3.198 \(\int \frac{\sec (e+f x) (a+a \sec (e+f x))^2}{(c+d \sec (e+f x))^2} \, dx\)

Optimal. Leaf size=117 \[ -\frac{2 a^2 \sqrt{c-d} (c+2 d) \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{d^2 f (c+d)^{3/2}}-\frac{a^2 (c-d) \tan (e+f x)}{d f (c+d) (c+d \sec (e+f x))}+\frac{a^2 \tanh ^{-1}(\sin (e+f x))}{d^2 f} \]

[Out]

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

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Rubi [A]  time = 0.25679, antiderivative size = 231, normalized size of antiderivative = 1.97, number of steps used = 8, number of rules used = 8, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.258, Rules used = {3987, 98, 157, 63, 217, 203, 93, 205} \[ \frac{2 a^3 \sqrt{c-d} (c+2 d) \tan (e+f x) \tan ^{-1}\left (\frac{\sqrt{c+d} \sqrt{a \sec (e+f x)+a}}{\sqrt{c-d} \sqrt{a-a \sec (e+f x)}}\right )}{d^2 f (c+d)^{3/2} \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}-\frac{a^2 (c-d) \tan (e+f x)}{d f (c+d) (c+d \sec (e+f x))}+\frac{2 a^3 \tan (e+f x) \tan ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a (\sec (e+f x)+1)}}\right )}{d^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*ArcTan[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a*(1 + Sec[e + f*x])]]*Tan[e + f*x])/(d^2*f*Sqrt[a - a*Sec[e + f*x
]]*Sqrt[a + a*Sec[e + f*x]]) + (2*a^3*Sqrt[c - d]*(c + 2*d)*ArcTan[(Sqrt[c + d]*Sqrt[a + a*Sec[e + f*x]])/(Sqr
t[c - d]*Sqrt[a - a*Sec[e + f*x]])]*Tan[e + f*x])/(d^2*(c + d)^(3/2)*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec
[e + f*x]]) - (a^2*(c - d)*Tan[e + f*x])/(d*(c + d)*f*(c + d*Sec[e + f*x]))

Rule 3987

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]
*(d_.) + (c_))^(n_), x_Symbol] :> Dist[(a^2*g*Cot[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x
]]), Subst[Int[((g*x)^(p - 1)*(a + b*x)^(m - 1/2)*(c + d*x)^n)/Sqrt[a - b*x], x], x, Csc[e + f*x]], x] /; Free
Q[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && (EqQ[p,
1] || IntegerQ[m - 1/2])

Rule 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] + Dist[1/(b*(b*e - a*
f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

Mathematica [C]  time = 1.47496, size = 312, normalized size = 2.67 \[ \frac{a^2 \sec ^4\left (\frac{1}{2} (e+f x)\right ) (\sec (e+f x)+1)^2 (c \cos (e+f x)+d) \left (\frac{2 \left (c^2+c d-2 d^2\right ) (\sin (e)+i \cos (e)) (c \cos (e+f x)+d) \tan ^{-1}\left (\frac{(\sin (e)+i \cos (e)) \left (\tan \left (\frac{f x}{2}\right ) (c \cos (e)-d)+c \sin (e)\right )}{\sqrt{c^2-d^2} \sqrt{(\cos (e)-i \sin (e))^2}}\right )}{(c+d) \sqrt{c^2-d^2} \sqrt{(\cos (e)-i \sin (e))^2}}+\frac{d (c-d) (d \sin (e)-c \sin (f x))}{c (c+d) \left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\sin \left (\frac{e}{2}\right )+\cos \left (\frac{e}{2}\right )\right )}-(c \cos (e+f x)+d) \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+(c \cos (e+f x)+d) \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )}{4 d^2 f (c+d \sec (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(d + c*Cos[e + f*x])*Sec[(e + f*x)/2]^4*(1 + Sec[e + f*x])^2*(-((d + c*Cos[e + f*x])*Log[Cos[(e + f*x)/2]
- Sin[(e + f*x)/2]]) + (d + c*Cos[e + f*x])*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + (2*(c^2 + c*d - 2*d^2)
*ArcTan[((I*Cos[e] + Sin[e])*(c*Sin[e] + (-d + c*Cos[e])*Tan[(f*x)/2]))/(Sqrt[c^2 - d^2]*Sqrt[(Cos[e] - I*Sin[
e])^2])]*(d + c*Cos[e + f*x])*(I*Cos[e] + Sin[e]))/((c + d)*Sqrt[c^2 - d^2]*Sqrt[(Cos[e] - I*Sin[e])^2]) + ((c
- d)*d*(d*Sin[e] - c*Sin[f*x]))/(c*(c + d)*(Cos[e/2] - Sin[e/2])*(Cos[e/2] + Sin[e/2]))))/(4*d^2*f*(c + d*Sec
[e + f*x])^2)

