3.91 \(\int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=156 \[ \frac {2 (5 A-8 B) \tan (c+d x)}{3 a^2 d}-\frac {(4 A-7 B) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}+\frac {(5 A-8 B) \tan (c+d x) \sec ^2(c+d x)}{3 a^2 d (\sec (c+d x)+1)}-\frac {(4 A-7 B) \tan (c+d x) \sec (c+d x)}{2 a^2 d}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[Out]

-1/2*(4*A-7*B)*arctanh(sin(d*x+c))/a^2/d+2/3*(5*A-8*B)*tan(d*x+c)/a^2/d-1/2*(4*A-7*B)*sec(d*x+c)*tan(d*x+c)/a^
2/d+1/3*(5*A-8*B)*sec(d*x+c)^2*tan(d*x+c)/a^2/d/(1+sec(d*x+c))+1/3*(A-B)*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*
x+c))^2

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Rubi [A]  time = 0.31, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4019, 3787, 3767, 8, 3768, 3770} \[ \frac {2 (5 A-8 B) \tan (c+d x)}{3 a^2 d}-\frac {(4 A-7 B) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}+\frac {(5 A-8 B) \tan (c+d x) \sec ^2(c+d x)}{3 a^2 d (\sec (c+d x)+1)}-\frac {(4 A-7 B) \tan (c+d x) \sec (c+d x)}{2 a^2 d}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^4*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^2,x]

[Out]

-((4*A - 7*B)*ArcTanh[Sin[c + d*x]])/(2*a^2*d) + (2*(5*A - 8*B)*Tan[c + d*x])/(3*a^2*d) - ((4*A - 7*B)*Sec[c +
d*x]*Tan[c + d*x])/(2*a^2*d) + ((5*A - 8*B)*Sec[c + d*x]^2*Tan[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) + ((A -
B)*Sec[c + d*x]^3*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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

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 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 ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx &=\frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {\int \frac {\sec ^3(c+d x) (3 a (A-B)-a (2 A-5 B) \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=\frac {(5 A-8 B) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}+\frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {\int \sec ^2(c+d x) \left (2 a^2 (5 A-8 B)-3 a^2 (4 A-7 B) \sec (c+d x)\right ) \, dx}{3 a^4}\\ &=\frac {(5 A-8 B) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}+\frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {(2 (5 A-8 B)) \int \sec ^2(c+d x) \, dx}{3 a^2}-\frac {(4 A-7 B) \int \sec ^3(c+d x) \, dx}{a^2}\\ &=-\frac {(4 A-7 B) \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac {(5 A-8 B) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}+\frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(4 A-7 B) \int \sec (c+d x) \, dx}{2 a^2}-\frac {(2 (5 A-8 B)) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 a^2 d}\\ &=-\frac {(4 A-7 B) \tanh ^{-1}(\sin (c+d x))}{2 a^2 d}+\frac {2 (5 A-8 B) \tan (c+d x)}{3 a^2 d}-\frac {(4 A-7 B) \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac {(5 A-8 B) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}+\frac {(A-B) \sec ^3(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 = 4.19, size = 496, normalized size = 3.18 \[ \frac {96 (4 A-7 B) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+\sec \left (\frac {c}{2}\right ) \sec (c) \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (-14 (A-B) \sin \left (\frac {d x}{2}\right )+(64 A-97 B) \sin \left (\frac {3 d x}{2}\right )-84 A \sin \left (c-\frac {d x}{2}\right )+42 A \sin \left (c+\frac {d x}{2}\right )-56 A \sin \left (2 c+\frac {d x}{2}\right )-6 A \sin \left (c+\frac {3 d x}{2}\right )+34 A \sin \left (2 c+\frac {3 d x}{2}\right )-36 A \sin \left (3 c+\frac {3 d x}{2}\right )+48 A \sin \left (c+\frac {5 d x}{2}\right )+6 A \sin \left (2 c+\frac {5 d x}{2}\right )+30 A \sin \left (3 c+\frac {5 d x}{2}\right )-12 A \sin \left (4 c+\frac {5 d x}{2}\right )+20 A \sin \left (2 c+\frac {7 d x}{2}\right )+6 A \sin \left (3 c+\frac {7 d x}{2}\right )+14 A \sin \left (4 c+\frac {7 d x}{2}\right )+126 B \sin \left (c-\frac {d x}{2}\right )-42 B \sin \left (c+\frac {d x}{2}\right )+98 B \sin \left (2 c+\frac {d x}{2}\right )+3 B \sin \left (c+\frac {3 d x}{2}\right )-37 B \sin \left (2 c+\frac {3 d x}{2}\right )+63 B \sin \left (3 c+\frac {3 d x}{2}\right )-75 B \sin \left (c+\frac {5 d x}{2}\right )-15 B \sin \left (2 c+\frac {5 d x}{2}\right )-39 B \sin \left (3 c+\frac {5 d x}{2}\right )+21 B \sin \left (4 c+\frac {5 d x}{2}\right )-32 B \sin \left (2 c+\frac {7 d x}{2}\right )-12 B \sin \left (3 c+\frac {7 d x}{2}\right )-20 B \sin \left (4 c+\frac {7 d x}{2}\right )\right )}{48 a^2 d (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^4*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^2,x]

