3.41 \(\int \frac{\sec (e+f x)}{(a+a \sec (e+f x)) (c-c \sec (e+f x))^4} \, dx\)

Optimal. Leaf size=120 \[ -\frac{4 \cot ^7(e+f x)}{7 a c^4 f}-\frac{\cot ^5(e+f x)}{5 a c^4 f}-\frac{4 \csc ^7(e+f x)}{7 a c^4 f}+\frac{9 \csc ^5(e+f x)}{5 a c^4 f}-\frac{2 \csc ^3(e+f x)}{a c^4 f}+\frac{\csc (e+f x)}{a c^4 f} \]

[Out]

-Cot[e + f*x]^5/(5*a*c^4*f) - (4*Cot[e + f*x]^7)/(7*a*c^4*f) + Csc[e + f*x]/(a*c^4*f) - (2*Csc[e + f*x]^3)/(a*
c^4*f) + (9*Csc[e + f*x]^5)/(5*a*c^4*f) - (4*Csc[e + f*x]^7)/(7*a*c^4*f)

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Rubi [A]  time = 0.229039, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 7, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.219, Rules used = {3958, 2606, 194, 2607, 30, 270, 14} \[ -\frac{4 \cot ^7(e+f x)}{7 a c^4 f}-\frac{\cot ^5(e+f x)}{5 a c^4 f}-\frac{4 \csc ^7(e+f x)}{7 a c^4 f}+\frac{9 \csc ^5(e+f x)}{5 a c^4 f}-\frac{2 \csc ^3(e+f x)}{a c^4 f}+\frac{\csc (e+f x)}{a c^4 f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]/((a + a*Sec[e + f*x])*(c - c*Sec[e + f*x])^4),x]

[Out]

-Cot[e + f*x]^5/(5*a*c^4*f) - (4*Cot[e + f*x]^7)/(7*a*c^4*f) + Csc[e + f*x]/(a*c^4*f) - (2*Csc[e + f*x]^3)/(a*
c^4*f) + (9*Csc[e + f*x]^5)/(5*a*c^4*f) - (4*Csc[e + f*x]^7)/(7*a*c^4*f)

Rule 3958

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)
)^(n_.), x_Symbol] :> Dist[(-(a*c))^m, Int[ExpandTrig[csc[e + f*x]*cot[e + f*x]^(2*m), (c + d*csc[e + f*x])^(n
- m), x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegersQ[m,
n] && GeQ[n - m, 0] && GtQ[m*n, 0]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \frac{\sec (e+f x)}{(a+a \sec (e+f x)) (c-c \sec (e+f x))^4} \, dx &=\frac{\int \left (a^3 \cot ^7(e+f x) \csc (e+f x)+3 a^3 \cot ^6(e+f x) \csc ^2(e+f x)+3 a^3 \cot ^5(e+f x) \csc ^3(e+f x)+a^3 \cot ^4(e+f x) \csc ^4(e+f x)\right ) \, dx}{a^4 c^4}\\ &=\frac{\int \cot ^7(e+f x) \csc (e+f x) \, dx}{a c^4}+\frac{\int \cot ^4(e+f x) \csc ^4(e+f x) \, dx}{a c^4}+\frac{3 \int \cot ^6(e+f x) \csc ^2(e+f x) \, dx}{a c^4}+\frac{3 \int \cot ^5(e+f x) \csc ^3(e+f x) \, dx}{a c^4}\\ &=-\frac{\operatorname{Subst}\left (\int \left (-1+x^2\right )^3 \, dx,x,\csc (e+f x)\right )}{a c^4 f}+\frac{\operatorname{Subst}\left (\int x^4 \left (1+x^2\right ) \, dx,x,-\cot (e+f x)\right )}{a c^4 f}+\frac{3 \operatorname{Subst}\left (\int x^6 \, dx,x,-\cot (e+f x)\right )}{a c^4 f}-\frac{3 \operatorname{Subst}\left (\int x^2 \left (-1+x^2\right )^2 \, dx,x,\csc (e+f x)\right )}{a c^4 f}\\ &=-\frac{3 \cot ^7(e+f x)}{7 a c^4 f}-\frac{\operatorname{Subst}\left (\int \left (-1+3 x^2-3 x^4+x^6\right ) \, dx,x,\csc (e+f x)\right )}{a c^4 f}+\frac{\operatorname{Subst}\left (\int \left (x^4+x^6\right ) \, dx,x,-\cot (e+f x)\right )}{a c^4 f}-\frac{3 \operatorname{Subst}\left (\int \left (x^2-2 x^4+x^6\right ) \, dx,x,\csc (e+f x)\right )}{a c^4 f}\\ &=-\frac{\cot ^5(e+f x)}{5 a c^4 f}-\frac{4 \cot ^7(e+f x)}{7 a c^4 f}+\frac{\csc (e+f x)}{a c^4 f}-\frac{2 \csc ^3(e+f x)}{a c^4 f}+\frac{9 \csc ^5(e+f x)}{5 a c^4 f}-\frac{4 \csc ^7(e+f x)}{7 a c^4 f}\\ \end{align*}

