\(\int \frac {\sec (e+f x) (c+d \sec (e+f x))^3}{a+a \sec (e+f x)} \, dx\) [211]

Optimal result

Integrand size = 31, antiderivative size = 117 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^3}{a+a \sec (e+f x)} \, dx=\frac {3 d \left (2 c^2-2 c d+d^2\right ) \text {arctanh}(\sin (e+f x))}{2 a f}+\frac {(c-d) (c+d \sec (e+f x))^2 \tan (e+f x)}{f (a+a \sec (e+f x))}-\frac {d \left (4 \left (c^2-3 c d+d^2\right )+(2 c-3 d) d \sec (e+f x)\right ) \tan (e+f x)}{2 a f} \]

[Out]

3/2*d*(2*c^2-2*c*d+d^2)*arctanh(sin(f*x+e))/a/f+(c-d)*(c+d*sec(f*x+e))^2*tan(f*x+e)/f/(a+a*sec(f*x+e))-1/2*d*(
4*c^2-12*c*d+4*d^2+(2*c-3*d)*d*sec(f*x+e))*tan(f*x+e)/a/f

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.46, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4072, 100, 152, 65, 223, 209} \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^3}{a+a \sec (e+f x)} \, dx=\frac {3 d \left (2 c^2-2 c d+d^2\right ) \tan (e+f x) \arctan \left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a (\sec (e+f x)+1)}}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}-\frac {d \tan (e+f x) \left (4 \left (c^2-3 c d+d^2\right )+d (2 c-3 d) \sec (e+f x)\right )}{2 a f}+\frac {(c-d) \tan (e+f x) (c+d \sec (e+f x))^2}{f (a \sec (e+f x)+a)} \]

[In]

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

[Out]

(3*d*(2*c^2 - 2*c*d + d^2)*ArcTan[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a*(1 + Sec[e + f*x])]]*Tan[e + f*x])/(f*Sqrt[a
- a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) + ((c - d)*(c + d*Sec[e + f*x])^2*Tan[e + f*x])/(f*(a + a*Sec[e +
f*x])) - (d*(4*(c^2 - 3*c*d + d^2) + (2*c - 3*d)*d*Sec[e + f*x])*Tan[e + f*x])/(2*a*f)

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 100

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((e + f*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] + Dist[1/(b*(b*e - a*
f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 152

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)
^(m + 1)*((c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d
*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1
)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)
^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 4072

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]
*(d_.) + (c_))^(n_), x_Symbol] :> Dist[a^2*g*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]
])), Subst[Int[(g*x)^(p - 1)*(a + b*x)^(m - 1/2)*((c + d*x)^n/Sqrt[a - b*x]), x], x, Csc[e + f*x]], x] /; Free
Q[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && (EqQ[p,
1] || IntegerQ[m - 1/2])

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (a^2 \tan (e+f x)\right ) \text {Subst}\left (\int \frac {(c+d x)^3}{\sqrt {a-a x} (a+a x)^{3/2}} \, dx,x,\sec (e+f x)\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}} \\ & = \frac {(c-d) (c+d \sec (e+f x))^2 \tan (e+f x)}{f (a+a \sec (e+f x))}+\frac {\tan (e+f x) \text {Subst}\left (\int \frac {(c+d x) \left (-a^2 (3 c-2 d) d+a^2 (2 c-3 d) d x\right )}{\sqrt {a-a x} \sqrt {a+a x}} \, dx,x,\sec (e+f x)\right )}{a f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}} \\ & = \frac {(c-d) (c+d \sec (e+f x))^2 \tan (e+f x)}{f (a+a \sec (e+f x))}-\frac {d \left (4 \left (c^2-3 c d+d^2\right )+(2 c-3 d) d \sec (e+f x)\right ) \tan (e+f x)}{2 a f}-\frac {\left (3 a d \left (2 c^2-2 c d+d^2\right ) \tan (e+f x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a-a x} \sqrt {a+a x}} \, dx,x,\sec (e+f x)\right )}{2 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}} \\ & = \frac {(c-d) (c+d \sec (e+f x))^2 \tan (e+f x)}{f (a+a \sec (e+f x))}-\frac {d \left (4 \left (c^2-3 c d+d^2\right )+(2 c-3 d) d \sec (e+f x)\right ) \tan (e+f x)}{2 a f}+\frac {\left (3 d \left (2 c^2-2 c d+d^2\right ) \tan (e+f x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {2 a-x^2}} \, dx,x,\sqrt {a-a \sec (e+f x)}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}} \\ & = \frac {(c-d) (c+d \sec (e+f x))^2 \tan (e+f x)}{f (a+a \sec (e+f x))}-\frac {d \left (4 \left (c^2-3 c d+d^2\right )+(2 c-3 d) d \sec (e+f x)\right ) \tan (e+f x)}{2 a f}+\frac {\left (3 d \left (2 c^2-2 c d+d^2\right ) \tan (e+f x)\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a+a \sec (e+f x)}}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}} \\ & = \frac {3 d \left (2 c^2-2 c d+d^2\right ) \arctan \left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a+a \sec (e+f x)}}\right ) \tan (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {(c-d) (c+d \sec (e+f x))^2 \tan (e+f x)}{f (a+a \sec (e+f x))}-\frac {d \left (4 \left (c^2-3 c d+d^2\right )+(2 c-3 d) d \sec (e+f x)\right ) \tan (e+f x)}{2 a f} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.60 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.67 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^3}{a+a \sec (e+f x)} \, dx=\frac {3 d \left (2 c^2-2 c d+d^2\right ) \text {arctanh}(\sin (e+f x))+2 (c-d)^3 \tan \left (\frac {1}{2} (e+f x)\right )+d^2 (6 c-2 d+d \sec (e+f x)) \tan (e+f x)}{2 a f} \]

