3.1.19 \(\int \frac {(a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^4} \, dx\) [19]

3.1.19.1 Optimal result
3.1.19.2 Mathematica [A] (verified)
3.1.19.3 Rubi [A] (verified)
3.1.19.4 Maple [A] (verified)
3.1.19.5 Fricas [A] (verification not implemented)
3.1.19.6 Sympy [F]
3.1.19.7 Maxima [B] (verification not implemented)
3.1.19.8 Giac [A] (verification not implemented)
3.1.19.9 Mupad [B] (verification not implemented)

3.1.19.1 Optimal result

Integrand size = 26, antiderivative size = 133 \[ \int \frac {(a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^4} \, dx=\frac {a^3 x}{c^4}-\frac {8 a^3 \tan (e+f x)}{7 c^4 f (1-\sec (e+f x))^4}+\frac {4 a^3 \tan (e+f x)}{35 c^4 f (1-\sec (e+f x))^3}-\frac {62 a^3 \tan (e+f x)}{105 c^4 f (1-\sec (e+f x))^2}-\frac {167 a^3 \tan (e+f x)}{105 c^4 f (1-\sec (e+f x))} \]

output
a^3*x/c^4-8/7*a^3*tan(f*x+e)/c^4/f/(1-sec(f*x+e))^4+4/35*a^3*tan(f*x+e)/c^ 
4/f/(1-sec(f*x+e))^3-62/105*a^3*tan(f*x+e)/c^4/f/(1-sec(f*x+e))^2-167/105* 
a^3*tan(f*x+e)/c^4/f/(1-sec(f*x+e))
 
3.1.19.2 Mathematica [A] (verified)

Time = 1.46 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.21 \[ \int \frac {(a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^4} \, dx=\frac {a^{5/2} \tan (e+f x) \left (\sqrt {a} \sqrt {c} \left (-337+276 \sec (e+f x)+50 \sec ^2(e+f x)-396 \sec ^3(e+f x)+167 \sec ^4(e+f x)\right )-840 \text {arctanh}\left (\frac {\sqrt {-a c \tan ^2(e+f x)}}{\sqrt {a} \sqrt {c}}\right ) \sec ^3(e+f x) \sin ^6\left (\frac {1}{2} (e+f x)\right ) \sqrt {-a c \tan ^2(e+f x)}\right )}{105 c^{9/2} f (-1+\sec (e+f x))^4 (1+\sec (e+f x))} \]

input
Integrate[(a + a*Sec[e + f*x])^3/(c - c*Sec[e + f*x])^4,x]
 
output
(a^(5/2)*Tan[e + f*x]*(Sqrt[a]*Sqrt[c]*(-337 + 276*Sec[e + f*x] + 50*Sec[e 
 + f*x]^2 - 396*Sec[e + f*x]^3 + 167*Sec[e + f*x]^4) - 840*ArcTanh[Sqrt[-( 
a*c*Tan[e + f*x]^2)]/(Sqrt[a]*Sqrt[c])]*Sec[e + f*x]^3*Sin[(e + f*x)/2]^6* 
Sqrt[-(a*c*Tan[e + f*x]^2)]))/(105*c^(9/2)*f*(-1 + Sec[e + f*x])^4*(1 + Se 
c[e + f*x]))
 
3.1.19.3 Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.92, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {3042, 4391, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(a \sec (e+f x)+a)^3}{(c-c \sec (e+f x))^4} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^3}{\left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^4}dx\)

\(\Big \downarrow \) 4391

\(\displaystyle \frac {\int \left (\frac {\sec ^3(e+f x) a^3}{(1-\sec (e+f x))^4}+\frac {3 \sec ^2(e+f x) a^3}{(1-\sec (e+f x))^4}+\frac {3 \sec (e+f x) a^3}{(1-\sec (e+f x))^4}+\frac {a^3}{(1-\sec (e+f x))^4}\right )dx}{c^4}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {167 a^3 \tan (e+f x)}{105 f (1-\sec (e+f x))}-\frac {62 a^3 \tan (e+f x)}{105 f (1-\sec (e+f x))^2}+\frac {4 a^3 \tan (e+f x)}{35 f (1-\sec (e+f x))^3}-\frac {8 a^3 \tan (e+f x)}{7 f (1-\sec (e+f x))^4}+a^3 x}{c^4}\)

input
Int[(a + a*Sec[e + f*x])^3/(c - c*Sec[e + f*x])^4,x]
 
output
(a^3*x - (8*a^3*Tan[e + f*x])/(7*f*(1 - Sec[e + f*x])^4) + (4*a^3*Tan[e + 
f*x])/(35*f*(1 - Sec[e + f*x])^3) - (62*a^3*Tan[e + f*x])/(105*f*(1 - Sec[ 
e + f*x])^2) - (167*a^3*Tan[e + f*x])/(105*f*(1 - Sec[e + f*x])))/c^4
 

