3.1.14 \(\int (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^2 \, dx\) [14]

Optimal. Leaf size=97 \[ a^3 c^2 x+\frac {3 a^3 c^2 \tanh ^{-1}(\sin (e+f x))}{8 f}-\frac {c^2 \left (8 a^3+3 a^3 \sec (e+f x)\right ) \tan (e+f x)}{8 f}+\frac {c^2 \left (4 a^3+3 a^3 \sec (e+f x)\right ) \tan ^3(e+f x)}{12 f} \]

[Out]

a^3*c^2*x+3/8*a^3*c^2*arctanh(sin(f*x+e))/f-1/8*c^2*(8*a^3+3*a^3*sec(f*x+e))*tan(f*x+e)/f+1/12*c^2*(4*a^3+3*a^
3*sec(f*x+e))*tan(f*x+e)^3/f

________________________________________________________________________________________

Rubi [A]
time = 0.09, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {3989, 3966, 3855} \begin {gather*} \frac {3 a^3 c^2 \tanh ^{-1}(\sin (e+f x))}{8 f}+\frac {c^2 \tan ^3(e+f x) \left (3 a^3 \sec (e+f x)+4 a^3\right )}{12 f}-\frac {c^2 \tan (e+f x) \left (3 a^3 \sec (e+f x)+8 a^3\right )}{8 f}+a^3 c^2 x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

a^3*c^2*x + (3*a^3*c^2*ArcTanh[Sin[e + f*x]])/(8*f) - (c^2*(8*a^3 + 3*a^3*Sec[e + f*x])*Tan[e + f*x])/(8*f) +
(c^2*(4*a^3 + 3*a^3*Sec[e + f*x])*Tan[e + f*x]^3)/(12*f)

Rule 3855

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

Rule 3966

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(-e)*(e*Cot
[c + d*x])^(m - 1)*((a*m + b*(m - 1)*Csc[c + d*x])/(d*m*(m - 1))), x] - Dist[e^2/m, Int[(e*Cot[c + d*x])^(m -
2)*(a*m + b*(m - 1)*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[m, 1]

Rule 3989

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Di
st[((-a)*c)^m, Int[Cot[e + f*x]^(2*m)*(c + d*Csc[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x]
&& EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && RationalQ[n] &&  !(IntegerQ[n] && GtQ[m - n, 0])

Rubi steps

\begin {align*} \int (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^2 \, dx &=\left (a^2 c^2\right ) \int (a+a \sec (e+f x)) \tan ^4(e+f x) \, dx\\ &=\frac {c^2 \left (4 a^3+3 a^3 \sec (e+f x)\right ) \tan ^3(e+f x)}{12 f}-\frac {1}{4} \left (a^2 c^2\right ) \int (4 a+3 a \sec (e+f x)) \tan ^2(e+f x) \, dx\\ &=-\frac {c^2 \left (8 a^3+3 a^3 \sec (e+f x)\right ) \tan (e+f x)}{8 f}+\frac {c^2 \left (4 a^3+3 a^3 \sec (e+f x)\right ) \tan ^3(e+f x)}{12 f}+\frac {1}{8} \left (a^2 c^2\right ) \int (8 a+3 a \sec (e+f x)) \, dx\\ &=a^3 c^2 x-\frac {c^2 \left (8 a^3+3 a^3 \sec (e+f x)\right ) \tan (e+f x)}{8 f}+\frac {c^2 \left (4 a^3+3 a^3 \sec (e+f x)\right ) \tan ^3(e+f x)}{12 f}+\frac {1}{8} \left (3 a^3 c^2\right ) \int \sec (e+f x) \, dx\\ &=a^3 c^2 x+\frac {3 a^3 c^2 \tanh ^{-1}(\sin (e+f x))}{8 f}-\frac {c^2 \left (8 a^3+3 a^3 \sec (e+f x)\right ) \tan (e+f x)}{8 f}+\frac {c^2 \left (4 a^3+3 a^3 \sec (e+f x)\right ) \tan ^3(e+f x)}{12 f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.86, size = 122, normalized size = 1.26 \begin {gather*} \frac {a^3 c^2 \sec ^4(e+f x) \left (72 e+72 f x+72 \tanh ^{-1}(\sin (e+f x)) \cos ^4(e+f x)+96 (e+f x) \cos (2 (e+f x))+24 e \cos (4 (e+f x))+24 f x \cos (4 (e+f x))+18 \sin (e+f x)-32 \sin (2 (e+f x))-30 \sin (3 (e+f x))-32 \sin (4 (e+f x))\right )}{192 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*c^2*Sec[e + f*x]^4*(72*e + 72*f*x + 72*ArcTanh[Sin[e + f*x]]*Cos[e + f*x]^4 + 96*(e + f*x)*Cos[2*(e + f*x
)] + 24*e*Cos[4*(e + f*x)] + 24*f*x*Cos[4*(e + f*x)] + 18*Sin[e + f*x] - 32*Sin[2*(e + f*x)] - 30*Sin[3*(e + f
*x)] - 32*Sin[4*(e + f*x)]))/(192*f)

