3.1.29 \(\int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^3} \, dx\) [29]

Optimal. Leaf size=132 \[ -\frac {a^3 \tanh ^{-1}(\sin (e+f x))}{c^3 f}-\frac {2 a (a+a \sec (e+f x))^2 \tan (e+f x)}{5 f (c-c \sec (e+f x))^3}+\frac {2 \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{3 c f (c-c \sec (e+f x))^2}-\frac {2 a^3 \tan (e+f x)}{f \left (c^3-c^3 \sec (e+f x)\right )} \]

[Out]

-a^3*arctanh(sin(f*x+e))/c^3/f-2/5*a*(a+a*sec(f*x+e))^2*tan(f*x+e)/f/(c-c*sec(f*x+e))^3+2/3*(a^3+a^3*sec(f*x+e
))*tan(f*x+e)/c/f/(c-c*sec(f*x+e))^2-2*a^3*tan(f*x+e)/f/(c^3-c^3*sec(f*x+e))

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Rubi [A]
time = 0.15, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {4042, 3855} \begin {gather*} -\frac {a^3 \tanh ^{-1}(\sin (e+f x))}{c^3 f}-\frac {2 a^3 \tan (e+f x)}{f \left (c^3-c^3 \sec (e+f x)\right )}+\frac {2 \tan (e+f x) \left (a^3 \sec (e+f x)+a^3\right )}{3 c f (c-c \sec (e+f x))^2}-\frac {2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{5 f (c-c \sec (e+f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3855

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

Rule 4042

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]

Rubi steps

\begin {align*} \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^3} \, dx &=-\frac {2 a (a+a \sec (e+f x))^2 \tan (e+f x)}{5 f (c-c \sec (e+f x))^3}-\frac {a \int \frac {\sec (e+f x) (a+a \sec (e+f x))^2}{(c-c \sec (e+f x))^2} \, dx}{c}\\ &=-\frac {2 a (a+a \sec (e+f x))^2 \tan (e+f x)}{5 f (c-c \sec (e+f x))^3}+\frac {2 \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{3 c f (c-c \sec (e+f x))^2}+\frac {a^2 \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{c-c \sec (e+f x)} \, dx}{c^2}\\ &=-\frac {2 a (a+a \sec (e+f x))^2 \tan (e+f x)}{5 f (c-c \sec (e+f x))^3}+\frac {2 \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{3 c f (c-c \sec (e+f x))^2}-\frac {2 a^3 \tan (e+f x)}{f \left (c^3-c^3 \sec (e+f x)\right )}-\frac {a^3 \int \sec (e+f x) \, dx}{c^3}\\ &=-\frac {a^3 \tanh ^{-1}(\sin (e+f x))}{c^3 f}-\frac {2 a (a+a \sec (e+f x))^2 \tan (e+f x)}{5 f (c-c \sec (e+f x))^3}+\frac {2 \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{3 c f (c-c \sec (e+f x))^2}-\frac {2 a^3 \tan (e+f x)}{f \left (c^3-c^3 \sec (e+f x)\right )}\\ \end {align*}

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Mathematica [A]
time = 0.13, size = 139, normalized size = 1.05 \begin {gather*} -\frac {a^3 \left (-\frac {26 \cot \left (\frac {1}{2} (e+f x)\right )}{15 f}+\frac {2 \cot \left (\frac {1}{2} (e+f x)\right ) \csc ^2\left (\frac {1}{2} (e+f x)\right )}{15 f}-\frac {2 \cot \left (\frac {1}{2} (e+f x)\right ) \csc ^4\left (\frac {1}{2} (e+f x)\right )}{5 f}-\frac {\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )}{f}+\frac {\log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )}{f}\right )}{c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^3*((-26*Cot[(e + f*x)/2])/(15*f) + (2*Cot[(e + f*x)/2]*Csc[(e + f*x)/2]^2)/(15*f) - (2*Cot[(e + f*x)/2]*C
sc[(e + f*x)/2]^4)/(5*f) - Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]]/f + Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2
]]/f))/c^3)

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Maple [A]
time = 0.21, size = 78, normalized size = 0.59

method	result	size
derivativedivides	\(\frac {2 a^{3} \left (\frac {\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2}+\frac {1}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}+\frac {1}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}+\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}-\frac {\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{2}\right )}{f \,c^{3}}\)	\(78\)
default	\(\frac {2 a^{3} \left (\frac {\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2}+\frac {1}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}+\frac {1}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}+\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}-\frac {\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{2}\right )}{f \,c^{3}}\)	\(78\)
risch	\(\frac {4 i a^{3} \left (15 \,{\mathrm e}^{4 i \left (f x +e \right )}-30 \,{\mathrm e}^{3 i \left (f x +e \right )}+100 \,{\mathrm e}^{2 i \left (f x +e \right )}-50 \,{\mathrm e}^{i \left (f x +e \right )}+13\right )}{15 f \,c^{3} \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{5}}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{c^{3} f}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{c^{3} f}\)	\(120\)
norman	\(\frac {-\frac {2 a^{3}}{5 c f}+\frac {8 a^{3} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15 c f}-\frac {6 a^{3} \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 c f}+\frac {22 a^{3} \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 c f}-\frac {16 a^{3} \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 c f}+\frac {2 a^{3} \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3} c^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}+\frac {a^{3} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{c^{3} f}-\frac {a^{3} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{c^{3} f}\)	\(199\)

