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

Optimal. Leaf size=95 \[ -\frac {a^2 (c-2 d) \tanh ^{-1}(\sin (e+f x))}{d^2 f}+\frac {2 a^2 (c-d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{d^2 f \sqrt {c+d}}+\frac {a^2 \tan (e+f x)}{d f} \]

[Out]

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

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Rubi [B]  time = 0.25, antiderivative size = 208, normalized size of antiderivative = 2.19, 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, 102, 157, 63, 217, 203, 93, 205} \[ -\frac {2 a^3 (c-2 d) \tan (e+f x) \tan ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a (\sec (e+f x)+1)}}\right )}{d^2 f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}-\frac {2 a^3 (c-d)^{3/2} \tan (e+f x) \tan ^{-1}\left (\frac {\sqrt {c+d} \sqrt {a \sec (e+f x)+a}}{\sqrt {c-d} \sqrt {a-a \sec (e+f x)}}\right )}{d^2 f \sqrt {c+d} \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}+\frac {a^2 \tan (e+f x)}{d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Tan[e + f*x])/(d*f) - (2*a^3*(c - 2*d)*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*(c - d)^(3/2)*ArcTan[(Sqrt[c + d]*S
qrt[a + a*Sec[e + f*x]])/(Sqrt[c - d]*Sqrt[a - a*Sec[e + f*x]])]*Tan[e + f*x])/(d^2*Sqrt[c + d]*f*Sqrt[a - a*S
ec[e + f*x]]*Sqrt[a + a*Sec[e + f*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 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 102

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m - 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 1)), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegersQ[2*m, 2*n, 2*p]

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 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 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]

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 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])

Rubi steps

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

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Mathematica [C]  time = 2.02, size = 329, normalized size = 3.46 \[ \frac {a^2 \cos (e+f x) \sec ^4\left (\frac {1}{2} (e+f x)\right ) (\sec (e+f x)+1)^2 (c \cos (e+f x)+d) \left (-\frac {2 i (c-d)^2 (\cos (e)-i \sin (e)) \tan ^{-1}\left (\frac {(\sin (e)+i \cos (e)) \left (\tan \left (\frac {f x}{2}\right ) (c \cos (e)-d)+c \sin (e)\right )}{\sqrt {c^2-d^2} \sqrt {(\cos (e)-i \sin (e))^2}}\right )}{\sqrt {c^2-d^2} \sqrt {(\cos (e)-i \sin (e))^2}}+(c-2 d) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-(c-2 d) \log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )+\frac {d \sin \left (\frac {f x}{2}\right )}{\left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )}+\frac {d \sin \left (\frac {f x}{2}\right )}{\left (\sin \left (\frac {e}{2}\right )+\cos \left (\frac {e}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )}\right )}{4 d^2 f (c+d \sec (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.63, size = 398, normalized size = 4.19 \[ \left [\frac {2 \, a^{2} d \sin \left (f x + e\right ) - {\left (a^{2} c - a^{2} d\right )} \sqrt {\frac {c - d}{c + d}} \cos \left (f x + e\right ) \log \left (\frac {2 \, c d \cos \left (f x + e\right ) - {\left (c^{2} - 2 \, d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (c^{2} + c d + {\left (c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {c - d}{c + d}} \sin \left (f x + e\right ) + 2 \, c^{2} - d^{2}}{c^{2} \cos \left (f x + e\right )^{2} + 2 \, c d \cos \left (f x + e\right ) + d^{2}}\right ) - {\left (a^{2} c - 2 \, a^{2} d\right )} \cos \left (f x + e\right ) \log \left (\sin \left (f x + e\right ) + 1\right ) + {\left (a^{2} c - 2 \, a^{2} d\right )} \cos \left (f x + e\right ) \log \left (-\sin \left (f x + e\right ) + 1\right )}{2 \, d^{2} f \cos \left (f x + e\right )}, \frac {2 \, a^{2} d \sin \left (f x + e\right ) + 2 \, {\left (a^{2} c - a^{2} d\right )} \sqrt {-\frac {c - d}{c + d}} \arctan \left (-\frac {{\left (d \cos \left (f x + e\right ) + c\right )} \sqrt {-\frac {c - d}{c + d}}}{{\left (c - d\right )} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right ) - {\left (a^{2} c - 2 \, a^{2} d\right )} \cos \left (f x + e\right ) \log \left (\sin \left (f x + e\right ) + 1\right ) + {\left (a^{2} c - 2 \, a^{2} d\right )} \cos \left (f x + e\right ) \log \left (-\sin \left (f x + e\right ) + 1\right )}{2 \, d^{2} f \cos \left (f x + e\right )}\right ] \]

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)),x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

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)),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)2/f*((-2*a^2*c^2+4*a^2*c*d-2*a^2*d^2)*1/2/d^2/sqrt(-c^2+d^2)*(
atan((c*tan((f*x+exp(1))/2)-d*tan((f*x+exp(1))/2))/sqrt(-c^2+d^2))+pi*sign(2*c-2*d)*floor((f*x+exp(1))/2/pi+1/
2))-(-a^2*c+2*a^2*d)*1/2/d^2*ln(abs(tan((f*x+exp(1))/2)-1))+(-a^2*c+2*a^2*d)*1/2/d^2*ln(abs(tan((f*x+exp(1))/2
)+1))-tan((f*x+exp(1))/2)*a^2/d/(tan((f*x+exp(1))/2)^2-1))

