3.1.1 \(\int \frac {(a+a \cos (e+f x))^2 \sec ^2(e+f x)}{-c+c \cos (e+f x)} \, dx\) [1]

Optimal. Leaf size=65 \[ -\frac {3 a^2 \tanh ^{-1}(\sin (e+f x))}{c f}+\frac {4 a^2 \sin (e+f x)}{c f (1-\cos (e+f x))}-\frac {a^2 \tan (e+f x)}{c f} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3031, 2727, 3855, 3852, 8} \begin {gather*} -\frac {a^2 \tan (e+f x)}{c f}-\frac {3 a^2 \tanh ^{-1}(\sin (e+f x))}{c f}+\frac {4 a^2 \sin (e+f x)}{c f (1-\cos (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a^2*ArcTanh[Sin[e + f*x]])/(c*f) + (4*a^2*Sin[e + f*x])/(c*f*(1 - Cos[e + f*x])) - (a^2*Tan[e + f*x])/(c*f
)

Rule 8

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

Rule 2727

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 3031

Int[((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) +
(f_.)*(x_)])^(n_), x_Symbol] :> Int[ExpandTrig[(g*sin[e + f*x])^p*(a + b*sin[e + f*x])^m*(c + d*sin[e + f*x])
^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && NeQ[b*c - a*d, 0] && (IntegersQ[m, n] || IntegersQ[m, p
] || IntegersQ[n, p]) && NeQ[p, 2]

Rule 3852

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 3855

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

Rubi steps

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(194\) vs. \(2(65)=130\).
time = 0.92, size = 194, normalized size = 2.98 \begin {gather*} \frac {2 a^2 \sin \left (\frac {1}{2} (e+f x)\right ) \left (4 \csc \left (\frac {e}{2}\right ) \sin \left (\frac {f x}{2}\right )+\sin \left (\frac {1}{2} (e+f x)\right ) \left (-3 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+3 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+\frac {\sin (f x)}{\left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {e}{2}\right )+\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )}\right )\right )}{c f (-1+\cos (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*Sin[(e + f*x)/2]*(4*Csc[e/2]*Sin[(f*x)/2] + Sin[(e + f*x)/2]*(-3*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2
]] + 3*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + Sin[f*x]/((Cos[e/2] - Sin[e/2])*(Cos[e/2] + Sin[e/2])*(Cos[(
e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])))))/(c*f*(-1 + Cos[e + f*x]))

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Maple [A]
time = 0.26, size = 82, normalized size = 1.26

method	result	size
derivativedivides	\(\frac {4 a^{2} \left (\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}+\frac {1}{4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-4}+\frac {3 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{4}+\frac {1}{4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+4}-\frac {3 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{4}\right )}{f c}\)	\(82\)
default	\(\frac {4 a^{2} \left (\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}+\frac {1}{4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-4}+\frac {3 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{4}+\frac {1}{4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+4}-\frac {3 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{4}\right )}{f c}\)	\(82\)
risch	\(\frac {2 i a^{2} \left (4 \,{\mathrm e}^{2 i \left (f x +e \right )}-{\mathrm e}^{i \left (f x +e \right )}+5\right )}{f c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{c f}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{c f}\)	\(112\)
norman	\(\frac {-\frac {4 a^{2}}{c f}-\frac {2 a^{2} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}+\frac {8 a^{2} \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}+\frac {6 a^{2} \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}+\frac {3 a^{2} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{c f}-\frac {3 a^{2} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{c f}\)	\(168\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 243 vs. \(2 (67) = 134\).
time = 0.27, size = 243, normalized size = 3.74 \begin {gather*} -\frac {a^{2} {\left (\frac {\frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 1}{\frac {c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {c \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c} - \frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c}\right )} + 2 \, a^{2} {\left (\frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c} - \frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c} - \frac {\cos \left (f x + e\right ) + 1}{c \sin \left (f x + e\right )}\right )} - \frac {a^{2} {\left (\cos \left (f x + e\right ) + 1\right )}}{c \sin \left (f x + e\right )}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.37, size = 118, normalized size = 1.82 \begin {gather*} -\frac {3 \, a^{2} \cos \left (f x + e\right ) \log \left (\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) - 3 \, a^{2} \cos \left (f x + e\right ) \log \left (-\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) - 10 \, a^{2} \cos \left (f x + e\right )^{2} - 8 \, a^{2} \cos \left (f x + e\right ) + 2 \, a^{2}}{2 \, c f \cos \left (f x + e\right ) \sin \left (f x + e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(3*a^2*cos(f*x + e)*log(sin(f*x + e) + 1)*sin(f*x + e) - 3*a^2*cos(f*x + e)*log(-sin(f*x + e) + 1)*sin(f*
x + e) - 10*a^2*cos(f*x + e)^2 - 8*a^2*cos(f*x + e) + 2*a^2)/(c*f*cos(f*x + e)*sin(f*x + e))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.49, size = 105, normalized size = 1.62 \begin {gather*} -\frac {\frac {3 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right )}{c} - \frac {3 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right )}{c} - \frac {2 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} c}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.62, size = 77, normalized size = 1.18 \begin {gather*} \frac {6\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-4\,a^2}{c\,f\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )}-\frac {6\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{c\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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