[Out]

(96*(4*A - 7*B)*Cos[(c + d*x)/2]^4*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c +
d*x)/2]]) + Cos[(c + d*x)/2]*Sec[c/2]*Sec[c]*Sec[c + d*x]^2*(-14*(A - B)*Sin[(d*x)/2] + (64*A - 97*B)*Sin[(3*
d*x)/2] - 84*A*Sin[c - (d*x)/2] + 126*B*Sin[c - (d*x)/2] + 42*A*Sin[c + (d*x)/2] - 42*B*Sin[c + (d*x)/2] - 56*
A*Sin[2*c + (d*x)/2] + 98*B*Sin[2*c + (d*x)/2] - 6*A*Sin[c + (3*d*x)/2] + 3*B*Sin[c + (3*d*x)/2] + 34*A*Sin[2*
c + (3*d*x)/2] - 37*B*Sin[2*c + (3*d*x)/2] - 36*A*Sin[3*c + (3*d*x)/2] + 63*B*Sin[3*c + (3*d*x)/2] + 48*A*Sin[
c + (5*d*x)/2] - 75*B*Sin[c + (5*d*x)/2] + 6*A*Sin[2*c + (5*d*x)/2] - 15*B*Sin[2*c + (5*d*x)/2] + 30*A*Sin[3*c
+ (5*d*x)/2] - 39*B*Sin[3*c + (5*d*x)/2] - 12*A*Sin[4*c + (5*d*x)/2] + 21*B*Sin[4*c + (5*d*x)/2] + 20*A*Sin[2
*c + (7*d*x)/2] - 32*B*Sin[2*c + (7*d*x)/2] + 6*A*Sin[3*c + (7*d*x)/2] - 12*B*Sin[3*c + (7*d*x)/2] + 14*A*Sin[
4*c + (7*d*x)/2] - 20*B*Sin[4*c + (7*d*x)/2]))/(48*a^2*d*(1 + Cos[c + d*x])^2)

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fricas [A]  time = 0.45, size = 228, normalized size = 1.46 \[ -\frac {3 \, {\left ({\left (4 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (4 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (4 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (4 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (4 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (4 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, {\left (5 \, A - 8 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (28 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{2} + 6 \, {\left (A - B\right )} \cos \left (d x + c\right ) + 3 \, B\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/12*(3*((4*A - 7*B)*cos(d*x + c)^4 + 2*(4*A - 7*B)*cos(d*x + c)^3 + (4*A - 7*B)*cos(d*x + c)^2)*log(sin(d*x
+ c) + 1) - 3*((4*A - 7*B)*cos(d*x + c)^4 + 2*(4*A - 7*B)*cos(d*x + c)^3 + (4*A - 7*B)*cos(d*x + c)^2)*log(-si
n(d*x + c) + 1) - 2*(4*(5*A - 8*B)*cos(d*x + c)^3 + (28*A - 43*B)*cos(d*x + c)^2 + 6*(A - B)*cos(d*x + c) + 3*
B)*sin(d*x + c))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2)

