3.27 \(\int \frac{\sec (e+f x) (a+a \sec (e+f x))^3}{c-c \sec (e+f x)} \, dx\)

Optimal. Leaf size=100 \[ -\frac{10 a^3 \tan (e+f x)}{c f}-\frac{15 a^3 \tanh ^{-1}(\sin (e+f x))}{2 c f}-\frac{5 a^3 \tan (e+f x) \sec (e+f x)}{2 c f}-\frac{2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{f (c-c \sec (e+f x))} \]

[Out]

(-15*a^3*ArcTanh[Sin[e + f*x]])/(2*c*f) - (10*a^3*Tan[e + f*x])/(c*f) - (5*a^3*Sec[e + f*x]*Tan[e + f*x])/(2*c
*f) - (2*a*(a + a*Sec[e + f*x])^2*Tan[e + f*x])/(f*(c - c*Sec[e + f*x]))

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Rubi [A]  time = 0.128658, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ -\frac{10 a^3 \tan (e+f x)}{c f}-\frac{15 a^3 \tanh ^{-1}(\sin (e+f x))}{2 c f}-\frac{5 a^3 \tan (e+f x) \sec (e+f x)}{2 c f}-\frac{2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{f (c-c \sec (e+f x))} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(a + a*Sec[e + f*x])^3)/(c - c*Sec[e + f*x]),x]

[Out]

(-15*a^3*ArcTanh[Sin[e + f*x]])/(2*c*f) - (10*a^3*Tan[e + f*x])/(c*f) - (5*a^3*Sec[e + f*x]*Tan[e + f*x])/(2*c
*f) - (2*a*(a + a*Sec[e + f*x])^2*Tan[e + f*x])/(f*(c - c*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) (a+a \sec (e+f x))^3}{c-c \sec (e+f x)} \, dx &=-\frac{2 a (a+a \sec (e+f x))^2 \tan (e+f x)}{f (c-c \sec (e+f x))}-\frac{(5 a) \int \sec (e+f x) (a+a \sec (e+f x))^2 \, dx}{c}\\ &=-\frac{2 a (a+a \sec (e+f x))^2 \tan (e+f x)}{f (c-c \sec (e+f x))}-\frac{(5 a) \int \sec (e+f x) \left (a^2+a^2 \sec ^2(e+f x)\right ) \, dx}{c}-\frac{\left (10 a^3\right ) \int \sec ^2(e+f x) \, dx}{c}\\ &=-\frac{5 a^3 \sec (e+f x) \tan (e+f x)}{2 c f}-\frac{2 a (a+a \sec (e+f x))^2 \tan (e+f x)}{f (c-c \sec (e+f x))}-\frac{\left (15 a^3\right ) \int \sec (e+f x) \, dx}{2 c}+\frac{\left (10 a^3\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (e+f x))}{c f}\\ &=-\frac{15 a^3 \tanh ^{-1}(\sin (e+f x))}{2 c f}-\frac{10 a^3 \tan (e+f x)}{c f}-\frac{5 a^3 \sec (e+f x) \tan (e+f x)}{2 c f}-\frac{2 a (a+a \sec (e+f x))^2 \tan (e+f x)}{f (c-c \sec (e+f x))}\\ \end{align*}

Mathematica [B]  time = 3.04039, size = 287, normalized size = 2.87 \[ \frac{a^3 \cos ^2(e+f x) \tan \left (\frac{1}{2} (e+f x)\right ) \sec ^4\left (\frac{1}{2} (e+f x)\right ) (\sec (e+f x)+1)^3 \left (32 \csc \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) \sec \left (\frac{1}{2} (e+f x)\right )+\tan \left (\frac{1}{2} (e+f x)\right ) \left (\frac{16 \sin (f x)}{\left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\sin \left (\frac{e}{2}\right )+\cos \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}+\frac{1}{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2}-\frac{1}{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2}-30 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+30 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )\right )}{16 f (c-c \sec (e+f x))} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(a + a*Sec[e + f*x])^3)/(c - c*Sec[e + f*x]),x]

[Out]

(a^3*Cos[e + f*x]^2*Sec[(e + f*x)/2]^4*(1 + Sec[e + f*x])^3*Tan[(e + f*x)/2]*(32*Csc[e/2]*Sec[(e + f*x)/2]*Sin
[(f*x)/2] + (-30*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + 30*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + (Cos
[(e + f*x)/2] - Sin[(e + f*x)/2])^(-2) - (Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^(-2) + (16*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])
))*Tan[(e + f*x)/2]))/(16*f*(c - c*Sec[e + f*x]))

