\(\int \frac {(c-c \sec (e+f x))^3}{(a+a \sec (e+f x))^2} \, dx\) [23]

Optimal result

Integrand size = 26, antiderivative size = 85 \[ \int \frac {(c-c \sec (e+f x))^3}{(a+a \sec (e+f x))^2} \, dx=\frac {c^3 x}{a^2}-\frac {c^3 \text {arctanh}(\sin (e+f x))}{a^2 f}-\frac {8 c^3 \tan (e+f x)}{3 a^2 f (1+\sec (e+f x))^2}+\frac {4 c^3 \tan (e+f x)}{3 a^2 f (1+\sec (e+f x))} \]

[Out]

c^3*x/a^2-c^3*arctanh(sin(f*x+e))/a^2/f-8/3*c^3*tan(f*x+e)/a^2/f/(1+sec(f*x+e))^2+4/3*c^3*tan(f*x+e)/a^2/f/(1+
sec(f*x+e))

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {3988, 3862, 4004, 3879, 3881, 3882, 3884, 4083, 3855} \[ \int \frac {(c-c \sec (e+f x))^3}{(a+a \sec (e+f x))^2} \, dx=-\frac {c^3 \text {arctanh}(\sin (e+f x))}{a^2 f}+\frac {4 c^3 \tan (e+f x)}{3 a^2 f (\sec (e+f x)+1)}-\frac {8 c^3 \tan (e+f x)}{3 a^2 f (\sec (e+f x)+1)^2}+\frac {c^3 x}{a^2} \]

[In]

Int[(c - c*Sec[e + f*x])^3/(a + a*Sec[e + f*x])^2,x]

[Out]

(c^3*x)/a^2 - (c^3*ArcTanh[Sin[e + f*x]])/(a^2*f) - (8*c^3*Tan[e + f*x])/(3*a^2*f*(1 + Sec[e + f*x])^2) + (4*c
^3*Tan[e + f*x])/(3*a^2*f*(1 + 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 3862

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

Rule 3879

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[-Cot[e + f*x]/(f*(b + a*
Csc[e + f*x])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 3881

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

Rule 3882

Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-Cot[e + f*x])*(
(a + b*Csc[e + f*x])^m/(f*(2*m + 1))), x] + Dist[m/(b*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1
), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rule 3884

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[b*Cot[e + f*x]*((
a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), x] - Dist[1/(a^2*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m
+ 1)*(a*m - b*(2*m + 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)
]

Rule 3988

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> Dis
t[c^n, Int[ExpandTrig[(1 + (d/c)*csc[e + f*x])^n, (a + b*csc[e + f*x])^m, x], x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0] && ILtQ[n, 0] && LtQ[m + n, 2]

Rule 4004

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[c*(x/a),
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 4083

