3.43 \(\int \frac{\sec (e+f x) (c-c \sec (e+f x))^4}{(a+a \sec (e+f x))^2} \, dx\)

Optimal. Leaf size=150 \[ -\frac{70 c^4 \tan (e+f x)}{3 a^2 f}+\frac{35 c^4 \tanh ^{-1}(\sin (e+f x))}{2 a^2 f}+\frac{35 c^4 \tan (e+f x) \sec (e+f x)}{6 a^2 f}-\frac{14 \tan (e+f x) \left (c^2-c^2 \sec (e+f x)\right )^2}{3 f \left (a^2 \sec (e+f x)+a^2\right )}+\frac{2 c \tan (e+f x) (c-c \sec (e+f x))^3}{3 f (a \sec (e+f x)+a)^2} \]

[Out]

(35*c^4*ArcTanh[Sin[e + f*x]])/(2*a^2*f) - (70*c^4*Tan[e + f*x])/(3*a^2*f) + (35*c^4*Sec[e + f*x]*Tan[e + f*x]
)/(6*a^2*f) + (2*c*(c - c*Sec[e + f*x])^3*Tan[e + f*x])/(3*f*(a + a*Sec[e + f*x])^2) - (14*(c^2 - c^2*Sec[e +
f*x])^2*Tan[e + f*x])/(3*f*(a^2 + a^2*Sec[e + f*x]))

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Rubi [A]  time = 0.215091, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {3957, 3788, 3767, 8, 4046, 3770} \[ -\frac{70 c^4 \tan (e+f x)}{3 a^2 f}+\frac{35 c^4 \tanh ^{-1}(\sin (e+f x))}{2 a^2 f}+\frac{35 c^4 \tan (e+f x) \sec (e+f x)}{6 a^2 f}-\frac{14 \tan (e+f x) \left (c^2-c^2 \sec (e+f x)\right )^2}{3 f \left (a^2 \sec (e+f x)+a^2\right )}+\frac{2 c \tan (e+f x) (c-c \sec (e+f x))^3}{3 f (a \sec (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(c - c*Sec[e + f*x])^4)/(a + a*Sec[e + f*x])^2,x]

[Out]

(35*c^4*ArcTanh[Sin[e + f*x]])/(2*a^2*f) - (70*c^4*Tan[e + f*x])/(3*a^2*f) + (35*c^4*Sec[e + f*x]*Tan[e + f*x]
)/(6*a^2*f) + (2*c*(c - c*Sec[e + f*x])^3*Tan[e + f*x])/(3*f*(a + a*Sec[e + f*x])^2) - (14*(c^2 - c^2*Sec[e +
f*x])^2*Tan[e + f*x])/(3*f*(a^2 + a^2*Sec[e + f*x]))

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 3788

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[(2*a*b)/
d, Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b
, d, e, f, n}, 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 4046

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> -Simp[(C*Cot[
e + f*x]*(b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x]
/; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

Rule 3770

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

Rubi steps

\begin{align*} \int \frac{\sec (e+f x) (c-c \sec (e+f x))^4}{(a+a \sec (e+f x))^2} \, dx &=\frac{2 c (c-c \sec (e+f x))^3 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac{(7 c) \int \frac{\sec (e+f x) (c-c \sec (e+f x))^3}{a+a \sec (e+f x)} \, dx}{3 a}\\ &=\frac{2 c (c-c \sec (e+f x))^3 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac{14 \left (c^2-c^2 \sec (e+f x)\right )^2 \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}+\frac{\left (35 c^2\right ) \int \sec (e+f x) (c-c \sec (e+f x))^2 \, dx}{3 a^2}\\ &=\frac{2 c (c-c \sec (e+f x))^3 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac{14 \left (c^2-c^2 \sec (e+f x)\right )^2 \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}+\frac{\left (35 c^2\right ) \int \sec (e+f x) \left (c^2+c^2 \sec ^2(e+f x)\right ) \, dx}{3 a^2}-\frac{\left (70 c^4\right ) \int \sec ^2(e+f x) \, dx}{3 a^2}\\ &=\frac{35 c^4 \sec (e+f x) \tan (e+f x)}{6 a^2 f}+\frac{2 c (c-c \sec (e+f x))^3 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac{14 \left (c^2-c^2 \sec (e+f x)\right )^2 \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}+\frac{\left (35 c^4\right ) \int \sec (e+f x) \, dx}{2 a^2}+\frac{\left (70 c^4\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (e+f x))}{3 a^2 f}\\ &=\frac{35 c^4 \tanh ^{-1}(\sin (e+f x))}{2 a^2 f}-\frac{70 c^4 \tan (e+f x)}{3 a^2 f}+\frac{35 c^4 \sec (e+f x) \tan (e+f x)}{6 a^2 f}+\frac{2 c (c-c \sec (e+f x))^3 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac{14 \left (c^2-c^2 \sec (e+f x)\right )^2 \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}\\ \end{align*}

