3.1.12 \(\int \sec (e+f x) (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^3 \, dx\) [12]

Optimal. Leaf size=94 \[ \frac {3 a^2 c^3 \tanh ^{-1}(\sin (e+f x))}{8 f}-\frac {3 a^2 c^3 \sec (e+f x) \tan (e+f x)}{8 f}+\frac {a^2 c^3 \sec (e+f x) \tan ^3(e+f x)}{4 f}-\frac {a^2 c^3 \tan ^5(e+f x)}{5 f} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {4043, 2691, 3855, 2687, 30} \begin {gather*} -\frac {a^2 c^3 \tan ^5(e+f x)}{5 f}+\frac {3 a^2 c^3 \tanh ^{-1}(\sin (e+f x))}{8 f}+\frac {a^2 c^3 \tan ^3(e+f x) \sec (e+f x)}{4 f}-\frac {3 a^2 c^3 \tan (e+f x) \sec (e+f x)}{8 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a^2*c^3*ArcTanh[Sin[e + f*x]])/(8*f) - (3*a^2*c^3*Sec[e + f*x]*Tan[e + f*x])/(8*f) + (a^2*c^3*Sec[e + f*x]*
Tan[e + f*x]^3)/(4*f) - (a^2*c^3*Tan[e + f*x]^5)/(5*f)

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 3855

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

Rule 4043

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)
)^(n_.), x_Symbol] :> Dist[((-a)*c)^m, Int[ExpandTrig[csc[e + f*x]*cot[e + f*x]^(2*m), (c + d*csc[e + f*x])^(n
- m), x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegersQ[m,
n] && GeQ[n - m, 0] && GtQ[m*n, 0]

Rubi steps

\begin {align*} \int \sec (e+f x) (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^3 \, dx &=\left (a^2 c^2\right ) \int \left (c \sec (e+f x) \tan ^4(e+f x)-c \sec ^2(e+f x) \tan ^4(e+f x)\right ) \, dx\\ &=\left (a^2 c^3\right ) \int \sec (e+f x) \tan ^4(e+f x) \, dx-\left (a^2 c^3\right ) \int \sec ^2(e+f x) \tan ^4(e+f x) \, dx\\ &=\frac {a^2 c^3 \sec (e+f x) \tan ^3(e+f x)}{4 f}-\frac {1}{4} \left (3 a^2 c^3\right ) \int \sec (e+f x) \tan ^2(e+f x) \, dx-\frac {\left (a^2 c^3\right ) \text {Subst}\left (\int x^4 \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {3 a^2 c^3 \sec (e+f x) \tan (e+f x)}{8 f}+\frac {a^2 c^3 \sec (e+f x) \tan ^3(e+f x)}{4 f}-\frac {a^2 c^3 \tan ^5(e+f x)}{5 f}+\frac {1}{8} \left (3 a^2 c^3\right ) \int \sec (e+f x) \, dx\\ &=\frac {3 a^2 c^3 \tanh ^{-1}(\sin (e+f x))}{8 f}-\frac {3 a^2 c^3 \sec (e+f x) \tan (e+f x)}{8 f}+\frac {a^2 c^3 \sec (e+f x) \tan ^3(e+f x)}{4 f}-\frac {a^2 c^3 \tan ^5(e+f x)}{5 f}\\ \end {align*}

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Mathematica [A]
time = 0.77, size = 82, normalized size = 0.87 \begin {gather*} \frac {a^2 c^3 \left (120 \tanh ^{-1}(\sin (e+f x))-\sec ^5(e+f x) (40 \sin (e+f x)+10 \sin (2 (e+f x))-20 \sin (3 (e+f x))+25 \sin (4 (e+f x))+4 \sin (5 (e+f x)))\right )}{320 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*c^3*(120*ArcTanh[Sin[e + f*x]] - Sec[e + f*x]^5*(40*Sin[e + f*x] + 10*Sin[2*(e + f*x)] - 20*Sin[3*(e + f*
x)] + 25*Sin[4*(e + f*x)] + 4*Sin[5*(e + f*x)])))/(320*f)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(191\) vs. \(2(86)=172\).
time = 0.27, size = 192, normalized size = 2.04