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \left (\frac {c^3}{(1+\sec (e+f x))^2}-\frac {3 c^3 \sec (e+f x)}{(1+\sec (e+f x))^2}+\frac {3 c^3 \sec ^2(e+f x)}{(1+\sec (e+f x))^2}-\frac {c^3 \sec ^3(e+f x)}{(1+\sec (e+f x))^2}\right ) \, dx}{a^2} \\ & = \frac {c^3 \int \frac {1}{(1+\sec (e+f x))^2} \, dx}{a^2}-\frac {c^3 \int \frac {\sec ^3(e+f x)}{(1+\sec (e+f x))^2} \, dx}{a^2}-\frac {\left (3 c^3\right ) \int \frac {\sec (e+f x)}{(1+\sec (e+f x))^2} \, dx}{a^2}+\frac {\left (3 c^3\right ) \int \frac {\sec ^2(e+f x)}{(1+\sec (e+f x))^2} \, dx}{a^2} \\ & = -\frac {8 c^3 \tan (e+f x)}{3 a^2 f (1+\sec (e+f x))^2}-\frac {c^3 \int \frac {-3+\sec (e+f x)}{1+\sec (e+f x)} \, dx}{3 a^2}-\frac {c^3 \int \frac {\sec (e+f x) (-2+3 \sec (e+f x))}{1+\sec (e+f x)} \, dx}{3 a^2}-\frac {c^3 \int \frac {\sec (e+f x)}{1+\sec (e+f x)} \, dx}{a^2}+\frac {\left (2 c^3\right ) \int \frac {\sec (e+f x)}{1+\sec (e+f x)} \, dx}{a^2} \\ & = \frac {c^3 x}{a^2}-\frac {8 c^3 \tan (e+f x)}{3 a^2 f (1+\sec (e+f x))^2}+\frac {c^3 \tan (e+f x)}{a^2 f (1+\sec (e+f x))}-\frac {c^3 \int \sec (e+f x) \, dx}{a^2}-\frac {\left (4 c^3\right ) \int \frac {\sec (e+f x)}{1+\sec (e+f x)} \, dx}{3 a^2}+\frac {\left (5 c^3\right ) \int \frac {\sec (e+f x)}{1+\sec (e+f x)} \, dx}{3 a^2} \\ & = \frac {c^3 x}{a^2}-\frac {c^3 \text {arctanh}(\sin (e+f x))}{a^2 f}-\frac {8 c^3 \tan (e+f x)}{3 a^2 f (1+\sec (e+f x))^2}+\frac {4 c^3 \tan (e+f x)}{3 a^2 f (1+\sec (e+f x))} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 1.37 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.22 \[ \int \frac {(c-c \sec (e+f x))^3}{(a+a \sec (e+f x))^2} \, dx=\frac {c^{5/2} \tan (e+f x) \left (4 \sqrt {2} \sqrt {a} \sqrt {c} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {3}{2},-\frac {1}{2},\frac {1}{2} (1+\sec (e+f x))\right ) \sqrt {1-\sec (e+f x)}-4 \sqrt {a} \sqrt {c} \left (-2+\sec (e+f x)+\sec ^2(e+f x)\right )-6 \text {arctanh}\left (\frac {\sqrt {-a c \tan ^2(e+f x)}}{\sqrt {a} \sqrt {c}}\right ) \cos ^2\left (\frac {1}{2} (e+f x)\right ) \sec (e+f x) \sqrt {-a c \tan ^2(e+f x)}\right )}{3 a^{5/2} f (-1+\sec (e+f x)) (1+\sec (e+f x))^2} \]

[In]

Integrate[(c - c*Sec[e + f*x])^3/(a + a*Sec[e + f*x])^2,x]

[Out]

(c^(5/2)*Tan[e + f*x]*(4*Sqrt[2]*Sqrt[a]*Sqrt[c]*Hypergeometric2F1[-3/2, -3/2, -1/2, (1 + Sec[e + f*x])/2]*Sqr
t[1 - Sec[e + f*x]] - 4*Sqrt[a]*Sqrt[c]*(-2 + Sec[e + f*x] + Sec[e + f*x]^2) - 6*ArcTanh[Sqrt[-(a*c*Tan[e + f*
x]^2)]/(Sqrt[a]*Sqrt[c])]*Cos[(e + f*x)/2]^2*Sec[e + f*x]*Sqrt[-(a*c*Tan[e + f*x]^2)]))/(3*a^(5/2)*f*(-1 + Sec
[e + f*x])*(1 + Sec[e + f*x])^2)

Maple [A] (verified)

Time = 0.66 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.68

method	result	size
parallelrisch	\(\frac {c^{3} \left (4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+3 f x +3 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )-3 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )\right )}{3 a^{2} f}\)	\(58\)
derivativedivides	\(\frac {4 c^{3} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}+\frac {\arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}-\frac {\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{4}+\frac {\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{4}\right )}{f \,a^{2}}\)	\(66\)
default	\(\frac {4 c^{3} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}+\frac {\arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}-\frac {\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{4}+\frac {\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{4}\right )}{f \,a^{2}}\)	\(66\)
risch	\(\frac {c^{3} x}{a^{2}}-\frac {8 i c^{3} \left (3 \,{\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{3 f \,a^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{3}}+\frac {c^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{a^{2} f}-\frac {c^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{a^{2} f}\)	\(95\)
norman	\(\frac {\frac {c^{3} x}{a}+\frac {c^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{a}+\frac {4 c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3 a f}-\frac {8 c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{3 a f}+\frac {4 c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{3 a f}-\frac {2 c^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{a}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{2} a}+\frac {c^{3} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{a^{2} f}-\frac {c^{3} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{a^{2} f}\)	\(180\)

[In]

int((c-c*sec(f*x+e))^3/(a+a*sec(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 173 vs. \(2 (81) = 162\).