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Maple [B]  time = 0.11, size = 330, normalized size = 2.8 \begin{align*}{\frac{{a}^{2}}{f{d}^{2}}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) }-{\frac{{a}^{2}}{f{d}^{2}}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) }+2\,{\frac{{a}^{2}c\tan \left ( 1/2\,fx+e/2 \right ) }{fd \left ( c+d \right ) \left ( \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}c- \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}d-c-d \right ) }}-2\,{\frac{{a}^{2}{c}^{2}}{f{d}^{2} \left ( c+d \right ) \sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}{\it Artanh} \left ({\frac{\tan \left ( 1/2\,fx+e/2 \right ) \left ( c-d \right ) }{\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}} \right ) }-2\,{\frac{{a}^{2}c}{fd \left ( c+d \right ) \sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}{\it Artanh} \left ({\frac{\tan \left ( 1/2\,fx+e/2 \right ) \left ( c-d \right ) }{\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}} \right ) }-2\,{\frac{{a}^{2}\tan \left ( 1/2\,fx+e/2 \right ) }{f \left ( c+d \right ) \left ( \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}c- \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}d-c-d \right ) }}+4\,{\frac{{a}^{2}}{f \left ( c+d \right ) \sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}{\it Artanh} \left ({\frac{\tan \left ( 1/2\,fx+e/2 \right ) \left ( c-d \right ) }{\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(a+a*sec(f*x+e))^2/(c+d*sec(f*x+e))^2,x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.987852, size = 1289, normalized size = 11.02 \begin{align*} \left [\frac{{\left (a^{2} c d + 2 \, a^{2} d^{2} +{\left (a^{2} c^{2} + 2 \, a^{2} c d\right )} \cos \left (f x + e\right )\right )} \sqrt{\frac{c - d}{c + d}} \log \left (\frac{2 \, c d \cos \left (f x + e\right ) -{\left (c^{2} - 2 \, d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \,{\left (c^{2} + c d +{\left (c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt{\frac{c - d}{c + d}} \sin \left (f x + e\right ) + 2 \, c^{2} - d^{2}}{c^{2} \cos \left (f x + e\right )^{2} + 2 \, c d \cos \left (f x + e\right ) + d^{2}}\right ) +{\left (a^{2} c d + a^{2} d^{2} +{\left (a^{2} c^{2} + a^{2} c d\right )} \cos \left (f x + e\right )\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) -{\left (a^{2} c d + a^{2} d^{2} +{\left (a^{2} c^{2} + a^{2} c d\right )} \cos \left (f x + e\right )\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \,{\left (a^{2} c d - a^{2} d^{2}\right )} \sin \left (f x + e\right )}{2 \,{\left ({\left (c^{2} d^{2} + c d^{3}\right )} f \cos \left (f x + e\right ) +{\left (c d^{3} + d^{4}\right )} f\right )}}, -\frac{2 \,{\left (a^{2} c d + 2 \, a^{2} d^{2} +{\left (a^{2} c^{2} + 2 \, a^{2} c d\right )} \cos \left (f x + e\right )\right )} \sqrt{-\frac{c - d}{c + d}} \arctan \left (-\frac{{\left (d \cos \left (f x + e\right ) + c\right )} \sqrt{-\frac{c - d}{c + d}}}{{\left (c - d\right )} \sin \left (f x + e\right )}\right ) -{\left (a^{2} c d + a^{2} d^{2} +{\left (a^{2} c^{2} + a^{2} c d\right )} \cos \left (f x + e\right )\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) +{\left (a^{2} c d + a^{2} d^{2} +{\left (a^{2} c^{2} + a^{2} c d\right )} \cos \left (f x + e\right )\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \,{\left (a^{2} c d - a^{2} d^{2}\right )} \sin \left (f x + e\right )}{2 \,{\left ({\left (c^{2} d^{2} + c d^{3}\right )} f \cos \left (f x + e\right ) +{\left (c d^{3} + d^{4}\right )} f\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((a^2*c*d + 2*a^2*d^2 + (a^2*c^2 + 2*a^2*c*d)*cos(f*x + e))*sqrt((c - d)/(c + d))*log((2*c*d*cos(f*x + e)
- (c^2 - 2*d^2)*cos(f*x + e)^2 - 2*(c^2 + c*d + (c*d + d^2)*cos(f*x + e))*sqrt((c - d)/(c + d))*sin(f*x + e)
+ 2*c^2 - d^2)/(c^2*cos(f*x + e)^2 + 2*c*d*cos(f*x + e) + d^2)) + (a^2*c*d + a^2*d^2 + (a^2*c^2 + a^2*c*d)*cos
(f*x + e))*log(sin(f*x + e) + 1) - (a^2*c*d + a^2*d^2 + (a^2*c^2 + a^2*c*d)*cos(f*x + e))*log(-sin(f*x + e) +
1) - 2*(a^2*c*d - a^2*d^2)*sin(f*x + e))/((c^2*d^2 + c*d^3)*f*cos(f*x + e) + (c*d^3 + d^4)*f), -1/2*(2*(a^2*c*
d + 2*a^2*d^2 + (a^2*c^2 + 2*a^2*c*d)*cos(f*x + e))*sqrt(-(c - d)/(c + d))*arctan(-(d*cos(f*x + e) + c)*sqrt(-
(c - d)/(c + d))/((c - d)*sin(f*x + e))) - (a^2*c*d + a^2*d^2 + (a^2*c^2 + a^2*c*d)*cos(f*x + e))*log(sin(f*x
+ e) + 1) + (a^2*c*d + a^2*d^2 + (a^2*c^2 + a^2*c*d)*cos(f*x + e))*log(-sin(f*x + e) + 1) + 2*(a^2*c*d - a^2*d
^2)*sin(f*x + e))/((c^2*d^2 + c*d^3)*f*cos(f*x + e) + (c*d^3 + d^4)*f)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral(sec(e + f*x)/(c**2 + 2*c*d*sec(e + f*x) + d**2*sec(e + f*x)**2), x) + Integral(2*sec(e + f*x)**
2/(c**2 + 2*c*d*sec(e + f*x) + d**2*sec(e + f*x)**2), x) + Integral(sec(e + f*x)**3/(c**2 + 2*c*d*sec(e + f*x)
+ d**2*sec(e + f*x)**2), x))