3.1.19.3.1 Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4391
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*( 
d_.) + (c_))^(n_), x_Symbol] :> Simp[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]
 
3.1.19.4 Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.50

method result size
parallelrisch \(-\frac {a^{3} \left (15 \cot \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}-42 \cot \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+70 \cot \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}-105 f x -210 \cot \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{105 c^{4} f}\) \(67\)
derivativedivides \(\frac {a^{3} \left (-\frac {1}{7 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}+\frac {2}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}-\frac {2}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}+\frac {2}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}+2 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{f \,c^{4}}\) \(76\)
default \(\frac {a^{3} \left (-\frac {1}{7 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}+\frac {2}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}-\frac {2}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}+\frac {2}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}+2 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{f \,c^{4}}\) \(76\)
risch \(\frac {a^{3} x}{c^{4}}+\frac {2 i a^{3} \left (735 \,{\mathrm e}^{6 i \left (f x +e \right )}-2520 \,{\mathrm e}^{5 i \left (f x +e \right )}+5635 \,{\mathrm e}^{4 i \left (f x +e \right )}-6160 \,{\mathrm e}^{3 i \left (f x +e \right )}+4557 \,{\mathrm e}^{2 i \left (f x +e \right )}-1624 \,{\mathrm e}^{i \left (f x +e \right )}+337\right )}{105 f \,c^{4} \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{7}}\) \(103\)
norman \(\frac {\frac {a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{c}+\frac {a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}}{c}-\frac {a^{3}}{7 c f}+\frac {24 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{35 c f}-\frac {169 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{105 c f}+\frac {56 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{15 c f}-\frac {14 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{3 c f}+\frac {2 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}}{c f}-\frac {2 a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{c}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{2} c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}\) \(211\)

input
int((a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^4,x,method=_RETURNVERBOSE)
 
output
-1/105*a^3*(15*cot(1/2*f*x+1/2*e)^7-42*cot(1/2*f*x+1/2*e)^5+70*cot(1/2*f*x 
+1/2*e)^3-105*f*x-210*cot(1/2*f*x+1/2*e))/c^4/f
 
3.1.19.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.29 \[ \int \frac {(a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^4} \, dx=\frac {337 \, a^{3} \cos \left (f x + e\right )^{4} - 276 \, a^{3} \cos \left (f x + e\right )^{3} - 50 \, a^{3} \cos \left (f x + e\right )^{2} + 396 \, a^{3} \cos \left (f x + e\right ) - 167 \, a^{3} + 105 \, {\left (a^{3} f x \cos \left (f x + e\right )^{3} - 3 \, a^{3} f x \cos \left (f x + e\right )^{2} + 3 \, a^{3} f x \cos \left (f x + e\right ) - a^{3} f x\right )} \sin \left (f x + e\right )}{105 \, {\left (c^{4} f \cos \left (f x + e\right )^{3} - 3 \, c^{4} f \cos \left (f x + e\right )^{2} + 3 \, c^{4} f \cos \left (f x + e\right ) - c^{4} f\right )} \sin \left (f x + e\right )} \]

input
integrate((a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^4,x, algorithm="fricas")
 
output
1/105*(337*a^3*cos(f*x + e)^4 - 276*a^3*cos(f*x + e)^3 - 50*a^3*cos(f*x + 
e)^2 + 396*a^3*cos(f*x + e) - 167*a^3 + 105*(a^3*f*x*cos(f*x + e)^3 - 3*a^ 
3*f*x*cos(f*x + e)^2 + 3*a^3*f*x*cos(f*x + e) - a^3*f*x)*sin(f*x + e))/((c 
^4*f*cos(f*x + e)^3 - 3*c^4*f*cos(f*x + e)^2 + 3*c^4*f*cos(f*x + e) - c^4* 
f)*sin(f*x + e))
 