________________________________________________________________________________________

Maple [A]
time = 0.08, size = 169, normalized size = 1.74

method	result	size
risch	\(a^{3} c^{2} x +\frac {i c^{2} a^{3} \left (15 \,{\mathrm e}^{7 i \left (f x +e \right )}-48 \,{\mathrm e}^{6 i \left (f x +e \right )}-9 \,{\mathrm e}^{5 i \left (f x +e \right )}-96 \,{\mathrm e}^{4 i \left (f x +e \right )}+9 \,{\mathrm e}^{3 i \left (f x +e \right )}-80 \,{\mathrm e}^{2 i \left (f x +e \right )}-15 \,{\mathrm e}^{i \left (f x +e \right )}-32\right )}{12 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4}}+\frac {3 c^{2} a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{8 f}-\frac {3 c^{2} a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{8 f}\)	\(162\)
derivativedivides	\(\frac {c^{2} a^{3} \left (-\left (-\frac {\left (\sec ^{3}\left (f x +e \right )\right )}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )-c^{2} a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (f x +e \right )\right )}{3}\right ) \tan \left (f x +e \right )-2 c^{2} a^{3} \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )-2 c^{2} a^{3} \tan \left (f x +e \right )+c^{2} a^{3} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )+c^{2} a^{3} \left (f x +e \right )}{f}\)	\(169\)
default	\(\frac {c^{2} a^{3} \left (-\left (-\frac {\left (\sec ^{3}\left (f x +e \right )\right )}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )-c^{2} a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (f x +e \right )\right )}{3}\right ) \tan \left (f x +e \right )-2 c^{2} a^{3} \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )-2 c^{2} a^{3} \tan \left (f x +e \right )+c^{2} a^{3} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )+c^{2} a^{3} \left (f x +e \right )}{f}\)	\(169\)
norman	\(\frac {a^{3} c^{2} x +a^{3} c^{2} x \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-4 a^{3} c^{2} x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+6 a^{3} c^{2} x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-4 a^{3} c^{2} x \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {11 c^{2} a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{4 f}+\frac {137 c^{2} a^{3} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{12 f}-\frac {71 c^{2} a^{3} \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{12 f}+\frac {5 c^{2} a^{3} \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4 f}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {3 c^{2} a^{3} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{8 f}+\frac {3 c^{2} a^{3} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{8 f}\)	\(238\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (96) = 192\).
time = 0.31, size = 219, normalized size = 2.26 \begin {gather*} \frac {16 \, {\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{3} c^{2} + 48 \, {\left (f x + e\right )} a^{3} c^{2} - 3 \, a^{3} c^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (f x + e\right )^{3} - 5 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1} - 3 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 3 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} + 24 \, a^{3} c^{2} {\left (\frac {2 \, \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1} - \log \left (\sin \left (f x + e\right ) + 1\right ) + \log \left (\sin \left (f x + e\right ) - 1\right )\right )} + 48 \, a^{3} c^{2} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) - 96 \, a^{3} c^{2} \tan \left (f x + e\right )}{48 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(16*(tan(f*x + e)^3 + 3*tan(f*x + e))*a^3*c^2 + 48*(f*x + e)*a^3*c^2 - 3*a^3*c^2*(2*(3*sin(f*x + e)^3 - 5
*sin(f*x + e))/(sin(f*x + e)^4 - 2*sin(f*x + e)^2 + 1) - 3*log(sin(f*x + e) + 1) + 3*log(sin(f*x + e) - 1)) +
24*a^3*c^2*(2*sin(f*x + e)/(sin(f*x + e)^2 - 1) - log(sin(f*x + e) + 1) + log(sin(f*x + e) - 1)) + 48*a^3*c^2*
log(sec(f*x + e) + tan(f*x + e)) - 96*a^3*c^2*tan(f*x + e))/f