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maple [B]  time = 0.62, size = 291, normalized size = 3.06 \[ \frac {2 a^{2} \arctanh \left (\frac {\tan \left (\frac {e}{2}+\frac {f x}{2}\right ) \left (c -d \right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right ) c^{2}}{f \,d^{2} \sqrt {\left (c +d \right ) \left (c -d \right )}}-\frac {4 a^{2} \arctanh \left (\frac {\tan \left (\frac {e}{2}+\frac {f x}{2}\right ) \left (c -d \right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right ) c}{f d \sqrt {\left (c +d \right ) \left (c -d \right )}}+\frac {2 a^{2} \arctanh \left (\frac {\tan \left (\frac {e}{2}+\frac {f x}{2}\right ) \left (c -d \right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{f \sqrt {\left (c +d \right ) \left (c -d \right )}}-\frac {a^{2}}{f d \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right )}+\frac {a^{2} \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right ) c}{f \,d^{2}}-\frac {2 a^{2} \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right )}{f d}-\frac {a^{2}}{f d \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right )}-\frac {a^{2} \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right ) c}{f \,d^{2}}+\frac {2 a^{2} \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right )}{f d} \]

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)),x)

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*c^2-4*d^2>0)', see `assume?`
for more details)Is 4*c^2-4*d^2 positive or negative?

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mupad [B]  time = 2.59, size = 529, normalized size = 5.57 \[ \frac {2\,a^2\,\left (\frac {\sin \left (e+f\,x\right )}{2}+2\,\cos \left (e+f\,x\right )\,\mathrm {atanh}\left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )\right )}{f\,\cos \left (e+f\,x\right )\,\left (c+d\right )}+\frac {2\,a^2\,\left (\frac {c\,\sin \left (e+f\,x\right )}{2}+c\,\cos \left (e+f\,x\right )\,\mathrm {atanh}\left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )\right )}{d\,f\,\cos \left (e+f\,x\right )\,\left (c+d\right )}-\frac {2\,a^2\,\left (c^2\,\cos \left (e+f\,x\right )\,\mathrm {atanh}\left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )+\cos \left (e+f\,x\right )\,\mathrm {atan}\left (\frac {\left (2\,c\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,{\left (c^4-2\,c^3\,d+2\,c\,d^3-d^4\right )}^{3/2}-2\,c^5\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {c^4-2\,c^3\,d+2\,c\,d^3-d^4}+5\,d^5\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {c^4-2\,c^3\,d+2\,c\,d^3-d^4}-c\,d^4\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {c^4-2\,c^3\,d+2\,c\,d^3-d^4}+4\,c^4\,d\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {c^4-2\,c^3\,d+2\,c\,d^3-d^4}-9\,c^2\,d^3\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {c^4-2\,c^3\,d+2\,c\,d^3-d^4}+3\,c^3\,d^2\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {c^4-2\,c^3\,d+2\,c\,d^3-d^4}\right )\,1{}\mathrm {i}}{d\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (c+d\right )\,\left (3\,c^4\,d-8\,c^3\,d^2+2\,c^2\,d^3+8\,c\,d^4-5\,d^5\right )}\right )\,\sqrt {\left (c+d\right )\,{\left (c-d\right )}^3}\,1{}\mathrm {i}\right )}{d^2\,f\,\cos \left (e+f\,x\right )\,\left (c+d\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a^2*(sin(e + f*x)/2 + 2*cos(e + f*x)*atanh(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2))))/(f*cos(e + f*x)*(c + d)
) + (2*a^2*((c*sin(e + f*x))/2 + c*cos(e + f*x)*atanh(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2))))/(d*f*cos(e + f*
x)*(c + d)) - (2*a^2*(c^2*cos(e + f*x)*atanh(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2)) + cos(e + f*x)*atan(((2*c*
sin(e/2 + (f*x)/2)*(2*c*d^3 - 2*c^3*d + c^4 - d^4)^(3/2) - 2*c^5*sin(e/2 + (f*x)/2)*(2*c*d^3 - 2*c^3*d + c^4 -
d^4)^(1/2) + 5*d^5*sin(e/2 + (f*x)/2)*(2*c*d^3 - 2*c^3*d + c^4 - d^4)^(1/2) - c*d^4*sin(e/2 + (f*x)/2)*(2*c*d
^3 - 2*c^3*d + c^4 - d^4)^(1/2) + 4*c^4*d*sin(e/2 + (f*x)/2)*(2*c*d^3 - 2*c^3*d + c^4 - d^4)^(1/2) - 9*c^2*d^3
*sin(e/2 + (f*x)/2)*(2*c*d^3 - 2*c^3*d + c^4 - d^4)^(1/2) + 3*c^3*d^2*sin(e/2 + (f*x)/2)*(2*c*d^3 - 2*c^3*d +
c^4 - d^4)^(1/2))*1i)/(d*cos(e/2 + (f*x)/2)*(c + d)*(8*c*d^4 + 3*c^4*d - 5*d^5 + 2*c^2*d^3 - 8*c^3*d^2)))*((c
+ d)*(c - d)^3)^(1/2)*1i))/(d^2*f*cos(e + f*x)*(c + d))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \left (\int \frac {\sec {\left (e + f x \right )}}{c + d \sec {\left (e + f x \right )}}\, dx + \int \frac {2 \sec ^{2}{\left (e + f x \right )}}{c + d \sec {\left (e + f x \right )}}\, dx + \int \frac {\sec ^{3}{\left (e + f x \right )}}{c + d \sec {\left (e + f x \right )}}\, dx\right ) \]

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)),x)

[Out]

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

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