[In]

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

[Out]

(3*d*(2*c^2 - 2*c*d + d^2)*ArcTanh[Sin[e + f*x]] + 2*(c - d)^3*Tan[(e + f*x)/2] + d^2*(6*c - 2*d + d*Sec[e + f
*x])*Tan[e + f*x])/(2*a*f)

Maple [A] (verified)

Time = 0.83 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.55

method	result	size
parallelrisch	\(\frac {-3 \left (c^{2}-c d +\frac {1}{2} d^{2}\right ) \left (1+\cos \left (2 f x +2 e \right )\right ) d \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )+3 \left (c^{2}-c d +\frac {1}{2} d^{2}\right ) \left (1+\cos \left (2 f x +2 e \right )\right ) d \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )+\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) \left (\left (c^{3}-3 c^{2} d +6 c \,d^{2}-2 d^{3}\right ) \cos \left (2 f x +2 e \right )+\left (6 c \,d^{2}-d^{3}\right ) \cos \left (f x +e \right )+c^{3}-3 c^{2} d +6 c \,d^{2}-d^{3}\right )}{a f \left (1+\cos \left (2 f x +2 e \right )\right )}\)	\(181\)
derivativedivides	\(\frac {c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-3 c^{2} d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c \,d^{2}-d^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\frac {d^{3}}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}+\frac {3 d \left (2 c^{2}-2 c d +d^{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{2}-\frac {3 d^{2} \left (2 c -d \right )}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {d^{3}}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {3 d \left (2 c^{2}-2 c d +d^{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2}-\frac {3 d^{2} \left (2 c -d \right )}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}}{f a}\)	\(208\)
default	\(\frac {c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-3 c^{2} d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c \,d^{2}-d^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\frac {d^{3}}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}+\frac {3 d \left (2 c^{2}-2 c d +d^{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{2}-\frac {3 d^{2} \left (2 c -d \right )}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {d^{3}}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {3 d \left (2 c^{2}-2 c d +d^{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2}-\frac {3 d^{2} \left (2 c -d \right )}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}}{f a}\)	\(208\)
norman	\(\frac {\frac {\left (c^{3}-3 c^{2} d +3 c \,d^{2}-d^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{a f}+\frac {\left (3 c^{3}-9 c^{2} d +21 c \,d^{2}-7 d^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{a f}-\frac {3 \left (c^{3}-3 c^{2} d +5 c \,d^{2}-2 d^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{a f}-\frac {\left (c^{3}-3 c^{2} d +9 c \,d^{2}-2 d^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{3}}-\frac {3 d \left (2 c^{2}-2 c d +d^{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2 a f}+\frac {3 d \left (2 c^{2}-2 c d +d^{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{2 a f}\)	\(245\)
risch	\(\frac {i \left (2 c^{3} {\mathrm e}^{4 i \left (f x +e \right )}-6 c^{2} d \,{\mathrm e}^{4 i \left (f x +e \right )}+6 c \,d^{2} {\mathrm e}^{4 i \left (f x +e \right )}-3 d^{3} {\mathrm e}^{4 i \left (f x +e \right )}+6 c \,d^{2} {\mathrm e}^{3 i \left (f x +e \right )}-3 d^{3} {\mathrm e}^{3 i \left (f x +e \right )}+4 c^{3} {\mathrm e}^{2 i \left (f x +e \right )}-12 c^{2} d \,{\mathrm e}^{2 i \left (f x +e \right )}+18 c \,d^{2} {\mathrm e}^{2 i \left (f x +e \right )}-5 d^{3} {\mathrm e}^{2 i \left (f x +e \right )}+6 c \,d^{2} {\mathrm e}^{i \left (f x +e \right )}-d^{3} {\mathrm e}^{i \left (f x +e \right )}+2 c^{3}-6 c^{2} d +12 c \,d^{2}-4 d^{3}\right )}{f a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )^{2}}-\frac {3 d \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) c^{2}}{a f}+\frac {3 d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) c}{a f}-\frac {3 d^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{2 a f}+\frac {3 d \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) c^{2}}{a f}-\frac {3 d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) c}{a f}+\frac {3 d^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{2 a f}\)	\(382\)

[In]

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

[Out]

