3.13 \(\int (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^3 \, dx\)

Optimal. Leaf size=68 \[ -\frac {a^3 c^3 \tan ^5(e+f x)}{5 f}+\frac {a^3 c^3 \tan ^3(e+f x)}{3 f}-\frac {a^3 c^3 \tan (e+f x)}{f}+a^3 c^3 x \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 68, 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 = {3904, 3473, 8} \[ -\frac {a^3 c^3 \tan ^5(e+f x)}{5 f}+\frac {a^3 c^3 \tan ^3(e+f x)}{3 f}-\frac {a^3 c^3 \tan (e+f x)}{f}+a^3 c^3 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^3*c^3*x - (a^3*c^3*Tan[e + f*x])/f + (a^3*c^3*Tan[e + f*x]^3)/(3*f) - (a^3*c^3*Tan[e + f*x]^5)/(5*f)

Rule 8

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

Rule 3473

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 3904

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))^3 \, dx &=-\left (\left (a^3 c^3\right ) \int \tan ^6(e+f x) \, dx\right )\\ &=-\frac {a^3 c^3 \tan ^5(e+f x)}{5 f}+\left (a^3 c^3\right ) \int \tan ^4(e+f x) \, dx\\ &=\frac {a^3 c^3 \tan ^3(e+f x)}{3 f}-\frac {a^3 c^3 \tan ^5(e+f x)}{5 f}-\left (a^3 c^3\right ) \int \tan ^2(e+f x) \, dx\\ &=-\frac {a^3 c^3 \tan (e+f x)}{f}+\frac {a^3 c^3 \tan ^3(e+f x)}{3 f}-\frac {a^3 c^3 \tan ^5(e+f x)}{5 f}+\left (a^3 c^3\right ) \int 1 \, dx\\ &=a^3 c^3 x-\frac {a^3 c^3 \tan (e+f x)}{f}+\frac {a^3 c^3 \tan ^3(e+f x)}{3 f}-\frac {a^3 c^3 \tan ^5(e+f x)}{5 f}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 61, normalized size = 0.90 \[ -a^3 c^3 \left (-\frac {\tan ^{-1}(\tan (e+f x))}{f}+\frac {\tan ^5(e+f x)}{5 f}-\frac {\tan ^3(e+f x)}{3 f}+\frac {\tan (e+f x)}{f}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*c^3*(-(ArcTan[Tan[e + f*x]]/f) + Tan[e + f*x]/f - Tan[e + f*x]^3/(3*f) + Tan[e + f*x]^5/(5*f)))

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fricas [A]  time = 0.46, size = 81, normalized size = 1.19 \[ \frac {15 \, a^{3} c^{3} f x \cos \left (f x + e\right )^{5} - {\left (23 \, a^{3} c^{3} \cos \left (f x + e\right )^{4} - 11 \, a^{3} c^{3} \cos \left (f x + e\right )^{2} + 3 \, a^{3} c^{3}\right )} \sin \left (f x + e\right )}{15 \, f \cos \left (f x + e\right )^{5}} \]

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))^3,x, algorithm="fricas")

[Out]

1/15*(15*a^3*c^3*f*x*cos(f*x + e)^5 - (23*a^3*c^3*cos(f*x + e)^4 - 11*a^3*c^3*cos(f*x + e)^2 + 3*a^3*c^3)*sin(
f*x + e))/(f*cos(f*x + e)^5)

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giac [A]  time = 0.55, size = 69, normalized size = 1.01 \[ -\frac {3 \, a^{3} c^{3} \tan \left (f x + e\right )^{5} - 5 \, a^{3} c^{3} \tan \left (f x + e\right )^{3} - 15 \, {\left (f x + e\right )} a^{3} c^{3} + 15 \, a^{3} c^{3} \tan \left (f x + e\right )}{15 \, f} \]

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))^3,x, algorithm="giac")

[Out]

-1/15*(3*a^3*c^3*tan(f*x + e)^5 - 5*a^3*c^3*tan(f*x + e)^3 - 15*(f*x + e)*a^3*c^3 + 15*a^3*c^3*tan(f*x + e))/f

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maple [A]  time = 1.06, size = 93, normalized size = 1.37 \[ \frac {-3 a^{3} c^{3} \tan \left (f x +e \right )+\left (f x +e \right ) a^{3} c^{3}-3 a^{3} c^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (f x +e \right )\right )}{3}\right ) \tan \left (f x +e \right )+a^{3} c^{3} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (f x +e \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (f x +e \right )\right )}{15}\right ) \tan \left (f x +e \right )}{f} \]

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))^3,x)

[Out]

1/f*(-3*a^3*c^3*tan(f*x+e)+(f*x+e)*a^3*c^3-3*a^3*c^3*(-2/3-1/3*sec(f*x+e)^2)*tan(f*x+e)+a^3*c^3*(-8/15-1/5*sec
(f*x+e)^4-4/15*sec(f*x+e)^2)*tan(f*x+e))

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maxima [A]  time = 0.32, size = 94, normalized size = 1.38 \[ -\frac {{\left (3 \, \tan \left (f x + e\right )^{5} + 10 \, \tan \left (f x + e\right )^{3} + 15 \, \tan \left (f x + e\right )\right )} a^{3} c^{3} - 15 \, {\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{3} c^{3} - 15 \, {\left (f x + e\right )} a^{3} c^{3} + 45 \, a^{3} c^{3} \tan \left (f x + e\right )}{15 \, f} \]

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))^3,x, algorithm="maxima")

[Out]

-1/15*((3*tan(f*x + e)^5 + 10*tan(f*x + e)^3 + 15*tan(f*x + e))*a^3*c^3 - 15*(tan(f*x + e)^3 + 3*tan(f*x + e))
*a^3*c^3 - 15*(f*x + e)*a^3*c^3 + 45*a^3*c^3*tan(f*x + e))/f

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mupad [B]  time = 4.96, size = 122, normalized size = 1.79 \[ a^3\,c^3\,x+\frac {2\,a^3\,c^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9-\frac {32\,a^3\,c^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7}{3}+\frac {356\,a^3\,c^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{15}-\frac {32\,a^3\,c^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{3}+2\,a^3\,c^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{f\,{\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )}^5} \]

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))^3,x)

[Out]

a^3*c^3*x + ((356*a^3*c^3*tan(e/2 + (f*x)/2)^5)/15 - (32*a^3*c^3*tan(e/2 + (f*x)/2)^3)/3 - (32*a^3*c^3*tan(e/2
+ (f*x)/2)^7)/3 + 2*a^3*c^3*tan(e/2 + (f*x)/2)^9 + 2*a^3*c^3*tan(e/2 + (f*x)/2))/(f*(tan(e/2 + (f*x)/2)^2 - 1
)^5)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - a^{3} c^{3} \left (\int \left (-1\right )\, dx + \int 3 \sec ^{2}{\left (e + f x \right )}\, dx + \int \left (- 3 \sec ^{4}{\left (e + f x \right )}\right )\, dx + \int \sec ^{6}{\left (e + f x \right )}\, dx\right ) \]

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))**3,x)

[Out]

-a**3*c**3*(Integral(-1, x) + Integral(3*sec(e + f*x)**2, x) + Integral(-3*sec(e + f*x)**4, x) + Integral(sec(
e + f*x)**6, x))

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