________________________________________________________________________________________

Fricas [A]
time = 3.37, size = 157, normalized size = 1.62 \begin {gather*} \frac {48 \, a^{3} c^{2} f x \cos \left (f x + e\right )^{4} + 9 \, a^{3} c^{2} \cos \left (f x + e\right )^{4} \log \left (\sin \left (f x + e\right ) + 1\right ) - 9 \, a^{3} c^{2} \cos \left (f x + e\right )^{4} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (32 \, a^{3} c^{2} \cos \left (f x + e\right )^{3} + 15 \, a^{3} c^{2} \cos \left (f x + e\right )^{2} - 8 \, a^{3} c^{2} \cos \left (f x + e\right ) - 6 \, a^{3} c^{2}\right )} \sin \left (f x + e\right )}{48 \, f \cos \left (f x + e\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(48*a^3*c^2*f*x*cos(f*x + e)^4 + 9*a^3*c^2*cos(f*x + e)^4*log(sin(f*x + e) + 1) - 9*a^3*c^2*cos(f*x + e)^
4*log(-sin(f*x + e) + 1) - 2*(32*a^3*c^2*cos(f*x + e)^3 + 15*a^3*c^2*cos(f*x + e)^2 - 8*a^3*c^2*cos(f*x + e) -
6*a^3*c^2)*sin(f*x + e))/(f*cos(f*x + e)^4)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{3} c^{2} \left (\int 1\, dx + \int \sec {\left (e + f x \right )}\, dx + \int \left (- 2 \sec ^{2}{\left (e + f x \right )}\right )\, dx + \int \left (- 2 \sec ^{3}{\left (e + f x \right )}\right )\, dx + \int \sec ^{4}{\left (e + f x \right )}\, dx + \int \sec ^{5}{\left (e + f x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**3*c**2*(Integral(1, x) + Integral(sec(e + f*x), x) + Integral(-2*sec(e + f*x)**2, x) + Integral(-2*sec(e +
f*x)**3, x) + Integral(sec(e + f*x)**4, x) + Integral(sec(e + f*x)**5, x))

________________________________________________________________________________________

Giac [A]
time = 0.55, size = 153, normalized size = 1.58 \begin {gather*} \frac {24 \, {\left (f x + e\right )} a^{3} c^{2} + 9 \, a^{3} c^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right ) - 9 \, a^{3} c^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right ) + \frac {2 \, {\left (15 \, a^{3} c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} - 71 \, a^{3} c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 137 \, a^{3} c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 33 \, a^{3} c^{2} \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 )}^{4}}}{24 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(24*(f*x + e)*a^3*c^2 + 9*a^3*c^2*log(abs(tan(1/2*f*x + 1/2*e) + 1)) - 9*a^3*c^2*log(abs(tan(1/2*f*x + 1/
2*e) - 1)) + 2*(15*a^3*c^2*tan(1/2*f*x + 1/2*e)^7 - 71*a^3*c^2*tan(1/2*f*x + 1/2*e)^5 + 137*a^3*c^2*tan(1/2*f*
x + 1/2*e)^3 - 33*a^3*c^2*tan(1/2*f*x + 1/2*e))/(tan(1/2*f*x + 1/2*e)^2 - 1)^4)/f

________________________________________________________________________________________

Mupad [B]
time = 2.13, size = 163, normalized size = 1.68 \begin {gather*} \frac {\frac {5\,a^3\,c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7}{4}-\frac {71\,a^3\,c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{12}+\frac {137\,a^3\,c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{12}-\frac {11\,a^3\,c^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{4}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}+a^3\,c^2\,x+\frac {3\,a^3\,c^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{4\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((137*a^3*c^2*tan(e/2 + (f*x)/2)^3)/12 - (71*a^3*c^2*tan(e/2 + (f*x)/2)^5)/12 + (5*a^3*c^2*tan(e/2 + (f*x)/2)^
7)/4 - (11*a^3*c^2*tan(e/2 + (f*x)/2))/4)/(f*(6*tan(e/2 + (f*x)/2)^4 - 4*tan(e/2 + (f*x)/2)^2 - 4*tan(e/2 + (f
*x)/2)^6 + tan(e/2 + (f*x)/2)^8 + 1)) + a^3*c^2*x + (3*a^3*c^2*atanh(tan(e/2 + (f*x)/2)))/(4*f)

________________________________________________________________________________________