3.1.15 \(\int (a+a \sec (e+f x))^3 (c-c \sec (e+f x)) \, dx\) [15]

Optimal. Leaf size=77 \[ a^3 c x+\frac {a^3 c \tanh ^{-1}(\sin (e+f x))}{f}-\frac {a^3 c \tan (e+f x)}{f}-\frac {a^3 c \sec (e+f x) \tan (e+f x)}{f}-\frac {a^3 c \tan ^3(e+f x)}{3 f} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3989, 3971, 3554, 8, 2691, 3855, 2687, 30} \begin {gather*} -\frac {a^3 c \tan ^3(e+f x)}{3 f}-\frac {a^3 c \tan (e+f x)}{f}+\frac {a^3 c \tanh ^{-1}(\sin (e+f x))}{f}-\frac {a^3 c \tan (e+f x) \sec (e+f x)}{f}+a^3 c x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2687

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2691

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a*Sec[e +
f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Dist[b^2*((n - 1)/(m + n - 1)), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3855

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

Rule 3971

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[ExpandI
ntegrand[(e*Cot[c + d*x])^m, (a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0]

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

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Mathematica [A]
time = 0.49, size = 101, normalized size = 1.31 \begin {gather*} \frac {a^3 c \sec ^3(e+f x) \left (9 (e+f x) \cos (e+f x)+12 \tanh ^{-1}(\sin (e+f x)) \cos ^3(e+f x)+3 e \cos (3 (e+f x))+3 f x \cos (3 (e+f x))-6 \sin (e+f x)-6 \sin (2 (e+f x))-2 \sin (3 (e+f x))\right )}{12 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*c*Sec[e + f*x]^3*(9*(e + f*x)*Cos[e + f*x] + 12*ArcTanh[Sin[e + f*x]]*Cos[e + f*x]^3 + 3*e*Cos[3*(e + f*x
)] + 3*f*x*Cos[3*(e + f*x)] - 6*Sin[e + f*x] - 6*Sin[2*(e + f*x)] - 2*Sin[3*(e + f*x)]))/(12*f)

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Maple [A]
time = 0.06, size = 96, normalized size = 1.25

method	result	size
derivativedivides	\(\frac {a^{3} c \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (f x +e \right )\right )}{3}\right ) \tan \left (f x +e \right )-2 a^{3} c \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )+2 a^{3} c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )+a^{3} c \left (f x +e \right )}{f}\)	\(96\)
default	\(\frac {a^{3} c \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (f x +e \right )\right )}{3}\right ) \tan \left (f x +e \right )-2 a^{3} c \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )+2 a^{3} c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )+a^{3} c \left (f x +e \right )}{f}\)	\(96\)
risch	\(a^{3} c x +\frac {2 i a^{3} c \left (3 \,{\mathrm e}^{5 i \left (f x +e \right )}-6 \,{\mathrm e}^{2 i \left (f x +e \right )}-3 \,{\mathrm e}^{i \left (f x +e \right )}-2\right )}{3 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{3}}+\frac {a^{3} c \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{f}-\frac {a^{3} c \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{f}\)	\(109\)
norman	\(\frac {a^{3} c x \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-a^{3} c x +\frac {4 a^{3} c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}-\frac {4 a^{3} c \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f}+3 a^{3} c x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-3 a^{3} c x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}+\frac {a^{3} c \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{f}-\frac {a^{3} c \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{f}\)	\(158\)

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)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 116, normalized size = 1.51 \begin {gather*} -\frac {2 \, {\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{3} c - 6 \, {\left (f x + e\right )} a^{3} c - 3 \, a^{3} c {\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 )} - 12 \, a^{3} c \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right )}{6 \, 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)),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 2.71, size = 127, normalized size = 1.65 \begin {gather*} \frac {6 \, a^{3} c f x \cos \left (f x + e\right )^{3} + 3 \, a^{3} c \cos \left (f x + e\right )^{3} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \, a^{3} c \cos \left (f x + e\right )^{3} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (2 \, a^{3} c \cos \left (f x + e\right )^{2} + 3 \, a^{3} c \cos \left (f x + e\right ) + a^{3} c\right )} \sin \left (f x + e\right )}{6 \, f \cos \left (f x + e\right )^{3}} \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)),x, algorithm="fricas")

[Out]

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

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

[Out]

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

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Giac [A]
time = 0.54, size = 104, normalized size = 1.35 \begin {gather*} \frac {3 \, {\left (f x + e\right )} a^{3} c + 3 \, a^{3} c \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right ) - 3 \, a^{3} c \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right ) - \frac {4 \, {\left (a^{3} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 3 \, a^{3} c \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 )}^{3}}}{3 \, 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)),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 1.55, size = 104, normalized size = 1.35 \begin {gather*} \frac {4\,a^3\,c\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )-\frac {4\,a^3\,c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{3}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )}+a^3\,c\,x+\frac {2\,a^3\,c\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{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)),x)

[Out]

(4*a^3*c*tan(e/2 + (f*x)/2) - (4*a^3*c*tan(e/2 + (f*x)/2)^3)/3)/(f*(3*tan(e/2 + (f*x)/2)^2 - 3*tan(e/2 + (f*x)
/2)^4 + tan(e/2 + (f*x)/2)^6 - 1)) + a^3*c*x + (2*a^3*c*atanh(tan(e/2 + (f*x)/2)))/f

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