3.1.36 \(\int \frac {\sec (e+f x) (c-c \sec (e+f x))^2}{a+a \sec (e+f x)} \, dx\) [36]

Optimal. Leaf size=74 \[ -\frac {3 c^2 \tanh ^{-1}(\sin (e+f x))}{a f}+\frac {3 c^2 \tan (e+f x)}{a f}+\frac {2 \left (c^2-c^2 \sec (e+f x)\right ) \tan (e+f x)}{f (a+a \sec (e+f x))} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {4042, 3872, 3855, 3852, 8} \begin {gather*} \frac {3 c^2 \tan (e+f x)}{a f}-\frac {3 c^2 \tanh ^{-1}(\sin (e+f x))}{a f}+\frac {2 \tan (e+f x) \left (c^2-c^2 \sec (e+f x)\right )}{f (a \sec (e+f x)+a)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

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]

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 4042

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))
^(n_.), x_Symbol] :> Simp[2*a*c*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((c + d*Csc[e + f*x])^(n - 1)/(b*f*(2*m +
1))), x] - Dist[d*((2*n - 1)/(b*(2*m + 1))), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x]
)^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[n, 0] && L
tQ[m, -2^(-1)] && IntegerQ[2*m]

Rubi steps

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(220\) vs. \(2(74)=148\).
time = 1.67, size = 220, normalized size = 2.97 \begin {gather*} \frac {2 c^2 \cos \left (\frac {1}{2} (e+f x)\right ) \sec (e+f x) \sin \left (\frac {1}{2} (e+f x)\right ) \left (4 \csc \left (\frac {1}{2} (e+f x)\right ) \sec \left (\frac {e}{2}\right ) \sin \left (\frac {f x}{2}\right )+\cot \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 )}{a f (1+\sec (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c^2*Cos[(e + f*x)/2]*Sec[e + f*x]*Sin[(e + f*x)/2]*(4*Csc[(e + f*x)/2]*Sec[e/2]*Sin[(f*x)/2] + Cot[(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])))))/(a*f*(1 + Sec[e + f*x]))

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Maple [A]
time = 0.14, size = 80, normalized size = 1.08

method	result	size
derivativedivides	\(\frac {4 c^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\frac {1}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}+\frac {3 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{4}-\frac {1}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {3 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{4}\right )}{f a}\)	\(80\)
default	\(\frac {4 c^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\frac {1}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}+\frac {3 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{4}-\frac {1}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {3 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{4}\right )}{f a}\)	\(80\)
risch	\(\frac {2 i c^{2} \left (4 \,{\mathrm e}^{2 i \left (f x +e \right )}+{\mathrm e}^{i \left (f x +e \right )}+5\right )}{f a \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}+\frac {3 c^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{a f}-\frac {3 c^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{a f}\)	\(110\)
norman	\(\frac {\frac {6 c^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a f}-\frac {10 c^{2} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a f}+\frac {4 c^{2} \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a f}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}+\frac {3 c^{2} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{a f}-\frac {3 c^{2} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{a f}\)	\(129\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/f*c^2/a*(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. 242 vs. \(2 (80) = 160\).
time = 0.29, size = 242, normalized size = 3.27 \begin {gather*} -\frac {c^{2} {\left (\frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a} - \frac {2 \, \sin \left (f x + e\right )}{{\left (a - \frac {a \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}} - \frac {\sin \left (f x + e\right )}{a {\left (\cos \left (f x + e\right ) + 1\right )}}\right )} + 2 \, c^{2} {\left (\frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a} - \frac {\sin \left (f x + e\right )}{a {\left (\cos \left (f x + e\right ) + 1\right )}}\right )} - \frac {c^{2} \sin \left (f x + e\right )}{a {\left (\cos \left (f x + e\right ) + 1\right )}}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.82, size = 129, normalized size = 1.74 \begin {gather*} -\frac {3 \, {\left (c^{2} \cos \left (f x + e\right )^{2} + c^{2} \cos \left (f x + e\right )\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \, {\left (c^{2} \cos \left (f x + e\right )^{2} + c^{2} \cos \left (f x + e\right )\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (5 \, c^{2} \cos \left (f x + e\right ) + c^{2}\right )} \sin \left (f x + e\right )}{2 \, {\left (a f \cos \left (f x + e\right )^{2} + a f \cos \left (f x + e\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(3*(c^2*cos(f*x + e)^2 + c^2*cos(f*x + e))*log(sin(f*x + e) + 1) - 3*(c^2*cos(f*x + e)^2 + c^2*cos(f*x +
e))*log(-sin(f*x + e) + 1) - 2*(5*c^2*cos(f*x + e) + c^2)*sin(f*x + e))/(a*f*cos(f*x + e)^2 + a*f*cos(f*x + e)
)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.56, size = 97, normalized size = 1.31 \begin {gather*} -\frac {\frac {3 \, c^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right )}{a} - \frac {3 \, c^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right )}{a} - \frac {4 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{a} + \frac {2 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} a}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.64, size = 77, normalized size = 1.04 \begin {gather*} \frac {4\,c^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{a\,f}+\frac {2\,c^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{f\,\left (a-a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\right )}-\frac {6\,c^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{a\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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