3.1.55 \(\int \frac {\sec (e+f x) (c-c \sec (e+f x))^3}{(a+a \sec (e+f x))^3} \, dx\) [55]

Optimal. Leaf size=131 \[ -\frac {c^3 \tanh ^{-1}(\sin (e+f x))}{a^3 f}+\frac {2 c^3 \tan (e+f x)}{f \left (a^3+a^3 \sec (e+f x)\right )}+\frac {2 c (c-c \sec (e+f x))^2 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}-\frac {2 \left (c^3-c^3 \sec (e+f x)\right ) \tan (e+f x)}{3 a f (a+a \sec (e+f x))^2} \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 131, 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 {c^3 \tanh ^{-1}(\sin (e+f x))}{a^3 f}+\frac {2 c^3 \tan (e+f x)}{f \left (a^3 \sec (e+f x)+a^3\right )}-\frac {2 \tan (e+f x) \left (c^3-c^3 \sec (e+f x)\right )}{3 a f (a \sec (e+f x)+a)^2}+\frac {2 c \tan (e+f x) (c-c \sec (e+f x))^2}{5 f (a \sec (e+f x)+a)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]
time = 0.12, size = 139, normalized size = 1.06 \begin {gather*} -\frac {c^3 \left (-\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}-\frac {26 \tan \left (\frac {1}{2} (e+f x)\right )}{15 f}+\frac {2 \sec ^2\left (\frac {1}{2} (e+f x)\right ) \tan \left (\frac {1}{2} (e+f x)\right )}{15 f}-\frac {2 \sec ^4\left (\frac {1}{2} (e+f x)\right ) \tan \left (\frac {1}{2} (e+f x)\right )}{5 f}\right )}{a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((c^3*(-(Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]]/f) + Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]/f - (26*Tan[(
e + f*x)/2])/(15*f) + (2*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/(15*f) - (2*Sec[(e + f*x)/2]^4*Tan[(e + f*x)/2])
/(5*f)))/a^3)

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Maple [A]
time = 0.20, size = 76, normalized size = 0.58

method	result	size
derivativedivides	\(\frac {2 c^{3} \left (\frac {\left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5}+\frac {\left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3}+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\frac {\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2}-\frac {\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{2}\right )}{f \,a^{3}}\)	\(76\)
default	\(\frac {2 c^{3} \left (\frac {\left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5}+\frac {\left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3}+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\frac {\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2}-\frac {\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{2}\right )}{f \,a^{3}}\)	\(76\)
risch	\(\frac {4 i c^{3} \left (15 \,{\mathrm e}^{4 i \left (f x +e \right )}+30 \,{\mathrm e}^{3 i \left (f x +e \right )}+100 \,{\mathrm e}^{2 i \left (f x +e \right )}+50 \,{\mathrm e}^{i \left (f x +e \right )}+13\right )}{15 f \,a^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{5}}+\frac {c^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{a^{3} f}-\frac {c^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{a^{3} f}\)	\(120\)
norman	\(\frac {\frac {16 c^{3} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 a f}-\frac {22 c^{3} \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 a f}+\frac {6 c^{3} \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 a f}-\frac {2 c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a f}-\frac {8 c^{3} \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15 a f}+\frac {2 c^{3} \left (\tan ^{11}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 a f}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3} a^{2}}+\frac {c^{3} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{a^{3} f}-\frac {c^{3} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{a^{3} f}\)	\(197\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(c-c*sec(f*x+e))^3/(a+a*sec(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

2/f*c^3/a^3*(1/5*tan(1/2*f*x+1/2*e)^5+1/3*tan(1/2*f*x+1/2*e)^3+tan(1/2*f*x+1/2*e)+1/2*ln(tan(1/2*f*x+1/2*e)-1)
-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. 330 vs. \(2 (138) = 276\).
time = 0.28, size = 330, normalized size = 2.52 \begin {gather*} \frac {c^{3} {\left (\frac {\frac {105 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {20 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{3}}\right )} + \frac {3 \, c^{3} {\left (\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}} + \frac {c^{3} {\left (\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}} - \frac {9 \, c^{3} {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))^3/(a+a*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 3.27, size = 206, normalized size = 1.57 \begin {gather*} -\frac {15 \, {\left (c^{3} \cos \left (f x + e\right )^{3} + 3 \, c^{3} \cos \left (f x + e\right )^{2} + 3 \, c^{3} \cos \left (f x + e\right ) + c^{3}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 15 \, {\left (c^{3} \cos \left (f x + e\right )^{3} + 3 \, c^{3} \cos \left (f x + e\right )^{2} + 3 \, c^{3} \cos \left (f x + e\right ) + c^{3}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 4 \, {\left (13 \, c^{3} \cos \left (f x + e\right )^{2} + 24 \, c^{3} \cos \left (f x + e\right ) + 23 \, c^{3}\right )} \sin \left (f x + e\right )}{30 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} + 3 \, a^{3} f \cos \left (f x + e\right ) + a^{3} f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))^3/(a+a*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

-1/30*(15*(c^3*cos(f*x + e)^3 + 3*c^3*cos(f*x + e)^2 + 3*c^3*cos(f*x + e) + c^3)*log(sin(f*x + e) + 1) - 15*(c
^3*cos(f*x + e)^3 + 3*c^3*cos(f*x + e)^2 + 3*c^3*cos(f*x + e) + c^3)*log(-sin(f*x + e) + 1) - 4*(13*c^3*cos(f*
x + e)^2 + 24*c^3*cos(f*x + e) + 23*c^3)*sin(f*x + e))/(a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 + 3*a^3*
f*cos(f*x + e) + a^3*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {c^{3} \left (\int \left (- \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}\right )\, 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 \left (- \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}\right )\, 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 )}{a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c**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))/a**3

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Giac [A]
time = 0.58, size = 109, normalized size = 0.83 \begin {gather*} -\frac {\frac {15 \, c^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right )}{a^{3}} - \frac {15 \, c^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right )}{a^{3}} - \frac {2 \, {\left (3 \, a^{12} c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 5 \, a^{12} c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 15 \, a^{12} c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{a^{15}}}{15 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))^3/(a+a*sec(f*x+e))^3,x, algorithm="giac")

[Out]

-1/15*(15*c^3*log(abs(tan(1/2*f*x + 1/2*e) + 1))/a^3 - 15*c^3*log(abs(tan(1/2*f*x + 1/2*e) - 1))/a^3 - 2*(3*a^
12*c^3*tan(1/2*f*x + 1/2*e)^5 + 5*a^12*c^3*tan(1/2*f*x + 1/2*e)^3 + 15*a^12*c^3*tan(1/2*f*x + 1/2*e))/a^15)/f

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Mupad [B]
time = 1.65, size = 61, normalized size = 0.47 \begin {gather*} \frac {2\,c^3\,\left (15\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )-15\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )+5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\right )}{15\,a^3\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*c^3*(15*tan(e/2 + (f*x)/2) - 15*atanh(tan(e/2 + (f*x)/2)) + 5*tan(e/2 + (f*x)/2)^3 + 3*tan(e/2 + (f*x)/2)^5
))/(15*a^3*f)

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