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Giac [B]  time = 1.38316, size = 323, normalized size = 2.76 \begin{align*} \frac{\frac{a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{d^{2}} - \frac{a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{d^{2}} + \frac{2 \,{\left (a^{2} c^{2} + a^{2} c d - 2 \, a^{2} d^{2}\right )}{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, c - 2 \, d\right ) + \arctan \left (\frac{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{\sqrt{-c^{2} + d^{2}}}\right )\right )}}{{\left (c d^{2} + d^{3}\right )} \sqrt{-c^{2} + d^{2}}} + \frac{2 \,{\left (a^{2} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - a^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c - d\right )}{\left (c d + d^{2}\right )}}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*log(abs(tan(1/2*f*x + 1/2*e) + 1))/d^2 - a^2*log(abs(tan(1/2*f*x + 1/2*e) - 1))/d^2 + 2*(a^2*c^2 + a^2*c*
d - 2*a^2*d^2)*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(2*c - 2*d) + arctan((c*tan(1/2*f*x + 1/2*e) - d*tan(1/2*f
*x + 1/2*e))/sqrt(-c^2 + d^2)))/((c*d^2 + d^3)*sqrt(-c^2 + d^2)) + 2*(a^2*c*tan(1/2*f*x + 1/2*e) - a^2*d*tan(1
/2*f*x + 1/2*e))/((c*tan(1/2*f*x + 1/2*e)^2 - d*tan(1/2*f*x + 1/2*e)^2 - c - d)*(c*d + d^2)))/f