3.1.19.6 Sympy [F]

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

input
integrate((a+a*sec(f*x+e))**3/(c-c*sec(f*x+e))**4,x)
 
output
a**3*(Integral(3*sec(e + f*x)/(sec(e + f*x)**4 - 4*sec(e + f*x)**3 + 6*sec 
(e + f*x)**2 - 4*sec(e + f*x) + 1), x) + Integral(3*sec(e + f*x)**2/(sec(e 
 + f*x)**4 - 4*sec(e + f*x)**3 + 6*sec(e + f*x)**2 - 4*sec(e + f*x) + 1), 
x) + Integral(sec(e + f*x)**3/(sec(e + f*x)**4 - 4*sec(e + f*x)**3 + 6*sec 
(e + f*x)**2 - 4*sec(e + f*x) + 1), x) + Integral(1/(sec(e + f*x)**4 - 4*s 
ec(e + f*x)**3 + 6*sec(e + f*x)**2 - 4*sec(e + f*x) + 1), x))/c**4
 
3.1.19.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 383 vs. \(2 (117) = 234\).

Time = 0.34 (sec) , antiderivative size = 383, normalized size of antiderivative = 2.88 \[ \int \frac {(a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^4} \, dx=\frac {5 \, a^{3} {\left (\frac {336 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c^{4}} + \frac {{\left (\frac {21 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {77 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {315 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - 3\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{7}}{c^{4} \sin \left (f x + e\right )^{7}}\right )} + \frac {3 \, a^{3} {\left (\frac {21 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {35 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {105 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - 15\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{7}}{c^{4} \sin \left (f x + e\right )^{7}} + \frac {9 \, a^{3} {\left (\frac {21 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {35 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {35 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - 5\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{7}}{c^{4} \sin \left (f x + e\right )^{7}} - \frac {a^{3} {\left (\frac {21 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {35 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {105 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + 15\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{7}}{c^{4} \sin \left (f x + e\right )^{7}}}{840 \, f} \]

input
integrate((a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^4,x, algorithm="maxima")
 
output
1/840*(5*a^3*(336*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c^4 + (21*sin(f* 
x + e)^2/(cos(f*x + e) + 1)^2 - 77*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 3 
15*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 3)*(cos(f*x + e) + 1)^7/(c^4*sin( 
f*x + e)^7)) + 3*a^3*(21*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 35*sin(f*x 
+ e)^4/(cos(f*x + e) + 1)^4 - 105*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 15 
)*(cos(f*x + e) + 1)^7/(c^4*sin(f*x + e)^7) + 9*a^3*(21*sin(f*x + e)^2/(co 
s(f*x + e) + 1)^2 - 35*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 35*sin(f*x + 
e)^6/(cos(f*x + e) + 1)^6 - 5)*(cos(f*x + e) + 1)^7/(c^4*sin(f*x + e)^7) - 
 a^3*(21*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 35*sin(f*x + e)^4/(cos(f*x 
+ e) + 1)^4 - 105*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 15)*(cos(f*x + e) 
+ 1)^7/(c^4*sin(f*x + e)^7))/f
 
3.1.19.8 Giac [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.66 \[ \int \frac {(a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^4} \, dx=\frac {\frac {105 \, {\left (f x + e\right )} a^{3}}{c^{4}} + \frac {210 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 70 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 42 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 15 \, a^{3}}{c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7}}}{105 \, f} \]

input
integrate((a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^4,x, algorithm="giac")
 
output
1/105*(105*(f*x + e)*a^3/c^4 + (210*a^3*tan(1/2*f*x + 1/2*e)^6 - 70*a^3*ta 
n(1/2*f*x + 1/2*e)^4 + 42*a^3*tan(1/2*f*x + 1/2*e)^2 - 15*a^3)/(c^4*tan(1/ 
2*f*x + 1/2*e)^7))/f
 
3.1.19.9 Mupad [B] (verification not implemented)

Time = 14.80 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.92 \[ \int \frac {(a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^4} \, dx=\frac {a^3\,\left (-\frac {{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7}{7}+\frac {2\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2}{5}-\frac {2\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4}{3}+2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+\left (e+f\,x\right )\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7\right )}{c^4\,f\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7} \]

input
int((a + a/cos(e + f*x))^3/(c - c/cos(e + f*x))^4,x)
 
output
(a^3*(2*cos(e/2 + (f*x)/2)*sin(e/2 + (f*x)/2)^6 - cos(e/2 + (f*x)/2)^7/7 + 
 sin(e/2 + (f*x)/2)^7*(e + f*x) - (2*cos(e/2 + (f*x)/2)^3*sin(e/2 + (f*x)/ 
2)^4)/3 + (2*cos(e/2 + (f*x)/2)^5*sin(e/2 + (f*x)/2)^2)/5))/(c^4*f*sin(e/2 
 + (f*x)/2)^7)