3.132 \(\int \frac {\sec ^2(c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=99 \[ \frac {(A+4 C) \tan (c+d x)}{3 a^2 d}-\frac {2 C \tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac {2 C \tan (c+d x)}{a^2 d (\sec (c+d x)+1)}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[Out]

-2*C*arctanh(sin(d*x+c))/a^2/d+1/3*(A+4*C)*tan(d*x+c)/a^2/d+2*C*tan(d*x+c)/a^2/d/(1+sec(d*x+c))-1/3*(A+C)*sec(
d*x+c)^2*tan(d*x+c)/d/(a+a*sec(d*x+c))^2

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Rubi [A]  time = 0.25, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4085, 4008, 3787, 3770, 3767, 8} \[ \frac {(A+4 C) \tan (c+d x)}{3 a^2 d}-\frac {2 C \tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac {2 C \tan (c+d x)}{a^2 d (\sec (c+d x)+1)}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*C*ArcTanh[Sin[c + d*x]])/(a^2*d) + ((A + 4*C)*Tan[c + d*x])/(3*a^2*d) + (2*C*Tan[c + d*x])/(a^2*d*(1 + Sec
[c + d*x])) - ((A + C)*Sec[c + d*x]^2*Tan[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2)

Rule 8

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

Rule 3767

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 3770

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

Rule 3787

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 4008

Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_
)), x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(b*f*(2*m + 1)), x] + Dist[1/(b^2*(2*
m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[A*b*m - a*B*m + b*B*(2*m + 1)*Csc[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rule 4085

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> -Simp[(a*(A + C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(a*f*(
2*m + 1)), x] + Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*C*n + A*b*(
2*m + n + 1) - (a*(A*(m + n + 1) - C*(m - n)))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x]
&& EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx &=-\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {\int \frac {\sec ^2(c+d x) (-a (A-2 C)-a (A+4 C) \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=\frac {2 C \tan (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {\int \sec (c+d x) \left (-6 a^2 C+a^2 (A+4 C) \sec (c+d x)\right ) \, dx}{3 a^4}\\ &=\frac {2 C \tan (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(2 C) \int \sec (c+d x) \, dx}{a^2}+\frac {(A+4 C) \int \sec ^2(c+d x) \, dx}{3 a^2}\\ &=-\frac {2 C \tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac {2 C \tan (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(A+4 C) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 a^2 d}\\ &=-\frac {2 C \tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac {(A+4 C) \tan (c+d x)}{3 a^2 d}+\frac {2 C \tan (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}\\ \end {align*}

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Mathematica [B]  time = 1.60, size = 280, normalized size = 2.83 \[ \frac {4 \cos \left (\frac {1}{2} (c+d x)\right ) \left (A+C \sec ^2(c+d x)\right ) \left ((A+C) \tan \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right )+(A+C) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+2 (A+7 C) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right )+6 C \cos ^3\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\sin (d x)}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}+2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{3 a^2 d (\sec (c+d x)+1)^2 (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Cos[(c + d*x)/2]*(A + C*Sec[c + d*x]^2)*((A + C)*Sec[c/2]*Sin[(d*x)/2] + 2*(A + 7*C)*Cos[(c + d*x)/2]^2*Sec
[c/2]*Sin[(d*x)/2] + 6*C*Cos[(c + d*x)/2]^3*(2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 2*Log[Cos[(c + d*x)/
2] + Sin[(c + d*x)/2]] + Sin[d*x]/((Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*
x)/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))) + (A + C)*Cos[(c + d*x)/2]*Tan[c/2]))/(3*a^2*d*(A + 2*C + A*Cos
[2*(c + d*x)])*(1 + Sec[c + d*x])^2)

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fricas [A]  time = 0.45, size = 167, normalized size = 1.69 \[ -\frac {3 \, {\left (C \cos \left (d x + c\right )^{3} + 2 \, C \cos \left (d x + c\right )^{2} + C \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (C \cos \left (d x + c\right )^{3} + 2 \, C \cos \left (d x + c\right )^{2} + C \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (A + 10 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (A + 7 \, C\right )} \cos \left (d x + c\right ) + 3 \, C\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(3*(C*cos(d*x + c)^3 + 2*C*cos(d*x + c)^2 + C*cos(d*x + c))*log(sin(d*x + c) + 1) - 3*(C*cos(d*x + c)^3 +
2*C*cos(d*x + c)^2 + C*cos(d*x + c))*log(-sin(d*x + c) + 1) - ((A + 10*C)*cos(d*x + c)^2 + 2*(A + 7*C)*cos(d*
x + c) + 3*C)*sin(d*x + c))/(a^2*d*cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))

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giac [A]  time = 0.29, size = 142, normalized size = 1.43 \[ -\frac {\frac {12 \, C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {12 \, C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac {12 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{2}} - \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(12*C*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 12*C*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^2 + 12*C*tan(1/2
*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*a^2) - (A*a^4*tan(1/2*d*x + 1/2*c)^3 + C*a^4*tan(1/2*d*x + 1/2*c)^
3 + 3*A*a^4*tan(1/2*d*x + 1/2*c) + 15*C*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

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maple [A]  time = 0.85, size = 164, normalized size = 1.66 \[ \frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{6 d \,a^{2}}+\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}+\frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}+\frac {5 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}-\frac {C}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) C}{d \,a^{2}}-\frac {C}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) C}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x)

[Out]

1/6/d/a^2*tan(1/2*d*x+1/2*c)^3*A+1/6/d/a^2*C*tan(1/2*d*x+1/2*c)^3+1/2/d/a^2*A*tan(1/2*d*x+1/2*c)+5/2/d/a^2*C*t
an(1/2*d*x+1/2*c)-1/d/a^2/(tan(1/2*d*x+1/2*c)-1)*C+2/d/a^2*ln(tan(1/2*d*x+1/2*c)-1)*C-1/d/a^2/(tan(1/2*d*x+1/2
*c)+1)*C-2/d/a^2*ln(tan(1/2*d*x+1/2*c)+1)*C

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maxima [B]  time = 0.42, size = 191, normalized size = 1.93 \[ \frac {C {\left (\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} - \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + \frac {A {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(C*((15*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 12*log(sin(d*x + c)/(
cos(d*x + c) + 1) + 1)/a^2 + 12*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2 + 12*sin(d*x + c)/((a^2 - a^2*sin
(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1))) + A*(3*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3
/(cos(d*x + c) + 1)^3)/a^2)/d

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mupad [B]  time = 2.70, size = 111, normalized size = 1.12 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A+C}{a^2}-\frac {A-3\,C}{2\,a^2}\right )}{d}-\frac {4\,C\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}-\frac {2\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A+C\right )}{6\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C/cos(c + d*x)^2)/(cos(c + d*x)^2*(a + a/cos(c + d*x))^2),x)

[Out]

(tan(c/2 + (d*x)/2)*((A + C)/a^2 - (A - 3*C)/(2*a^2)))/d - (4*C*atanh(tan(c/2 + (d*x)/2)))/(a^2*d) - (2*C*tan(
c/2 + (d*x)/2))/(d*(a^2*tan(c/2 + (d*x)/2)^2 - a^2)) + (tan(c/2 + (d*x)/2)^3*(A + C))/(6*a^2*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {A \sec ^{2}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{4}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**2,x)

[Out]

(Integral(A*sec(c + d*x)**2/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x)**4/(sec(c + d
*x)**2 + 2*sec(c + d*x) + 1), x))/a**2

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