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

Optimal. Leaf size=119 \[ -\frac {5 c^3 \tan (e+f x)}{a^2 f}+\frac {5 c^3 \tanh ^{-1}(\sin (e+f x))}{a^2 f}-\frac {10 \tan (e+f x) \left (c^3-c^3 \sec (e+f x)\right )}{3 f \left (a^2 \sec (e+f x)+a^2\right )}+\frac {2 c \tan (e+f x) (c-c \sec (e+f x))^2}{3 f (a \sec (e+f x)+a)^2} \]

[Out]

5*c^3*arctanh(sin(f*x+e))/a^2/f-5*c^3*tan(f*x+e)/a^2/f+2/3*c*(c-c*sec(f*x+e))^2*tan(f*x+e)/f/(a+a*sec(f*x+e))^
2-10/3*(c^3-c^3*sec(f*x+e))*tan(f*x+e)/f/(a^2+a^2*sec(f*x+e))

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Rubi [A]  time = 0.19, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {3957, 3787, 3770, 3767, 8} \[ -\frac {5 c^3 \tan (e+f x)}{a^2 f}+\frac {5 c^3 \tanh ^{-1}(\sin (e+f x))}{a^2 f}-\frac {10 \tan (e+f x) \left (c^3-c^3 \sec (e+f x)\right )}{3 f \left (a^2 \sec (e+f x)+a^2\right )}+\frac {2 c \tan (e+f x) (c-c \sec (e+f x))^2}{3 f (a \sec (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*c^3*ArcTanh[Sin[e + f*x]])/(a^2*f) - (5*c^3*Tan[e + f*x])/(a^2*f) + (2*c*(c - c*Sec[e + f*x])^2*Tan[e + f*x
])/(3*f*(a + a*Sec[e + f*x])^2) - (10*(c^3 - c^3*Sec[e + f*x])*Tan[e + f*x])/(3*f*(a^2 + a^2*Sec[e + f*x]))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3770

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

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3957

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))^2} \, dx &=\frac {2 c (c-c \sec (e+f x))^2 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac {(5 c) \int \frac {\sec (e+f x) (c-c \sec (e+f x))^2}{a+a \sec (e+f x)} \, dx}{3 a}\\ &=\frac {2 c (c-c \sec (e+f x))^2 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac {10 \left (c^3-c^3 \sec (e+f x)\right ) \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}+\frac {\left (5 c^2\right ) \int \sec (e+f x) (c-c \sec (e+f x)) \, dx}{a^2}\\ &=\frac {2 c (c-c \sec (e+f x))^2 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac {10 \left (c^3-c^3 \sec (e+f x)\right ) \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}+\frac {\left (5 c^3\right ) \int \sec (e+f x) \, dx}{a^2}-\frac {\left (5 c^3\right ) \int \sec ^2(e+f x) \, dx}{a^2}\\ &=\frac {5 c^3 \tanh ^{-1}(\sin (e+f x))}{a^2 f}+\frac {2 c (c-c \sec (e+f x))^2 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac {10 \left (c^3-c^3 \sec (e+f x)\right ) \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}+\frac {\left (5 c^3\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (e+f x))}{a^2 f}\\ &=\frac {5 c^3 \tanh ^{-1}(\sin (e+f x))}{a^2 f}-\frac {5 c^3 \tan (e+f x)}{a^2 f}+\frac {2 c (c-c \sec (e+f x))^2 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac {10 \left (c^3-c^3 \sec (e+f x)\right ) \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}\\ \end {align*}

