3.92 \(\int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=108 \[ -\frac {(A-4 B) \tan (c+d x)}{3 a^2 d}+\frac {(A-2 B) \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac {(A-2 B) \tan (c+d x)}{a^2 d (\sec (c+d x)+1)}+\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[Out]

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

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Rubi [A]  time = 0.26, antiderivative size = 108, normalized size of antiderivative = 1.00, 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 = {4019, 4008, 3787, 3770, 3767, 8} \[ -\frac {(A-4 B) \tan (c+d x)}{3 a^2 d}+\frac {(A-2 B) \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac {(A-2 B) \tan (c+d x)}{a^2 d (\sec (c+d x)+1)}+\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^3*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^2,x]

[Out]

((A - 2*B)*ArcTanh[Sin[c + d*x]])/(a^2*d) - ((A - 4*B)*Tan[c + d*x])/(3*a^2*d) - ((A - 2*B)*Tan[c + d*x])/(a^2
*d*(1 + Sec[c + d*x])) + ((A - B)*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 4019

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

Rubi steps

\begin {align*} \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx &=\frac {(A-B) \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) (2 a (A-B)-a (A-4 B) \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=-\frac {(A-2 B) \tan (c+d x)}{a^2 d (1+\sec (c+d x))}+\frac {(A-B) \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 (-3 a^2 (A-2 B)+a^2 (A-4 B) \sec (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac {(A-2 B) \tan (c+d x)}{a^2 d (1+\sec (c+d x))}+\frac {(A-B) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(A-4 B) \int \sec ^2(c+d x) \, dx}{3 a^2}+\frac {(A-2 B) \int \sec (c+d x) \, dx}{a^2}\\ &=\frac {(A-2 B) \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac {(A-2 B) \tan (c+d x)}{a^2 d (1+\sec (c+d x))}+\frac {(A-B) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {(A-4 B) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 a^2 d}\\ &=\frac {(A-2 B) \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac {(A-4 B) \tan (c+d x)}{3 a^2 d}-\frac {(A-2 B) \tan (c+d x)}{a^2 d (1+\sec (c+d x))}+\frac {(A-B) \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 = 1.94, size = 292, normalized size = 2.70 \[ \frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) (A+B \sec (c+d x)) \left (-(A-B) \tan \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right )+(B-A) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+\cos ^3\left (\frac {1}{2} (c+d x)\right ) \left (\frac {6 B \sin (d x)}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}-6 (A-2 B) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )-2 (4 A-7 B) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right )\right )}{3 a^2 d (\sec (c+d x)+1)^2 (A \cos (c+d x)+B)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^3*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^2,x]

[Out]

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

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fricas [A]  time = 0.47, size = 195, normalized size = 1.81 \[ \frac {3 \, {\left ({\left (A - 2 \, B\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (A - 2 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (A - 2 \, B\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (A - 2 \, B\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (A - 2 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (A - 2 \, B\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, {\left (2 \, A - 5 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (5 \, A - 14 \, B\right )} \cos \left (d x + c\right ) - 3 \, B\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/6*(3*((A - 2*B)*cos(d*x + c)^3 + 2*(A - 2*B)*cos(d*x + c)^2 + (A - 2*B)*cos(d*x + c))*log(sin(d*x + c) + 1)
- 3*((A - 2*B)*cos(d*x + c)^3 + 2*(A - 2*B)*cos(d*x + c)^2 + (A - 2*B)*cos(d*x + c))*log(-sin(d*x + c) + 1) -
2*(2*(2*A - 5*B)*cos(d*x + c)^2 + (5*A - 14*B)*cos(d*x + c) - 3*B)*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 = 151, normalized size = 1.40 \[ \frac {\frac {6 \, {\left (A - 2 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {6 \, {\left (A - 2 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} - \frac {12 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{2}} - \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/6*(6*(A - 2*B)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 6*(A - 2*B)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^2 -
12*B*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 - B*a^4*tan(1/2*
d*x + 1/2*c)^3 + 9*A*a^4*tan(1/2*d*x + 1/2*c) - 15*B*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

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maple [A]  time = 0.70, size = 205, normalized size = 1.90 \[ -\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{6 d \,a^{2}}+\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}-\frac {3 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}+\frac {5 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}-\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{2}}+\frac {2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B}{d \,a^{2}}-\frac {B}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{2}}-\frac {2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B}{d \,a^{2}}-\frac {B}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(A+B*sec(d*x+c))/(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*B*tan(1/2*d*x+1/2*c)^3-3/2/d/a^2*A*tan(1/2*d*x+1/2*c)+5/2/d/a^2*B*
tan(1/2*d*x+1/2*c)-1/d/a^2*A*ln(tan(1/2*d*x+1/2*c)-1)+2/d/a^2*ln(tan(1/2*d*x+1/2*c)-1)*B-1/d/a^2/(tan(1/2*d*x+
1/2*c)-1)*B+1/d/a^2*A*ln(tan(1/2*d*x+1/2*c)+1)-2/d/a^2*ln(tan(1/2*d*x+1/2*c)+1)*B-1/d/a^2/(tan(1/2*d*x+1/2*c)+
1)*B

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maxima [B]  time = 0.34, size = 244, normalized size = 2.26 \[ \frac {B {\left (\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} - \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - A {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

1/6*(B*((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*((9*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^
3/(cos(d*x + c) + 1)^3)/a^2 - 6*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 6*log(sin(d*x + c)/(cos(d*x + c
) + 1) - 1)/a^2))/d

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mupad [B]  time = 1.93, size = 120, normalized size = 1.11 \[ \frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A-2\,B\right )}{a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-B\right )}{6\,a^2\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A-B}{a^2}+\frac {A-3\,B}{2\,a^2}\right )}{d}-\frac {2\,B\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B/cos(c + d*x))/(cos(c + d*x)^3*(a + a/cos(c + d*x))^2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {A \sec ^{3}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{4}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**2,x)

[Out]

(Integral(A*sec(c + d*x)**3/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + Integral(B*sec(c + d*x)**4/(sec(c + d
*x)**2 + 2*sec(c + d*x) + 1), x))/a**2

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