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Maple [A]  time = 0.076, size = 166, normalized size = 1.7 \begin{align*}{\frac{{a}^{3}}{2\,fc} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-2}}+{\frac{7\,{a}^{3}}{2\,fc} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-1}}-{\frac{15\,{a}^{3}}{2\,fc}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) }-{\frac{{a}^{3}}{2\,fc} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-2}}+{\frac{7\,{a}^{3}}{2\,fc} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-1}}+{\frac{15\,{a}^{3}}{2\,fc}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) }+8\,{\frac{{a}^{3}}{fc\tan \left ( 1/2\,fx+e/2 \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))^3/(c-c*sec(f*x+e)),x)

[Out]

1/2/f*a^3/c/(tan(1/2*f*x+1/2*e)+1)^2+7/2/f*a^3/c/(tan(1/2*f*x+1/2*e)+1)-15/2/f*a^3/c*ln(tan(1/2*f*x+1/2*e)+1)-
1/2/f*a^3/c/(tan(1/2*f*x+1/2*e)-1)^2+7/2/f*a^3/c/(tan(1/2*f*x+1/2*e)-1)+15/2/f*a^3/c*ln(tan(1/2*f*x+1/2*e)-1)+
8/f*a^3/c/tan(1/2*f*x+1/2*e)

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Maxima [B]  time = 1.01378, size = 522, normalized size = 5.22 \begin{align*} -\frac{a^{3}{\left (\frac{2 \,{\left (\frac{5 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{2 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 1\right )}}{\frac{c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{2 \, c \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{c \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} + \frac{3 \, \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c} - \frac{3 \, \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c}\right )} + 6 \, a^{3}{\left (\frac{\frac{3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 1}{\frac{c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{c \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c} - \frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c}\right )} + 6 \, a^{3}{\left (\frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c} - \frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c} - \frac{\cos \left (f x + e\right ) + 1}{c \sin \left (f x + e\right )}\right )} - \frac{2 \, a^{3}{\left (\cos \left (f x + e\right ) + 1\right )}}{c \sin \left (f x + e\right )}}{2 \, 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))^3/(c-c*sec(f*x+e)),x, algorithm="maxima")

[Out]

-1/2*(a^3*(2*(5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 2*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 1)/(c*sin(f*x +
e)/(cos(f*x + e) + 1) - 2*c*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + c*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 3*l
og(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/c - 3*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/c) + 6*a^3*((3*sin(f*x
+ e)^2/(cos(f*x + e) + 1)^2 - 1)/(c*sin(f*x + e)/(cos(f*x + e) + 1) - c*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) +
log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/c - log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/c) + 6*a^3*(log(sin(f*x
+ e)/(cos(f*x + e) + 1) + 1)/c - log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/c - (cos(f*x + e) + 1)/(c*sin(f*x +
e))) - 2*a^3*(cos(f*x + e) + 1)/(c*sin(f*x + e)))/f

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Fricas [A]  time = 0.484043, size = 320, normalized size = 3.2 \begin{align*} -\frac{15 \, a^{3} \cos \left (f x + e\right )^{2} \log \left (\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) - 15 \, a^{3} \cos \left (f x + e\right )^{2} \log \left (-\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) - 48 \, a^{3} \cos \left (f x + e\right )^{3} - 34 \, a^{3} \cos \left (f x + e\right )^{2} + 16 \, a^{3} \cos \left (f x + e\right ) + 2 \, a^{3}}{4 \, c f \cos \left (f x + e\right )^{2} \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))^3/(c-c*sec(f*x+e)),x, algorithm="fricas")

[Out]

-1/4*(15*a^3*cos(f*x + e)^2*log(sin(f*x + e) + 1)*sin(f*x + e) - 15*a^3*cos(f*x + e)^2*log(-sin(f*x + e) + 1)*
sin(f*x + e) - 48*a^3*cos(f*x + e)^3 - 34*a^3*cos(f*x + e)^2 + 16*a^3*cos(f*x + e) + 2*a^3)/(c*f*cos(f*x + e)^
2*sin(f*x + e))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))**3/(c-c*sec(f*x+e)),x)

[Out]

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

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Giac [A]  time = 1.27316, size = 167, normalized size = 1.67 \begin{align*} -\frac{\frac{15 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{c} - \frac{15 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{c} - \frac{16 \, a^{3}}{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )} - \frac{2 \,{\left (7 \, a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 9 \, a^{3} \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} c}}{2 \, 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))^3/(c-c*sec(f*x+e)),x, algorithm="giac")

[Out]

-1/2*(15*a^3*log(abs(tan(1/2*f*x + 1/2*e) + 1))/c - 15*a^3*log(abs(tan(1/2*f*x + 1/2*e) - 1))/c - 16*a^3/(c*ta
n(1/2*f*x + 1/2*e)) - 2*(7*a^3*tan(1/2*f*x + 1/2*e)^3 - 9*a^3*tan(1/2*f*x + 1/2*e))/((tan(1/2*f*x + 1/2*e)^2 -
1)^2*c))/f