f x)) (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{7 f}+\frac {1}{7} (8 c) \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^{3/2} \, dx\\ &=-\frac {16 c^2 (a+a \sec (e+f x)) \sqrt {c-c \sec (e+f x)} \tan (e+f x)}{35 f}-\frac {2 c (a+a \sec (e+f x)) (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{7 f}+\frac {1}{35} \left (32 c^2\right ) \int \sec (e+f x) (a+a \sec (e+f x)) \sqrt {c-c \sec (e+f x)} \, dx\\ &=-\frac {64 c^3 (a+a \sec (e+f x)) \tan (e+f x)}{105 f \sqrt {c-c \sec (e+f x)}}-\frac {16 c^2 (a+a \sec (e+f x)) \sqrt {c-c \sec (e+f x)} \tan (e+f x)}{35 f}-\frac {2 c (a+a \sec (e+f x)) (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{7 f}\\ \end {align*}

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Mathematica [A]  time = 0.49, size = 76, normalized size = 0.62 \[ \frac {2 a c^2 \cos ^2\left (\frac {1}{2} (e+f x)\right ) (-108 \cos (e+f x)+71 \cos (2 (e+f x))+101) \cot \left (\frac {1}{2} (e+f x)\right ) \sec ^3(e+f x) \sqrt {c-c \sec (e+f x)}}{105 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*c^2*Cos[(e + f*x)/2]^2*(101 - 108*Cos[e + f*x] + 71*Cos[2*(e + f*x)])*Cot[(e + f*x)/2]*Sec[e + f*x]^3*Sqr
t[c - c*Sec[e + f*x]])/(105*f)

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fricas [A]  time = 0.45, size = 105, normalized size = 0.86 \[ \frac {2 \, {\left (71 \, a c^{2} \cos \left (f x + e\right )^{4} + 88 \, a c^{2} \cos \left (f x + e\right )^{3} - 22 \, a c^{2} \cos \left (f x + e\right )^{2} - 24 \, a c^{2} \cos \left (f x + e\right ) + 15 \, a c^{2}\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{105 \, f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))*(c-c*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

2/105*(71*a*c^2*cos(f*x + e)^4 + 88*a*c^2*cos(f*x + e)^3 - 22*a*c^2*cos(f*x + e)^2 - 24*a*c^2*cos(f*x + e) + 1
5*a*c^2)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))/(f*cos(f*x + e)^3*sin(f*x + e))

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giac [A]  time = 3.41, size = 86, normalized size = 0.70 \[ \frac {16 \, \sqrt {2} {\left (35 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{2} c^{2} + 42 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )} c^{3} + 15 \, c^{4}\right )} a c^{2}}{105 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{\frac {7}{2}} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))*(c-c*sec(f*x+e))^(5/2),x, algorithm="giac")

[Out]

16/105*sqrt(2)*(35*(c*tan(1/2*f*x + 1/2*e)^2 - c)^2*c^2 + 42*(c*tan(1/2*f*x + 1/2*e)^2 - c)*c^3 + 15*c^4)*a*c^
2/((c*tan(1/2*f*x + 1/2*e)^2 - c)^(7/2)*f)

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maple [A]  time = 1.18, size = 73, normalized size = 0.60 \[ \frac {2 a \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \left (\sin ^{3}\left (f x +e \right )\right ) \left (71 \left (\cos ^{2}\left (f x +e \right )\right )-54 \cos \left (f x +e \right )+15\right )}{105 f \left (-1+\cos \left (f x +e \right )\right )^{4} \cos \left (f x +e \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(a+a*sec(f*x+e))*(c-c*sec(f*x+e))^(5/2),x)

[Out]

2/105*a/f*(c*(-1+cos(f*x+e))/cos(f*x+e))^(5/2)*sin(f*x+e)^3*(71*cos(f*x+e)^2-54*cos(f*x+e)+15)/(-1+cos(f*x+e))
^4/cos(f*x+e)

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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))*(c-c*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [B]  time = 6.00, size = 384, normalized size = 3.15 \[ \frac {\sqrt {c-\frac {c}{\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {a\,c^2\,2{}\mathrm {i}}{f}+\frac {a\,c^2\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,142{}\mathrm {i}}{105\,f}\right )}{{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-1}+\frac {\sqrt {c-\frac {c}{\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {a\,c^2\,16{}\mathrm {i}}{7\,f}-\frac {a\,c^2\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,16{}\mathrm {i}}{7\,f}\right )}{\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-1\right )\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^3}-\frac {\sqrt {c-\frac {c}{\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {a\,c^2\,8{}\mathrm {i}}{5\,f}-\frac {a\,c^2\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,184{}\mathrm {i}}{35\,f}\right )}{\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-1\right )\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {\sqrt {c-\frac {c}{\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {a\,c^2\,4{}\mathrm {i}}{3\,f}+\frac {a\,c^2\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,244{}\mathrm {i}}{105\,f}\right )}{\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-1\right )\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + a/cos(e + f*x))*(c - c/cos(e + f*x))^(5/2))/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*c^2*2i)/f + (a*c^2*exp(e*1i + f*x*1i)*142i)
/(105*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*c^2*1
6i)/(7*f) - (a*c^2*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*c^2*8i)/(5*f) - (a*c^2*exp(e*1i + f*x*1i)*184i
)/(35*f)))/((exp(e*1i + f*x*1i) - 1)*(exp(e*2i + f*x*2i) + 1)^2) - ((c - c/(exp(- e*1i - f*x*1i)/2 + exp(e*1i
+ f*x*1i)/2))^(1/2)*((a*c^2*4i)/(3*f) + (a*c^2*exp(e*1i + f*x*1i)*244i)/(105*f)))/((exp(e*1i + f*x*1i) - 1)*(e
xp(e*2i + f*x*2i) + 1))

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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))*(c-c*sec(f*x+e))**(5/2),x)

[Out]

Timed out

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