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

Optimal. Leaf size=68 \[ \frac {d (2 c-d) \tanh ^{-1}(\sin (e+f x))}{a f}+\frac {(c-d)^2 \tan (e+f x)}{f (a \sec (e+f x)+a)}+\frac {d^2 \tan (e+f x)}{a f} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 125, normalized size of antiderivative = 1.84, number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3987, 89, 80, 63, 217, 203} \[ \frac {2 d (2 c-d) \tan (e+f x) \tan ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a (\sec (e+f x)+1)}}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}+\frac {(c-d)^2 \tan (e+f x)}{f (a \sec (e+f x)+a)}+\frac {d^2 \tan (e+f x)}{a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*Tan[e + f*x])/(a*f) + ((c - d)^2*Tan[e + f*x])/(f*(a + a*Sec[e + f*x])) + (2*(2*c - d)*d*ArcTan[Sqrt[a -
a*Sec[e + f*x]]/Sqrt[a*(1 + Sec[e + f*x])]]*Tan[e + f*x])/(f*Sqrt[a - a*Sec[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 80

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

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

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

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Mathematica [B]  time = 1.73, size = 237, normalized size = 3.49 \[ \frac {2 \cos \left (\frac {1}{2} (e+f x)\right ) \cos (e+f x) (c+d \sec (e+f x))^2 \left ((c-d)^2 \sec \left (\frac {e}{2}\right ) \sin \left (\frac {f x}{2}\right )+d \cos \left (\frac {1}{2} (e+f x)\right ) \left (\frac {d \sin (f x)}{\left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\sin \left (\frac {e}{2}\right )+\cos \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )}-(2 c-d) \left (\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )\right )\right )}{a f (\sec (e+f x)+1) (c \cos (e+f x)+d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.44, size = 155, normalized size = 2.28 \[ \frac {{\left ({\left (2 \, c d - d^{2}\right )} \cos \left (f x + e\right )^{2} + {\left (2 \, c d - d^{2}\right )} \cos \left (f x + e\right )\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - {\left ({\left (2 \, c d - d^{2}\right )} \cos \left (f x + e\right )^{2} + {\left (2 \, c d - d^{2}\right )} \cos \left (f x + e\right )\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (d^{2} + {\left (c^{2} - 2 \, c d + 2 \, d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{2 \, {\left (a f \cos \left (f x + e\right )^{2} + a f \cos \left (f x + e\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(((2*c*d - d^2)*cos(f*x + e)^2 + (2*c*d - d^2)*cos(f*x + e))*log(sin(f*x + e) + 1) - ((2*c*d - d^2)*cos(f*
x + e)^2 + (2*c*d - d^2)*cos(f*x + e))*log(-sin(f*x + e) + 1) + 2*(d^2 + (c^2 - 2*c*d + 2*d^2)*cos(f*x + e))*s
in(f*x + e))/(a*f*cos(f*x + e)^2 + a*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)*(c+d*sec(f*x+e))^2/(a+a*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*((tan((f*x+exp(1))/2)*c^2-2*tan((f*x+exp(1))/2)*c*d+tan((f
*x+exp(1))/2)*d^2)*1/2/a-tan((f*x+exp(1))/2)*d^2/a/(tan((f*x+exp(1))/2)^2-1)-(2*c*d-d^2)*1/2/a*ln(abs(tan((f*x
+exp(1))/2)-1))+(2*c*d-d^2)*1/2/a*ln(abs(tan((f*x+exp(1))/2)+1)))

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maple [B]  time = 0.64, size = 196, normalized size = 2.88 \[ \frac {\tan \left (\frac {e}{2}+\frac {f x}{2}\right ) c^{2}}{a f}-\frac {2 c d \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{a f}+\frac {\tan \left (\frac {e}{2}+\frac {f x}{2}\right ) d^{2}}{a f}-\frac {d^{2}}{a f \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right )}-\frac {2 d \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right ) c}{a f}+\frac {d^{2} \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right )}{a f}-\frac {d^{2}}{a f \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right )}+\frac {2 d \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right ) c}{a f}-\frac {d^{2} \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right )}{a f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.44, size = 223, normalized size = 3.28 \[ -\frac {d^{2} {\left (\frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a} - \frac {2 \, \sin \left (f x + e\right )}{{\left (a - \frac {a \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}} - \frac {\sin \left (f x + e\right )}{a {\left (\cos \left (f x + e\right ) + 1\right )}}\right )} - 2 \, c d {\left (\frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a} - \frac {\sin \left (f x + e\right )}{a {\left (\cos \left (f x + e\right ) + 1\right )}}\right )} - \frac {c^{2} \sin \left (f x + e\right )}{a {\left (\cos \left (f x + e\right ) + 1\right )}}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(d^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*d*(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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mupad [B]  time = 1.82, size = 85, normalized size = 1.25 \[ \frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,{\left (c-d\right )}^2}{a\,f}+\frac {2\,d^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{f\,\left (a-a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\right )}+\frac {2\,d\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )\,\left (2\,c-d\right )}{a\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(e/2 + (f*x)/2)*(c - d)^2)/(a*f) + (2*d^2*tan(e/2 + (f*x)/2))/(f*(a - a*tan(e/2 + (f*x)/2)^2)) + (2*d*atan
h(tan(e/2 + (f*x)/2))*(2*c - d))/(a*f)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {c^{2} \sec {\left (e + f x \right )}}{\sec {\left (e + f x \right )} + 1}\, dx + \int \frac {d^{2} \sec ^{3}{\left (e + f x \right )}}{\sec {\left (e + f x \right )} + 1}\, dx + \int \frac {2 c d \sec ^{2}{\left (e + f x \right )}}{\sec {\left (e + f x \right )} + 1}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(c**2*sec(e + f*x)/(sec(e + f*x) + 1), x) + Integral(d**2*sec(e + f*x)**3/(sec(e + f*x) + 1), x) + In
tegral(2*c*d*sec(e + f*x)**2/(sec(e + f*x) + 1), x))/a

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