(-3*(c^2-c*d+1/2*d^2)*(1+cos(2*f*x+2*e))*d*ln(tan(1/2*f*x+1/2*e)-1)+3*(c^2-c*d+1/2*d^2)*(1+cos(2*f*x+2*e))*d*l
n(tan(1/2*f*x+1/2*e)+1)+tan(1/2*f*x+1/2*e)*((c^3-3*c^2*d+6*c*d^2-2*d^3)*cos(2*f*x+2*e)+(6*c*d^2-d^3)*cos(f*x+e
)+c^3-3*c^2*d+6*c*d^2-d^3))/a/f/(1+cos(2*f*x+2*e))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.85 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^3}{a+a \sec (e+f x)} \, dx=\frac {3 \, {\left ({\left (2 \, c^{2} d - 2 \, c d^{2} + d^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (2 \, c^{2} d - 2 \, c d^{2} + d^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \, {\left ({\left (2 \, c^{2} d - 2 \, c d^{2} + d^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (2 \, c^{2} d - 2 \, c d^{2} + d^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (d^{3} + 2 \, {\left (c^{3} - 3 \, c^{2} d + 6 \, c d^{2} - 2 \, d^{3}\right )} \cos \left (f x + e\right )^{2} + {\left (6 \, c d^{2} - d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{4 \, {\left (a f \cos \left (f x + e\right )^{3} + a f \cos \left (f x + e\right )^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 388 vs. \(2 (114) = 228\).

Time = 0.22 (sec) , antiderivative size = 388, normalized size of antiderivative = 3.32 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^3}{a+a \sec (e+f x)} \, dx=-\frac {d^{3} {\left (\frac {2 \, {\left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {3 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a - \frac {2 \, a \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}} - \frac {3 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a} + \frac {3 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a} + \frac {2 \, \sin \left (f x + e\right )}{a {\left (\cos \left (f x + e\right ) + 1\right )}}\right )} + 6 \, c d^{2} {\left (\frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a} - \frac {2 \, \sin \left (f x + e\right )}{{\left (a - \frac {a \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}} - \frac {\sin \left (f x + e\right )}{a {\left (\cos \left (f x + e\right ) + 1\right )}}\right )} - 6 \, c^{2} d {\left (\frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a} - \frac {\sin \left (f x + e\right )}{a {\left (\cos \left (f x + e\right ) + 1\right )}}\right )} - \frac {2 \, c^{3} \sin \left (f x + e\right )}{a {\left (\cos \left (f x + e\right ) + 1\right )}}}{2 \, f} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.87 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^3}{a+a \sec (e+f x)} \, dx=\frac {\frac {3 \, {\left (2 \, c^{2} d - 2 \, c d^{2} + d^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right )}{a} - \frac {3 \, {\left (2 \, c^{2} d - 2 \, c d^{2} + d^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right )}{a} + \frac {2 \, {\left (c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 3 \, c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{a} - \frac {2 \, {\left (6 \, c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 3 \, d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 6 \, c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{2} a}}{2 \, f} \]

[In]

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

[Out]

1/2*(3*(2*c^2*d - 2*c*d^2 + d^3)*log(abs(tan(1/2*f*x + 1/2*e) + 1))/a - 3*(2*c^2*d - 2*c*d^2 + d^3)*log(abs(ta
n(1/2*f*x + 1/2*e) - 1))/a + 2*(c^3*tan(1/2*f*x + 1/2*e) - 3*c^2*d*tan(1/2*f*x + 1/2*e) + 3*c*d^2*tan(1/2*f*x
+ 1/2*e) - d^3*tan(1/2*f*x + 1/2*e))/a - 2*(6*c*d^2*tan(1/2*f*x + 1/2*e)^3 - 3*d^3*tan(1/2*f*x + 1/2*e)^3 - 6*
c*d^2*tan(1/2*f*x + 1/2*e) + d^3*tan(1/2*f*x + 1/2*e))/((tan(1/2*f*x + 1/2*e)^2 - 1)^2*a))/f

Mupad [B] (verification not implemented)

Time = 13.68 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.19 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^3}{a+a \sec (e+f x)} \, dx=\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (6\,c\,d^2-d^3\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (6\,c\,d^2-3\,d^3\right )}{f\,\left (a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-2\,a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+a\right )}+\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,{\left (c-d\right )}^3}{a\,f}+\frac {3\,d\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )\,\left (2\,c^2-2\,c\,d+d^2\right )}{a\,f} \]

[In]

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

[Out]

(tan(e/2 + (f*x)/2)*(6*c*d^2 - d^3) - tan(e/2 + (f*x)/2)^3*(6*c*d^2 - 3*d^3))/(f*(a - 2*a*tan(e/2 + (f*x)/2)^2
+ a*tan(e/2 + (f*x)/2)^4)) + (tan(e/2 + (f*x)/2)*(c - d)^3)/(a*f) + (3*d*atanh(tan(e/2 + (f*x)/2))*(2*c^2 - 2
*c*d + d^2))/(a*f)