method	result	size
risch	\(\frac {i c^{3} a^{2} \left (25 \,{\mathrm e}^{9 i \left (f x +e \right )}-40 \,{\mathrm e}^{8 i \left (f x +e \right )}+10 \,{\mathrm e}^{7 i \left (f x +e \right )}-80 \,{\mathrm e}^{4 i \left (f x +e \right )}-10 \,{\mathrm e}^{3 i \left (f x +e \right )}-25 \,{\mathrm e}^{i \left (f x +e \right )}-8\right )}{20 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{5}}+\frac {3 c^{3} a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{8 f}-\frac {3 c^{3} a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{8 f}\)	\(143\)
norman	\(\frac {\frac {3 c^{3} a^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{4 f}-\frac {7 c^{3} a^{2} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 f}+\frac {32 c^{3} a^{2} \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 f}+\frac {7 c^{3} a^{2} \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 f}-\frac {3 c^{3} a^{2} \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4 f}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {3 c^{3} a^{2} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{8 f}+\frac {3 c^{3} a^{2} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{8 f}\)	\(173\)
derivativedivides	\(\frac {c^{3} a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (f x +e \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (f x +e \right )\right )}{15}\right ) \tan \left (f x +e \right )+c^{3} a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (f x +e \right )\right )}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )-2 c^{3} a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (f x +e \right )\right )}{3}\right ) \tan \left (f x +e \right )-2 c^{3} a^{2} \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )-c^{3} a^{2} \tan \left (f x +e \right )+c^{3} a^{2} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}\)	\(192\)
default	\(\frac {c^{3} a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (f x +e \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (f x +e \right )\right )}{15}\right ) \tan \left (f x +e \right )+c^{3} a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (f x +e \right )\right )}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )-2 c^{3} a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (f x +e \right )\right )}{3}\right ) \tan \left (f x +e \right )-2 c^{3} a^{2} \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )-c^{3} a^{2} \tan \left (f x +e \right )+c^{3} a^{2} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}\)	\(192\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 245 vs. \(2 (92) = 184\).
time = 0.30, size = 245, normalized size = 2.61 \begin {gather*} -\frac {16 \, {\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^{2} c^{3} - 160 \, {\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{2} c^{3} + 15 \, a^{2} c^{3} {\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 )} - 120 \, a^{2} c^{3} {\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 )} - 240 \, a^{2} c^{3} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) + 240 \, a^{2} c^{3} \tan \left (f x + e\right )}{240 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*(16*(3*tan(f*x + e)^5 + 10*tan(f*x + e)^3 + 15*tan(f*x + e))*a^2*c^3 - 160*(tan(f*x + e)^3 + 3*tan(f*x
+ e))*a^2*c^3 + 15*a^2*c^3*(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)) - 120*a^2*c^3*(2*sin(f*x + e)/(sin(f*x + e)^2 - 1) - log(sin(
f*x + e) + 1) + log(sin(f*x + e) - 1)) - 240*a^2*c^3*log(sec(f*x + e) + tan(f*x + e)) + 240*a^2*c^3*tan(f*x +
e))/f