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Mathematica [B]  time = 4.32, size = 485, normalized size = 4.08 \[ \frac {c^3 (\cos (e+f x)-1)^3 \cot \left (\frac {1}{2} (e+f x)\right ) \csc ^2\left (\frac {1}{2} (e+f x)\right ) \left (-\frac {1}{16} \sec ^3\left (\frac {e}{2}\right ) \sin \left (\frac {f x}{2}\right ) (-80 \cos (e+f x)-40 \cos (2 (e+f x))+66 \cos (2 e+f x)+23 \cos (e+2 f x)+17 \cos (3 e+2 f x)+40 \cos (e)+78 \cos (f x)-40) \csc ^5\left (\frac {1}{2} (e+f x)\right )-26 \sec \left (\frac {e}{2}\right ) \sin \left (\frac {f x}{2}\right ) \cot ^4\left (\frac {1}{2} (e+f x)\right ) \csc \left (\frac {1}{2} (e+f x)\right )+20 \sec \left (\frac {e}{2}\right ) \sin \left (\frac {f x}{2}\right ) \cot ^2\left (\frac {1}{2} (e+f x)\right ) \csc \left (\frac {1}{2} (e+f x)\right )+2 \left (\sin \left (\frac {3 e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \sec ^3\left (\frac {e}{2}\right ) \cot \left (\frac {1}{2} (e+f x)\right ) \csc ^2\left (\frac {1}{2} (e+f x)\right )-15 \cos (e) \sec ^2\left (\frac {e}{2}\right ) \cot ^5\left (\frac {1}{2} (e+f x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )-\left (\tan ^2\left (\frac {e}{2}\right )-1\right ) \cot ^3\left (\frac {1}{2} (e+f x)\right ) \left (15 \left (\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )-4 \tan \left (\frac {e}{2}\right ) \csc ^2\left (\frac {1}{2} (e+f x)\right )\right )\right )}{6 a^2 f \left (\tan \left (\frac {e}{2}\right )-1\right ) \left (\tan \left (\frac {e}{2}\right )+1\right ) (\cos (e+f x)+1)^2 \left (\cot \left (\frac {1}{2} (e+f x)\right )-1\right ) \left (\cot \left (\frac {1}{2} (e+f x)\right )+1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^3*(-1 + Cos[e + f*x])^3*Cot[(e + f*x)/2]*Csc[(e + f*x)/2]^2*(-15*Cos[e]*Cot[(e + f*x)/2]^5*(Log[Cos[(e + f*
x)/2] - Sin[(e + f*x)/2]] - Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]])*Sec[e/2]^2 + 2*Cot[(e + f*x)/2]*Csc[(e +
f*x)/2]^2*Sec[e/2]^3*(-Sin[e/2] + Sin[(3*e)/2]) + 20*Cot[(e + f*x)/2]^2*Csc[(e + f*x)/2]*Sec[e/2]*Sin[(f*x)/2
] - 26*Cot[(e + f*x)/2]^4*Csc[(e + f*x)/2]*Sec[e/2]*Sin[(f*x)/2] - ((-40 + 40*Cos[e] + 78*Cos[f*x] - 80*Cos[e
+ f*x] - 40*Cos[2*(e + f*x)] + 66*Cos[2*e + f*x] + 23*Cos[e + 2*f*x] + 17*Cos[3*e + 2*f*x])*Csc[(e + f*x)/2]^5
*Sec[e/2]^3*Sin[(f*x)/2])/16 - Cot[(e + f*x)/2]^3*(15*(Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - Log[Cos[(e +
f*x)/2] + Sin[(e + f*x)/2]]) - 4*Csc[(e + f*x)/2]^2*Tan[e/2])*(-1 + Tan[e/2]^2)))/(6*a^2*f*(1 + Cos[e + f*x])
^2*(-1 + Cot[(e + f*x)/2])*(1 + Cot[(e + f*x)/2])*(-1 + Tan[e/2])*(1 + Tan[e/2]))

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fricas [A]  time = 0.49, size = 178, normalized size = 1.50 \[ \frac {15 \, {\left (c^{3} \cos \left (f x + e\right )^{3} + 2 \, c^{3} \cos \left (f x + e\right )^{2} + c^{3} \cos \left (f x + e\right )\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 15 \, {\left (c^{3} \cos \left (f x + e\right )^{3} + 2 \, c^{3} \cos \left (f x + e\right )^{2} + c^{3} \cos \left (f x + e\right )\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (23 \, c^{3} \cos \left (f x + e\right )^{2} + 34 \, c^{3} \cos \left (f x + e\right ) + 3 \, c^{3}\right )} \sin \left (f x + e\right )}{6 \, {\left (a^{2} f \cos \left (f x + e\right )^{3} + 2 \, a^{2} f \cos \left (f x + e\right )^{2} + a^{2} f \cos \left (f x + e\right )\right )}} \]

