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

Optimal. Leaf size=121 \[ \frac{7 c^4 \tan ^3(e+f x)}{3 a f}+\frac{28 c^4 \tan (e+f x)}{a f}-\frac{35 c^4 \tanh ^{-1}(\sin (e+f x))}{2 a f}-\frac{21 c^4 \tan (e+f x) \sec (e+f x)}{2 a f}+\frac{2 c \tan (e+f x) (c-c \sec (e+f x))^3}{f (a \sec (e+f x)+a)} \]

[Out]

(-35*c^4*ArcTanh[Sin[e + f*x]])/(2*a*f) + (28*c^4*Tan[e + f*x])/(a*f) - (21*c^4*Sec[e + f*x]*Tan[e + f*x])/(2*
a*f) + (2*c*(c - c*Sec[e + f*x])^3*Tan[e + f*x])/(f*(a + a*Sec[e + f*x])) + (7*c^4*Tan[e + f*x]^3)/(3*a*f)

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Rubi [A]  time = 0.155514, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {3957, 3791, 3770, 3767, 8, 3768} \[ \frac{7 c^4 \tan ^3(e+f x)}{3 a f}+\frac{28 c^4 \tan (e+f x)}{a f}-\frac{35 c^4 \tanh ^{-1}(\sin (e+f x))}{2 a f}-\frac{21 c^4 \tan (e+f x) \sec (e+f x)}{2 a f}+\frac{2 c \tan (e+f x) (c-c \sec (e+f x))^3}{f (a \sec (e+f x)+a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-35*c^4*ArcTanh[Sin[e + f*x]])/(2*a*f) + (28*c^4*Tan[e + f*x])/(a*f) - (21*c^4*Sec[e + f*x]*Tan[e + f*x])/(2*
a*f) + (2*c*(c - c*Sec[e + f*x])^3*Tan[e + f*x])/(f*(a + a*Sec[e + f*x])) + (7*c^4*Tan[e + f*x]^3)/(3*a*f)

Rule 3957

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

Rule 3791

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[Expand
Trig[(a + b*csc[e + f*x])^m*(d*csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[m, 0] && RationalQ[n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rubi steps

\begin{align*} \int \frac{\sec (e+f x) (c-c \sec (e+f x))^4}{a+a \sec (e+f x)} \, dx &=\frac{2 c (c-c \sec (e+f x))^3 \tan (e+f x)}{f (a+a \sec (e+f x))}-\frac{(7 c) \int \sec (e+f x) (c-c \sec (e+f x))^3 \, dx}{a}\\ &=\frac{2 c (c-c \sec (e+f x))^3 \tan (e+f x)}{f (a+a \sec (e+f x))}-\frac{(7 c) \int \left (c^3 \sec (e+f x)-3 c^3 \sec ^2(e+f x)+3 c^3 \sec ^3(e+f x)-c^3 \sec ^4(e+f x)\right ) \, dx}{a}\\ &=\frac{2 c (c-c \sec (e+f x))^3 \tan (e+f x)}{f (a+a \sec (e+f x))}-\frac{\left (7 c^4\right ) \int \sec (e+f x) \, dx}{a}+\frac{\left (7 c^4\right ) \int \sec ^4(e+f x) \, dx}{a}+\frac{\left (21 c^4\right ) \int \sec ^2(e+f x) \, dx}{a}-\frac{\left (21 c^4\right ) \int \sec ^3(e+f x) \, dx}{a}\\ &=-\frac{7 c^4 \tanh ^{-1}(\sin (e+f x))}{a f}-\frac{21 c^4 \sec (e+f x) \tan (e+f x)}{2 a f}+\frac{2 c (c-c \sec (e+f x))^3 \tan (e+f x)}{f (a+a \sec (e+f x))}-\frac{\left (21 c^4\right ) \int \sec (e+f x) \, dx}{2 a}-\frac{\left (7 c^4\right ) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (e+f x)\right )}{a f}-\frac{\left (21 c^4\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (e+f x))}{a f}\\ &=-\frac{35 c^4 \tanh ^{-1}(\sin (e+f x))}{2 a f}+\frac{28 c^4 \tan (e+f x)}{a f}-\frac{21 c^4 \sec (e+f x) \tan (e+f x)}{2 a f}+\frac{2 c (c-c \sec (e+f x))^3 \tan (e+f x)}{f (a+a \sec (e+f x))}+\frac{7 c^4 \tan ^3(e+f x)}{3 a f}\\ \end{align*}