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Fricas [A]
time = 2.37, size = 155, normalized size = 1.65 \begin {gather*} \frac {15 \, a^{2} c^{3} \cos \left (f x + e\right )^{5} \log \left (\sin \left (f x + e\right ) + 1\right ) - 15 \, a^{2} c^{3} \cos \left (f x + e\right )^{5} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (8 \, a^{2} c^{3} \cos \left (f x + e\right )^{4} + 25 \, a^{2} c^{3} \cos \left (f x + e\right )^{3} - 16 \, a^{2} c^{3} \cos \left (f x + e\right )^{2} - 10 \, a^{2} c^{3} \cos \left (f x + e\right ) + 8 \, a^{2} c^{3}\right )} \sin \left (f x + e\right )}{80 \, f \cos \left (f x + e\right )^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/80*(15*a^2*c^3*cos(f*x + e)^5*log(sin(f*x + e) + 1) - 15*a^2*c^3*cos(f*x + e)^5*log(-sin(f*x + e) + 1) - 2*(
8*a^2*c^3*cos(f*x + e)^4 + 25*a^2*c^3*cos(f*x + e)^3 - 16*a^2*c^3*cos(f*x + e)^2 - 10*a^2*c^3*cos(f*x + e) + 8
*a^2*c^3)*sin(f*x + e))/(f*cos(f*x + e)^5)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - a^{2} c^{3} \left (\int \left (- \sec {\left (e + f x \right )}\right )\, dx + \int \sec ^{2}{\left (e + f x \right )}\, dx + \int 2 \sec ^{3}{\left (e + f x \right )}\, dx + \int \left (- 2 \sec ^{4}{\left (e + f x \right )}\right )\, dx + \int \left (- \sec ^{5}{\left (e + f x \right )}\right )\, dx + \int \sec ^{6}{\left (e + f x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**2*c**3*(Integral(-sec(e + f*x), x) + Integral(sec(e + f*x)**2, x) + Integral(2*sec(e + f*x)**3, x) + Integ
ral(-2*sec(e + f*x)**4, x) + Integral(-sec(e + f*x)**5, x) + Integral(sec(e + f*x)**6, x))

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Giac [A]
time = 0.74, size = 159, normalized size = 1.69 \begin {gather*} \frac {15 \, a^{2} c^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right ) - 15 \, a^{2} c^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right ) - \frac {2 \, {\left (15 \, a^{2} c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{9} - 70 \, a^{2} c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} - 128 \, a^{2} c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 70 \, a^{2} c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 15 \, a^{2} c^{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 )}^{5}}}{40 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40*(15*a^2*c^3*log(abs(tan(1/2*f*x + 1/2*e) + 1)) - 15*a^2*c^3*log(abs(tan(1/2*f*x + 1/2*e) - 1)) - 2*(15*a^
2*c^3*tan(1/2*f*x + 1/2*e)^9 - 70*a^2*c^3*tan(1/2*f*x + 1/2*e)^7 - 128*a^2*c^3*tan(1/2*f*x + 1/2*e)^5 + 70*a^2
*c^3*tan(1/2*f*x + 1/2*e)^3 - 15*a^2*c^3*tan(1/2*f*x + 1/2*e))/(tan(1/2*f*x + 1/2*e)^2 - 1)^5)/f

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Mupad [B]
time = 6.50, size = 187, normalized size = 1.99 \begin {gather*} \frac {-\frac {3\,a^2\,c^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9}{4}+\frac {7\,a^2\,c^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7}{2}+\frac {32\,a^2\,c^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{5}-\frac {7\,a^2\,c^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{2}+\frac {3\,a^2\,c^3\,\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 )}^{10}-5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )}+\frac {3\,a^2\,c^3\,\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))^2*(c - c/cos(e + f*x))^3)/cos(e + f*x),x)

[Out]

((32*a^2*c^3*tan(e/2 + (f*x)/2)^5)/5 - (7*a^2*c^3*tan(e/2 + (f*x)/2)^3)/2 + (7*a^2*c^3*tan(e/2 + (f*x)/2)^7)/2
- (3*a^2*c^3*tan(e/2 + (f*x)/2)^9)/4 + (3*a^2*c^3*tan(e/2 + (f*x)/2))/4)/(f*(5*tan(e/2 + (f*x)/2)^2 - 10*tan(
e/2 + (f*x)/2)^4 + 10*tan(e/2 + (f*x)/2)^6 - 5*tan(e/2 + (f*x)/2)^8 + tan(e/2 + (f*x)/2)^10 - 1)) + (3*a^2*c^3
*atanh(tan(e/2 + (f*x)/2)))/(4*f)

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