Time = 0.28 (sec) , antiderivative size = 173, normalized size of antiderivative = 2.04 \[ \int \frac {(c-c \sec (e+f x))^3}{(a+a \sec (e+f x))^2} \, dx=\frac {6 \, c^{3} f x \cos \left (f x + e\right )^{2} + 12 \, c^{3} f x \cos \left (f x + e\right ) + 6 \, c^{3} f x - 3 \, {\left (c^{3} \cos \left (f x + e\right )^{2} + 2 \, c^{3} \cos \left (f x + e\right ) + c^{3}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) + 3 \, {\left (c^{3} \cos \left (f x + e\right )^{2} + 2 \, c^{3} \cos \left (f x + e\right ) + c^{3}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 8 \, {\left (c^{3} \cos \left (f x + e\right ) - c^{3}\right )} \sin \left (f x + e\right )}{6 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} + 2 \, a^{2} f \cos \left (f x + e\right ) + a^{2} f\right )}} \]

[In]

integrate((c-c*sec(f*x+e))^3/(a+a*sec(f*x+e))^2,x, algorithm="fricas")

[Out]

1/6*(6*c^3*f*x*cos(f*x + e)^2 + 12*c^3*f*x*cos(f*x + e) + 6*c^3*f*x - 3*(c^3*cos(f*x + e)^2 + 2*c^3*cos(f*x +
e) + c^3)*log(sin(f*x + e) + 1) + 3*(c^3*cos(f*x + e)^2 + 2*c^3*cos(f*x + e) + c^3)*log(-sin(f*x + e) + 1) - 8
*(c^3*cos(f*x + e) - c^3)*sin(f*x + e))/(a^2*f*cos(f*x + e)^2 + 2*a^2*f*cos(f*x + e) + a^2*f)

Sympy [F]

\[ \int \frac {(c-c \sec (e+f x))^3}{(a+a \sec (e+f x))^2} \, dx=- \frac {c^{3} \left (\int \frac {3 \sec {\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\, dx + \int \left (- \frac {3 \sec ^{2}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\right )\, dx + \int \frac {\sec ^{3}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\, dx + \int \left (- \frac {1}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\right )\, dx\right )}{a^{2}} \]

[In]

integrate((c-c*sec(f*x+e))**3/(a+a*sec(f*x+e))**2,x)

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 268 vs. \(2 (81) = 162\).

Time = 0.29 (sec) , antiderivative size = 268, normalized size of antiderivative = 3.15 \[ \int \frac {(c-c \sec (e+f x))^3}{(a+a \sec (e+f x))^2} \, dx=\frac {c^{3} {\left (\frac {\frac {9 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}{a^{2}} - \frac {6 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{2}} + \frac {6 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{2}}\right )} - c^{3} {\left (\frac {\frac {9 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a^{2}}\right )} + \frac {3 \, c^{3} {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2}} - \frac {3 \, c^{3} {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, f} \]

[In]

integrate((c-c*sec(f*x+e))^3/(a+a*sec(f*x+e))^2,x, algorithm="maxima")

[Out]

1/6*(c^3*((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) - c^3*((9*sin(f*x + e)/(cos(f*x +
e) + 1) - sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2 - 12*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^2) + 3*c^3*
(3*sin(f*x + e)/(cos(f*x + e) + 1) + sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2 - 3*c^3*(3*sin(f*x + e)/(cos(f*x
+ e) + 1) - sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2)/f

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.94 \[ \int \frac {(c-c \sec (e+f x))^3}{(a+a \sec (e+f x))^2} \, dx=\frac {\frac {4 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3}}{a^{2}} + \frac {3 \, {\left (f x + e\right )} c^{3}}{a^{2}} - \frac {3 \, c^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right )}{a^{2}} + \frac {3 \, c^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right )}{a^{2}}}{3 \, f} \]

[In]

integrate((c-c*sec(f*x+e))^3/(a+a*sec(f*x+e))^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.22 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.54 \[ \int \frac {(c-c \sec (e+f x))^3}{(a+a \sec (e+f x))^2} \, dx=\frac {c^3\,x}{a^2}-\frac {c^3\,\left (2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )-\frac {4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{3}\right )}{a^2\,f} \]

[In]

int((c - c/cos(e + f*x))^3/(a + a/cos(e + f*x))^2,x)

[Out]

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