Mathematica [B]  time = 1.94691, size = 349, normalized size = 2.33 \[ \frac{c^4 \sin ^3\left (\frac{1}{2} (e+f x)\right ) \cos \left (\frac{1}{2} (e+f x)\right ) \sec ^2(e+f x) \left (-32 \tan \left (\frac{e}{2}\right ) \cot \left (\frac{1}{2} (e+f x)\right ) \csc ^2\left (\frac{1}{2} (e+f x)\right )-32 \sec \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) \csc ^3\left (\frac{1}{2} (e+f x)\right )+3 \cot ^3\left (\frac{1}{2} (e+f x)\right ) \left (-\frac{24 \sin (f x)}{\left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\sin \left (\frac{e}{2}\right )+\cos \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}+\frac{1}{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2}-\frac{1}{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2}-70 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+70 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )-256 \sec \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) \cot ^2\left (\frac{1}{2} (e+f x)\right ) \csc \left (\frac{1}{2} (e+f x)\right )\right )}{3 a^2 f (\sec (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(c - c*Sec[e + f*x])^4)/(a + a*Sec[e + f*x])^2,x]

[Out]

(c^4*Cos[(e + f*x)/2]*Sec[e + f*x]^2*Sin[(e + f*x)/2]^3*(-256*Cot[(e + f*x)/2]^2*Csc[(e + f*x)/2]*Sec[e/2]*Sin
[(f*x)/2] - 32*Csc[(e + f*x)/2]^3*Sec[e/2]*Sin[(f*x)/2] + 3*Cot[(e + f*x)/2]^3*(-70*Log[Cos[(e + f*x)/2] - Sin
[(e + f*x)/2]] + 70*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + (Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^(-2) - (C
os[(e + f*x)/2] + Sin[(e + f*x)/2])^(-2) - (24*Sin[f*x])/((Cos[e/2] - Sin[e/2])*(Cos[e/2] + Sin[e/2])*(Cos[(e
+ f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))) - 32*Cot[(e + f*x)/2]*Csc[(e + f*x)/2]^2
*Tan[e/2]))/(3*a^2*f*(1 + Sec[e + f*x])^2)

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Maple [A]  time = 0.092, size = 186, normalized size = 1.2 \begin{align*} -{\frac{8\,{c}^{4}}{3\,f{a}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}-24\,{\frac{{c}^{4}\tan \left ( 1/2\,fx+e/2 \right ) }{f{a}^{2}}}-{\frac{{c}^{4}}{2\,f{a}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-2}}+{\frac{13\,{c}^{4}}{2\,f{a}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-1}}+{\frac{35\,{c}^{4}}{2\,f{a}^{2}}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) }+{\frac{{c}^{4}}{2\,f{a}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-2}}+{\frac{13\,{c}^{4}}{2\,f{a}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-1}}-{\frac{35\,{c}^{4}}{2\,f{a}^{2}}\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))^2,x)

[Out]

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

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Maxima [B]  time = 1.02256, size = 717, normalized size = 4.78 \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))^2,x, algorithm="maxima")

[Out]

