3.1.6 \(\int \frac {(a+a \sec (e+f x))^2}{c-c \sec (e+f x)} \, dx\) [6]

Optimal. Leaf size=56 \[ \frac {a^2 x}{c}-\frac {a^2 \tanh ^{-1}(\sin (e+f x))}{c f}-\frac {4 a^2 \tan (e+f x)}{c f (1-\sec (e+f x))} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3988, 3862, 8, 3879, 3874, 3855} \begin {gather*} -\frac {a^2 \tanh ^{-1}(\sin (e+f x))}{c f}-\frac {4 a^2 \tan (e+f x)}{c f (1-\sec (e+f x))}+\frac {a^2 x}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 3855

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

Rule 3862

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

Rule 3874

Int[csc[(e_.) + (f_.)*(x_)]^2/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[Csc[e + f*x],
x], x] - Dist[a/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x]

Rule 3879

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[-Cot[e + f*x]/(f*(b + a*
Csc[e + f*x])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 3988

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> Dis
t[c^n, Int[ExpandTrig[(1 + (d/c)*csc[e + f*x])^n, (a + b*csc[e + f*x])^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] && IGtQ[m, 0] && ILtQ[n, 0] && LtQ[m + n, 2]

Rubi steps

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(169\) vs. \(2(56)=112\).
time = 0.31, size = 169, normalized size = 3.02 \begin {gather*} \frac {a^2 \csc \left (\frac {e}{2}\right ) \left (-\cos \left (\frac {f x}{2}\right ) \left (f x+\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )+\cos \left (e+\frac {f x}{2}\right ) \left (f x+\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )+8 \sin \left (\frac {f x}{2}\right )\right ) \sin \left (\frac {1}{2} (e+f x)\right )}{c f (-1+\cos (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Csc[e/2]*(-(Cos[(f*x)/2]*(f*x + Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - Log[Cos[(e + f*x)/2] + Sin[(e
+ f*x)/2]])) + Cos[e + (f*x)/2]*(f*x + Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - Log[Cos[(e + f*x)/2] + Sin[(
e + f*x)/2]]) + 8*Sin[(f*x)/2])*Sin[(e + f*x)/2])/(c*f*(-1 + Cos[e + f*x]))

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Maple [A]
time = 0.12, size = 64, normalized size = 1.14

method	result	size
derivativedivides	\(\frac {4 a^{2} \left (\frac {\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{4}+\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}+\frac {\arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}-\frac {\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{4}\right )}{f c}\)	\(64\)
default	\(\frac {4 a^{2} \left (\frac {\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{4}+\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}+\frac {\arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}-\frac {\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{4}\right )}{f c}\)	\(64\)
risch	\(\frac {a^{2} x}{c}+\frac {8 i a^{2}}{f c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{c f}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{c f}\)	\(82\)
norman	\(\frac {\frac {a^{2} x \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}-\frac {4 a^{2}}{c f}+\frac {4 a^{2} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}-\frac {a^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}+\frac {a^{2} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{c f}-\frac {a^{2} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{c f}\)	\(145\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/f*a^2/c*(1/4*ln(tan(1/2*f*x+1/2*e)-1)+1/tan(1/2*f*x+1/2*e)+1/2*arctan(tan(1/2*f*x+1/2*e))-1/4*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. 165 vs. \(2 (57) = 114\).
time = 0.51, size = 165, normalized size = 2.95 \begin {gather*} \frac {a^{2} {\left (\frac {2 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c} + \frac {\cos \left (f x + e\right ) + 1}{c \sin \left (f x + e\right )}\right )} - 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 {2 \, 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*sec(f*x+e))^2/(c-c*sec(f*x+e)),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 3.15, size = 94, normalized size = 1.68 \begin {gather*} \frac {2 \, a^{2} f x \sin \left (f x + e\right ) - a^{2} \log \left (\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) + a^{2} \log \left (-\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) + 8 \, a^{2} \cos \left (f x + e\right ) + 8 \, a^{2}}{2 \, c f \sin \left (f x + e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*a^2*f*x*sin(f*x + e) - a^2*log(sin(f*x + e) + 1)*sin(f*x + e) + a^2*log(-sin(f*x + e) + 1)*sin(f*x + e)
+ 8*a^2*cos(f*x + e) + 8*a^2)/(c*f*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 {2 \sec {\left (e + f x \right )}}{\sec {\left (e + f x \right )} - 1}\, dx + \int \frac {\sec ^{2}{\left (e + f x \right )}}{\sec {\left (e + f x \right )} - 1}\, dx + \int \frac {1}{\sec {\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*sec(f*x+e))**2/(c-c*sec(f*x+e)),x)

[Out]

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

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Giac [A]
time = 0.50, size = 77, normalized size = 1.38 \begin {gather*} \frac {\frac {{\left (f x + e\right )} a^{2}}{c} - \frac {a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right )}{c} + \frac {a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right )}{c} + \frac {4 \, a^{2}}{c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.48, size = 46, normalized size = 0.82 \begin {gather*} \frac {a^2\,x}{c}-\frac {a^2\,\left (2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )-\frac {4}{\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/(c - c/cos(e + f*x)),x)

[Out]

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

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