Mathematica [B]  time = 6.4092, size = 1036, normalized size = 8.56 \[ \text{result too large to display} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(35*Cos[e + f*x]^3*Cot[e/2 + (f*x)/2]^2*Csc[e/2 + (f*x)/2]^6*Log[Cos[e/2 + (f*x)/2] - Sin[e/2 + (f*x)/2]]*(c -
c*Sec[e + f*x])^4)/(16*f*(a + a*Sec[e + f*x])) - (35*Cos[e + f*x]^3*Cot[e/2 + (f*x)/2]^2*Csc[e/2 + (f*x)/2]^6
*Log[Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2]]*(c - c*Sec[e + f*x])^4)/(16*f*(a + a*Sec[e + f*x])) + (2*Cos[e +
f*x]^3*Cot[e/2 + (f*x)/2]*Csc[e/2 + (f*x)/2]^7*Sec[e/2]*(c - c*Sec[e + f*x])^4*Sin[(f*x)/2])/(f*(a + a*Sec[e
+ f*x])) + (Cos[e + f*x]^3*Cot[e/2 + (f*x)/2]^2*Csc[e/2 + (f*x)/2]^6*(c - c*Sec[e + f*x])^4*Sin[(f*x)/2])/(48*
f*(a + a*Sec[e + f*x])*(Cos[e/2] - Sin[e/2])*(Cos[e/2 + (f*x)/2] - Sin[e/2 + (f*x)/2])^3) + (Cos[e + f*x]^3*Co
t[e/2 + (f*x)/2]^2*Csc[e/2 + (f*x)/2]^6*(c - c*Sec[e + f*x])^4*(-7*Cos[e/2] + 8*Sin[e/2]))/(48*f*(a + a*Sec[e
+ f*x])*(Cos[e/2] - Sin[e/2])*(Cos[e/2 + (f*x)/2] - Sin[e/2 + (f*x)/2])^2) + (35*Cos[e + f*x]^3*Cot[e/2 + (f*x
)/2]^2*Csc[e/2 + (f*x)/2]^6*(c - c*Sec[e + f*x])^4*Sin[(f*x)/2])/(24*f*(a + a*Sec[e + f*x])*(Cos[e/2] - Sin[e/
2])*(Cos[e/2 + (f*x)/2] - Sin[e/2 + (f*x)/2])) + (Cos[e + f*x]^3*Cot[e/2 + (f*x)/2]^2*Csc[e/2 + (f*x)/2]^6*(c
- c*Sec[e + f*x])^4*Sin[(f*x)/2])/(48*f*(a + a*Sec[e + f*x])*(Cos[e/2] + Sin[e/2])*(Cos[e/2 + (f*x)/2] + Sin[e
/2 + (f*x)/2])^3) + (Cos[e + f*x]^3*Cot[e/2 + (f*x)/2]^2*Csc[e/2 + (f*x)/2]^6*(c - c*Sec[e + f*x])^4*(7*Cos[e/
2] + 8*Sin[e/2]))/(48*f*(a + a*Sec[e + f*x])*(Cos[e/2] + Sin[e/2])*(Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2])^2
) + (35*Cos[e + f*x]^3*Cot[e/2 + (f*x)/2]^2*Csc[e/2 + (f*x)/2]^6*(c - c*Sec[e + f*x])^4*Sin[(f*x)/2])/(24*f*(a
+ a*Sec[e + f*x])*(Cos[e/2] + Sin[e/2])*(Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2]))

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Maple [A]  time = 0.086, size = 212, normalized size = 1.8 \begin{align*} 16\,{\frac{{c}^{4}\tan \left ( 1/2\,fx+e/2 \right ) }{fa}}-{\frac{{c}^{4}}{3\,fa} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-3}}+3\,{\frac{{c}^{4}}{fa \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}}-{\frac{29\,{c}^{4}}{2\,fa} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-1}}-{\frac{35\,{c}^{4}}{2\,fa}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) }-{\frac{{c}^{4}}{3\,fa} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-3}}-3\,{\frac{{c}^{4}}{fa \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{2}}}-{\frac{29\,{c}^{4}}{2\,fa} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-1}}+{\frac{35\,{c}^{4}}{2\,fa}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(c-c*sec(f*x+e))^4/(a+a*sec(f*x+e)),x)

[Out]

16/f*c^4/a*tan(1/2*f*x+1/2*e)-1/3/f*c^4/a/(tan(1/2*f*x+1/2*e)+1)^3+3/f*c^4/a/(tan(1/2*f*x+1/2*e)+1)^2-29/2/f*c
^4/a/(tan(1/2*f*x+1/2*e)+1)-35/2/f*c^4/a*ln(tan(1/2*f*x+1/2*e)+1)-1/3/f*c^4/a/(tan(1/2*f*x+1/2*e)-1)^3-3/f*c^4
/a/(tan(1/2*f*x+1/2*e)-1)^2-29/2/f*c^4/a/(tan(1/2*f*x+1/2*e)-1)+35/2/f*c^4/a*ln(tan(1/2*f*x+1/2*e)-1)

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Maxima [B]  time = 1.05054, size = 798, normalized size = 6.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))^4/(a+a*sec(f*x+e)),x, algorithm="maxima")

[Out]