-1/6*(c^4*(6*(3*sin(f*x + e)/(cos(f*x + e) + 1) - 5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/(a^2 - 2*a^2*sin(f*x
+ e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^4/(cos(f*x + e) + 1)^4) + (21*sin(f*x + e)/(cos(f*x + e) + 1) +
sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2 - 21*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a^2 + 21*log(sin(f*x +
e)/(cos(f*x + e) + 1) - 1)/a^2) + 4*c^4*((15*sin(f*x + e)/(cos(f*x + e) + 1) + sin(f*x + e)^3/(cos(f*x + e) +
1)^3)/a^2 - 12*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a^2 + 12*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/a^2
+ 12*sin(f*x + e)/((a^2 - a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2)*(cos(f*x + e) + 1))) + 6*c^4*((9*sin(f*x +
e)/(cos(f*x + e) + 1) + sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2 - 6*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)
/a^2 + 6*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/a^2) + 4*c^4*(3*sin(f*x + e)/(cos(f*x + e) + 1) + sin(f*x +
e)^3/(cos(f*x + e) + 1)^3)/a^2 - c^4*(3*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^3/(cos(f*x + e) + 1)^3)
/a^2)/f

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Fricas [A]  time = 0.494155, size = 487, normalized size = 3.25 \begin{align*} \frac{105 \,{\left (c^{4} \cos \left (f x + e\right )^{4} + 2 \, c^{4} \cos \left (f x + e\right )^{3} + c^{4} \cos \left (f x + e\right )^{2}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 105 \,{\left (c^{4} \cos \left (f x + e\right )^{4} + 2 \, c^{4} \cos \left (f x + e\right )^{3} + c^{4} \cos \left (f x + e\right )^{2}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \,{\left (164 \, c^{4} \cos \left (f x + e\right )^{3} + 229 \, c^{4} \cos \left (f x + e\right )^{2} + 30 \, c^{4} \cos \left (f x + e\right ) - 3 \, c^{4}\right )} \sin \left (f x + e\right )}{12 \,{\left (a^{2} f \cos \left (f x + e\right )^{4} + 2 \, a^{2} f \cos \left (f x + e\right )^{3} + a^{2} f \cos \left (f x + e\right )^{2}\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))^2,x, algorithm="fricas")

[Out]

1/12*(105*(c^4*cos(f*x + e)^4 + 2*c^4*cos(f*x + e)^3 + c^4*cos(f*x + e)^2)*log(sin(f*x + e) + 1) - 105*(c^4*co
s(f*x + e)^4 + 2*c^4*cos(f*x + e)^3 + c^4*cos(f*x + e)^2)*log(-sin(f*x + e) + 1) - 2*(164*c^4*cos(f*x + e)^3 +
229*c^4*cos(f*x + e)^2 + 30*c^4*cos(f*x + e) - 3*c^4)*sin(f*x + e))/(a^2*f*cos(f*x + e)^4 + 2*a^2*f*cos(f*x +
e)^3 + a^2*f*cos(f*x + e)^2)

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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 ^{2}{\left (e + f x \right )} + 2 \sec{\left (e + f x \right )} + 1}\, dx + \int - \frac{4 \sec ^{2}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{6 \sec ^{3}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec{\left (e + f x \right )} + 1}\, dx + \int - \frac{4 \sec ^{4}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{\sec ^{5}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec{\left (e + f x \right )} + 1}\, dx\right )}{a^{2}} \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))**2,x)

[Out]

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

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Giac [A]  time = 1.27273, size = 198, normalized size = 1.32 \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^{2}} - \frac{105 \, c^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{a^{2}} + \frac{6 \,{\left (13 \, c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 11 \, 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 )}^{2} a^{2}} - \frac{16 \,{\left (a^{4} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 9 \, a^{4} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{a^{6}}}{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))^2,x, algorithm="giac")

[Out]

1/6*(105*c^4*log(abs(tan(1/2*f*x + 1/2*e) + 1))/a^2 - 105*c^4*log(abs(tan(1/2*f*x + 1/2*e) - 1))/a^2 + 6*(13*c
^4*tan(1/2*f*x + 1/2*e)^3 - 11*c^4*tan(1/2*f*x + 1/2*e))/((tan(1/2*f*x + 1/2*e)^2 - 1)^2*a^2) - 16*(a^4*c^4*ta
n(1/2*f*x + 1/2*e)^3 + 9*a^4*c^4*tan(1/2*f*x + 1/2*e))/a^6)/f