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))^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 335 vs. \(2 (140) = 280\).
time = 0.31, size = 335, normalized size = 2.54 \begin {gather*} -\frac {a^{3} {\left (\frac {60 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c^{3}} - \frac {60 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c^{3}} - \frac {{\left (\frac {20 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {105 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 3\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{c^{3} \sin \left (f x + e\right )^{5}}\right )} - \frac {3 \, a^{3} {\left (\frac {10 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 3\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{c^{3} \sin \left (f x + e\right )^{5}} + \frac {a^{3} {\left (\frac {10 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 3\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{c^{3} \sin \left (f x + e\right )^{5}} + \frac {9 \, a^{3} {\left (\frac {5 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 1\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{c^{3} \sin \left (f x + e\right )^{5}}}{60 \, f} \end {gather*}

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))^3,x, algorithm="maxima")

[Out]

-1/60*(a^3*(60*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/c^3 - 60*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/c^3
- (20*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 105*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 3)*(cos(f*x + e) + 1)^5/
(c^3*sin(f*x + e)^5)) - 3*a^3*(10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4
+ 3)*(cos(f*x + e) + 1)^5/(c^3*sin(f*x + e)^5) + a^3*(10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 15*sin(f*x + e
)^4/(cos(f*x + e) + 1)^4 - 3)*(cos(f*x + e) + 1)^5/(c^3*sin(f*x + e)^5) + 9*a^3*(5*sin(f*x + e)^4/(cos(f*x + e
) + 1)^4 - 1)*(cos(f*x + e) + 1)^5/(c^3*sin(f*x + e)^5))/f

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Fricas [A]
time = 2.34, size = 190, normalized size = 1.44 \begin {gather*} \frac {52 \, a^{3} \cos \left (f x + e\right )^{3} - 44 \, a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3} \cos \left (f x + e\right ) + 92 \, a^{3} - 15 \, {\left (a^{3} \cos \left (f x + e\right )^{2} - 2 \, a^{3} \cos \left (f x + e\right ) + a^{3}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) + 15 \, {\left (a^{3} \cos \left (f x + e\right )^{2} - 2 \, a^{3} \cos \left (f x + e\right ) + a^{3}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right )}{30 \, {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) + c^{3} f\right )} \sin \left (f x + e\right )} \end {gather*}

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))^3,x, algorithm="fricas")

[Out]

1/30*(52*a^3*cos(f*x + e)^3 - 44*a^3*cos(f*x + e)^2 - 4*a^3*cos(f*x + e) + 92*a^3 - 15*(a^3*cos(f*x + e)^2 - 2
*a^3*cos(f*x + e) + a^3)*log(sin(f*x + e) + 1)*sin(f*x + e) + 15*(a^3*cos(f*x + e)^2 - 2*a^3*cos(f*x + e) + a^
3)*log(-sin(f*x + e) + 1)*sin(f*x + e))/((c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) + c^3*f)*sin(f*x + e))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {a^{3} \left (\int \frac {\sec {\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} - 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} - 1}\, dx + \int \frac {3 \sec ^{2}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} - 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} - 1}\, dx + \int \frac {3 \sec ^{3}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} - 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} - 1}\, dx + \int \frac {\sec ^{4}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} - 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} - 1}\, dx\right )}{c^{3}} \end {gather*}

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))**3,x)

[Out]

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

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Giac [A]
time = 0.52, size = 102, normalized size = 0.77 \begin {gather*} -\frac {\frac {15 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right )}{c^{3}} - \frac {15 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right )}{c^{3}} - \frac {2 \, {\left (15 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 5 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, a^{3}\right )}}{c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5}}}{15 \, f} \end {gather*}

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))^3,x, algorithm="giac")

[Out]

-1/15*(15*a^3*log(abs(tan(1/2*f*x + 1/2*e) + 1))/c^3 - 15*a^3*log(abs(tan(1/2*f*x + 1/2*e) - 1))/c^3 - 2*(15*a
^3*tan(1/2*f*x + 1/2*e)^4 + 5*a^3*tan(1/2*f*x + 1/2*e)^2 + 3*a^3)/(c^3*tan(1/2*f*x + 1/2*e)^5))/f

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Mupad [B]
time = 1.78, size = 78, normalized size = 0.59 \begin {gather*} \frac {2\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+\frac {2\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2}{3}+\frac {2\,a^3}{5}}{c^3\,f\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}-\frac {2\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{c^3\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(e + f*x))^3/(cos(e + f*x)*(c - c/cos(e + f*x))^3),x)

[Out]

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

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