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giac [A]  time = 0.29, size = 198, normalized size = 1.27 \[ -\frac {\frac {3 \, {\left (4 \, A - 7 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {3 \, {\left (4 \, A - 7 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac {6 \, {\left (2 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{2}} - \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 21 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

-1/6*(3*(4*A - 7*B)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 3*(4*A - 7*B)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/
a^2 + 6*(2*A*tan(1/2*d*x + 1/2*c)^3 - 5*B*tan(1/2*d*x + 1/2*c)^3 - 2*A*tan(1/2*d*x + 1/2*c) + 3*B*tan(1/2*d*x
+ 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*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
+ 15*A*a^4*tan(1/2*d*x + 1/2*c) - 21*B*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

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maple [B]  time = 0.68, size = 294, normalized size = 1.88 \[ \frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{6 d \,a^{2}}-\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}+\frac {5 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}-\frac {7 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}-\frac {A}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {5 B}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {2 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{2}}-\frac {7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B}{2 d \,a^{2}}+\frac {B}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {A}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {5 B}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{2}}+\frac {7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B}{2 d \,a^{2}}-\frac {B}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(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+5/2/d/a^2*A*tan(1/2*d*x+1/2*c)-7/2/d/a^2*B*t
an(1/2*d*x+1/2*c)-1/d/a^2*A/(tan(1/2*d*x+1/2*c)-1)+5/2/d/a^2/(tan(1/2*d*x+1/2*c)-1)*B+2/d/a^2*A*ln(tan(1/2*d*x
+1/2*c)-1)-7/2/d/a^2*ln(tan(1/2*d*x+1/2*c)-1)*B+1/2/d/a^2/(tan(1/2*d*x+1/2*c)-1)^2*B-1/d/a^2*A/(tan(1/2*d*x+1/
2*c)+1)+5/2/d/a^2/(tan(1/2*d*x+1/2*c)+1)*B-2/d/a^2*A*ln(tan(1/2*d*x+1/2*c)+1)+7/2/d/a^2*ln(tan(1/2*d*x+1/2*c)+
1)*B-1/2/d/a^2/(tan(1/2*d*x+1/2*c)+1)^2*B

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maxima [B]  time = 0.36, size = 336, normalized size = 2.15 \[ -\frac {B {\left (\frac {6 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} - \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {21 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {21 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )} - A {\left (\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} - \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/6*(B*(6*(3*sin(d*x + c)/(cos(d*x + c) + 1) - 5*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^2 - 2*a^2*sin(d*x +
c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (21*sin(d*x + c)/(cos(d*x + c) + 1) + s
in(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 21*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 21*log(sin(d*x + c
)/(cos(d*x + c) + 1) - 1)/a^2) - A*((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))))/d

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mupad [B]  time = 1.93, size = 166, normalized size = 1.06 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A-B\right )}{2\,a^2}+\frac {2\,A-4\,B}{2\,a^2}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,A-5\,B\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A-3\,B\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-B\right )}{6\,a^2\,d}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (4\,A-7\,B\right )}{a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B/cos(c + d*x))/(cos(c + d*x)^4*(a + a/cos(c + d*x))^2),x)

[Out]

(tan(c/2 + (d*x)/2)*((3*(A - B))/(2*a^2) + (2*A - 4*B)/(2*a^2)))/d - (tan(c/2 + (d*x)/2)^3*(2*A - 5*B) - tan(c
/2 + (d*x)/2)*(2*A - 3*B))/(d*(a^2*tan(c/2 + (d*x)/2)^4 - 2*a^2*tan(c/2 + (d*x)/2)^2 + a^2)) + (tan(c/2 + (d*x
)/2)^3*(A - B))/(6*a^2*d) - (atanh(tan(c/2 + (d*x)/2))*(4*A - 7*B))/(a^2*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {A \sec ^{4}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{5}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**2,x)

[Out]

(Integral(A*sec(c + d*x)**4/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + Integral(B*sec(c + d*x)**5/(sec(c + d
*x)**2 + 2*sec(c + d*x) + 1), x))/a**2

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