1/6*(c^4*(2*(9*sin(f*x + e)/(cos(f*x + e) + 1) - 16*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 15*sin(f*x + e)^5/(c
os(f*x + e) + 1)^5)/(a - 3*a*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 3*a*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - a
*sin(f*x + e)^6/(cos(f*x + e) + 1)^6) - 9*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a + 9*log(sin(f*x + e)/(cos
(f*x + e) + 1) - 1)/a + 6*sin(f*x + e)/(a*(cos(f*x + e) + 1))) + 12*c^4*(2*(sin(f*x + e)/(cos(f*x + e) + 1) -
3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/(a - 2*a*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a*sin(f*x + e)^4/(cos(f*
x + e) + 1)^4) - 3*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a + 3*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/a +
2*sin(f*x + e)/(a*(cos(f*x + e) + 1))) - 36*c^4*(log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a - log(sin(f*x + e
)/(cos(f*x + e) + 1) - 1)/a - 2*sin(f*x + e)/((a - a*sin(f*x + e)^2/(cos(f*x + e) + 1)^2)*(cos(f*x + e) + 1))
- sin(f*x + e)/(a*(cos(f*x + e) + 1))) - 24*c^4*(log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a - log(sin(f*x + e)
/(cos(f*x + e) + 1) - 1)/a - sin(f*x + e)/(a*(cos(f*x + e) + 1))) + 6*c^4*sin(f*x + e)/(a*(cos(f*x + e) + 1)))
/f

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Fricas [A]  time = 0.492197, size = 386, normalized size = 3.19 \begin{align*} -\frac{105 \,{\left (c^{4} \cos \left (f x + e\right )^{4} + c^{4} \cos \left (f x + e\right )^{3}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 105 \,{\left (c^{4} \cos \left (f x + e\right )^{4} + c^{4} \cos \left (f x + e\right )^{3}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \,{\left (166 \, c^{4} \cos \left (f x + e\right )^{3} + 55 \, c^{4} \cos \left (f x + e\right )^{2} - 13 \, c^{4} \cos \left (f x + e\right ) + 2 \, c^{4}\right )} \sin \left (f x + e\right )}{12 \,{\left (a f \cos \left (f x + e\right )^{4} + a f \cos \left (f x + e\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))^4/(a+a*sec(f*x+e)),x, algorithm="fricas")

[Out]

-1/12*(105*(c^4*cos(f*x + e)^4 + c^4*cos(f*x + e)^3)*log(sin(f*x + e) + 1) - 105*(c^4*cos(f*x + e)^4 + c^4*cos
(f*x + e)^3)*log(-sin(f*x + e) + 1) - 2*(166*c^4*cos(f*x + e)^3 + 55*c^4*cos(f*x + e)^2 - 13*c^4*cos(f*x + e)
+ 2*c^4)*sin(f*x + e))/(a*f*cos(f*x + e)^4 + a*f*cos(f*x + e)^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{c^{4} \left (\int \frac{\sec{\left (e + f x \right )}}{\sec{\left (e + f x \right )} + 1}\, dx + \int - \frac{4 \sec ^{2}{\left (e + f x \right )}}{\sec{\left (e + f x \right )} + 1}\, dx + \int \frac{6 \sec ^{3}{\left (e + f x \right )}}{\sec{\left (e + f x \right )} + 1}\, dx + \int - \frac{4 \sec ^{4}{\left (e + f x \right )}}{\sec{\left (e + f x \right )} + 1}\, dx + \int \frac{\sec ^{5}{\left (e + f x \right )}}{\sec{\left (e + f x \right )} + 1}\, dx\right )}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))**4/(a+a*sec(f*x+e)),x)

[Out]

c**4*(Integral(sec(e + f*x)/(sec(e + f*x) + 1), x) + Integral(-4*sec(e + f*x)**2/(sec(e + f*x) + 1), x) + Inte
gral(6*sec(e + f*x)**3/(sec(e + f*x) + 1), x) + Integral(-4*sec(e + f*x)**4/(sec(e + f*x) + 1), x) + Integral(
sec(e + f*x)**5/(sec(e + f*x) + 1), x))/a

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Giac [A]  time = 1.32156, size = 188, normalized size = 1.55 \begin{align*} -\frac{\frac{105 \, c^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{a} - \frac{105 \, c^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{a} - \frac{96 \, c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{a} + \frac{2 \,{\left (87 \, c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 136 \, c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 57 \, c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )}^{3} a}}{6 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))^4/(a+a*sec(f*x+e)),x, algorithm="giac")

[Out]

-1/6*(105*c^4*log(abs(tan(1/2*f*x + 1/2*e) + 1))/a - 105*c^4*log(abs(tan(1/2*f*x + 1/2*e) - 1))/a - 96*c^4*tan
(1/2*f*x + 1/2*e)/a + 2*(87*c^4*tan(1/2*f*x + 1/2*e)^5 - 136*c^4*tan(1/2*f*x + 1/2*e)^3 + 57*c^4*tan(1/2*f*x +
1/2*e))/((tan(1/2*f*x + 1/2*e)^2 - 1)^3*a))/f