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))^2,x, algorithm="fricas")

[Out]

1/6*(15*(c^3*cos(f*x + e)^3 + 2*c^3*cos(f*x + e)^2 + c^3*cos(f*x + e))*log(sin(f*x + e) + 1) - 15*(c^3*cos(f*x
+ e)^3 + 2*c^3*cos(f*x + e)^2 + c^3*cos(f*x + e))*log(-sin(f*x + e) + 1) - 2*(23*c^3*cos(f*x + e)^2 + 34*c^3*
cos(f*x + e) + 3*c^3)*sin(f*x + e))/(a^2*f*cos(f*x + e)^3 + 2*a^2*f*cos(f*x + e)^2 + a^2*f*cos(f*x + e))

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

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))^2,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)-2/f*((2/3*tan((f*x+exp(1))/2)^3*c^3*a^4+4*tan((f*x+exp(1))/2)
*c^3*a^4)/a^6-tan((f*x+exp(1))/2)*c^3/a^2/(tan((f*x+exp(1))/2)^2-1)+5*c^3*1/2/a^2*ln(abs(tan((f*x+exp(1))/2)-1
))-5*c^3*1/2/a^2*ln(abs(tan((f*x+exp(1))/2)+1)))

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maple [A]  time = 0.68, size = 136, normalized size = 1.14 \[ -\frac {4 c^{3} \left (\tan ^{3}\left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{3 f \,a^{2}}-\frac {8 c^{3} \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{f \,a^{2}}+\frac {c^{3}}{f \,a^{2} \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right )}-\frac {5 c^{3} \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right )}{f \,a^{2}}+\frac {c^{3}}{f \,a^{2} \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right )}+\frac {5 c^{3} \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right )}{f \,a^{2}} \]

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))^2,x)

[Out]

-4/3/f*c^3/a^2*tan(1/2*e+1/2*f*x)^3-8/f*c^3/a^2*tan(1/2*e+1/2*f*x)+1/f*c^3/a^2/(tan(1/2*e+1/2*f*x)-1)-5/f*c^3/
a^2*ln(tan(1/2*e+1/2*f*x)-1)+1/f*c^3/a^2/(tan(1/2*e+1/2*f*x)+1)+5/f*c^3/a^2*ln(tan(1/2*e+1/2*f*x)+1)

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maxima [B]  time = 0.34, size = 341, normalized size = 2.87 \[ -\frac {c^{3} {\left (\frac {\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{2}} + \frac {12 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{2}} + \frac {12 \, \sin \left (f x + e\right )}{{\left (a^{2} - \frac {a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}}\right )} + 3 \, c^{3} {\left (\frac {\frac {9 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}{a^{2}} - \frac {6 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{2}} + \frac {6 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{2}}\right )} + \frac {3 \, c^{3} {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2}} - \frac {c^{3} {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, f} \]

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))^2,x, algorithm="maxima")

[Out]

-1/6*(c^3*((15*sin(f*x + e)/(cos(f*x + e) + 1) + sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2 - 12*log(sin(f*x + e
)/(cos(f*x + e) + 1) + 1)/a^2 + 12*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/a^2 + 12*sin(f*x + e)/((a^2 - a^2*
sin(f*x + e)^2/(cos(f*x + e) + 1)^2)*(cos(f*x + e) + 1))) + 3*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) + 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
- 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

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mupad [B]  time = 1.65, size = 104, normalized size = 0.87 \[ \frac {10\,c^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{a^2\,f}-\frac {4\,c^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{3\,a^2\,f}-\frac {8\,c^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{a^2\,f}+\frac {2\,c^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{f\,\left (a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-a^2\right )} \]

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))^2),x)

[Out]

(10*c^3*atanh(tan(e/2 + (f*x)/2)))/(a^2*f) - (4*c^3*tan(e/2 + (f*x)/2)^3)/(3*a^2*f) - (8*c^3*tan(e/2 + (f*x)/2
))/(a^2*f) + (2*c^3*tan(e/2 + (f*x)/2))/(f*(a^2*tan(e/2 + (f*x)/2)^2 - a^2))

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

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))**2,x)

[Out]

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

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