Mathematica [A]  time = 0.877835, size = 145, normalized size = 1.21 \[ \frac{\csc (e) (-1946 \sin (e+f x)+1946 \sin (2 (e+f x))-834 \sin (3 (e+f x))+139 \sin (4 (e+f x))-1400 \sin (2 e+f x)+616 \sin (e+2 f x)+840 \sin (3 e+2 f x)-344 \sin (2 e+3 f x)-280 \sin (4 e+3 f x)+104 \sin (3 e+4 f x)+840 \sin (e)-56 \sin (f x)) \csc ^6\left (\frac{1}{2} (e+f x)\right ) \csc (e+f x)}{17920 a c^4 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]/((a + a*Sec[e + f*x])*(c - c*Sec[e + f*x])^4),x]

[Out]

(Csc[e]*Csc[(e + f*x)/2]^6*Csc[e + f*x]*(840*Sin[e] - 56*Sin[f*x] - 1946*Sin[e + f*x] + 1946*Sin[2*(e + f*x)]
- 834*Sin[3*(e + f*x)] + 139*Sin[4*(e + f*x)] - 1400*Sin[2*e + f*x] + 616*Sin[e + 2*f*x] + 840*Sin[3*e + 2*f*x
] - 344*Sin[2*e + 3*f*x] - 280*Sin[4*e + 3*f*x] + 104*Sin[3*e + 4*f*x]))/(17920*a*c^4*f)

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Maple [A]  time = 0.059, size = 74, normalized size = 0.6 \begin{align*}{\frac{1}{16\,fa{c}^{4}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -2\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{-3}+4\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{-1}+{\frac{4}{5} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-5}}-{\frac{1}{7} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-7}} \right ) } \end{align*}

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))^4,x)

[Out]

1/16/f/a/c^4*(tan(1/2*f*x+1/2*e)-2/tan(1/2*f*x+1/2*e)^3+4/tan(1/2*f*x+1/2*e)+4/5/tan(1/2*f*x+1/2*e)^5-1/7/tan(
1/2*f*x+1/2*e)^7)

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Maxima [A]  time = 1.01181, size = 158, normalized size = 1.32 \begin{align*} \frac{\frac{{\left (\frac{28 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{70 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{140 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - 5\right )}{\left (\cos \left (f x + e\right ) + 1\right )}^{7}}{a c^{4} \sin \left (f x + e\right )^{7}} + \frac{35 \, \sin \left (f x + e\right )}{a c^{4}{\left (\cos \left (f x + e\right ) + 1\right )}}}{560 \, f} \end{align*}

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))^4,x, algorithm="maxima")

[Out]

1/560*((28*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 70*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 140*sin(f*x + e)^6/(
cos(f*x + e) + 1)^6 - 5)*(cos(f*x + e) + 1)^7/(a*c^4*sin(f*x + e)^7) + 35*sin(f*x + e)/(a*c^4*(cos(f*x + e) +
1)))/f

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Fricas [A]  time = 0.451931, size = 255, normalized size = 2.12 \begin{align*} \frac{13 \, \cos \left (f x + e\right )^{4} - 4 \, \cos \left (f x + e\right )^{3} - 20 \, \cos \left (f x + e\right )^{2} + 24 \, \cos \left (f x + e\right ) - 8}{35 \,{\left (a c^{4} f \cos \left (f x + e\right )^{3} - 3 \, a c^{4} f \cos \left (f x + e\right )^{2} + 3 \, a c^{4} f \cos \left (f x + e\right ) - a c^{4} f\right )} \sin \left (f x + e\right )} \end{align*}

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))^4,x, algorithm="fricas")

[Out]

1/35*(13*cos(f*x + e)^4 - 4*cos(f*x + e)^3 - 20*cos(f*x + e)^2 + 24*cos(f*x + e) - 8)/((a*c^4*f*cos(f*x + e)^3
- 3*a*c^4*f*cos(f*x + e)^2 + 3*a*c^4*f*cos(f*x + e) - a*c^4*f)*sin(f*x + e))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\sec{\left (e + f x \right )}}{\sec ^{5}{\left (e + f x \right )} - 3 \sec ^{4}{\left (e + f x \right )} + 2 \sec ^{3}{\left (e + f x \right )} + 2 \sec ^{2}{\left (e + f x \right )} - 3 \sec{\left (e + f x \right )} + 1}\, dx}{a c^{4}} \end{align*}

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))**4,x)

[Out]

Integral(sec(e + f*x)/(sec(e + f*x)**5 - 3*sec(e + f*x)**4 + 2*sec(e + f*x)**3 + 2*sec(e + f*x)**2 - 3*sec(e +
f*x) + 1), x)/(a*c**4)

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Giac [A]  time = 1.21872, size = 117, normalized size = 0.98 \begin{align*} \frac{\frac{35 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{a c^{4}} + \frac{140 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 70 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 28 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 5}{a c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7}}}{560 \, f} \end{align*}

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))^4,x, algorithm="giac")

[Out]

1/560*(35*tan(1/2*f*x + 1/2*e)/(a*c^4) + (140*tan(1/2*f*x + 1/2*e)^6 - 70*tan(1/2*f*x + 1/2*e)^4 + 28*tan(1/2*
f*x + 1/2*e)^2 - 5)/(a*c^4*tan(1/2*f*x + 1/2*e)^7))/f