3.99 \(\int \frac {\sec ^5(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=202 \[ \frac {8 (9 A-19 B) \tan (c+d x)}{15 a^3 d}-\frac {(6 A-13 B) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}+\frac {4 (9 A-19 B) \tan (c+d x) \sec ^2(c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}-\frac {(6 A-13 B) \tan (c+d x) \sec (c+d x)}{2 a^3 d}+\frac {(A-B) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac {(6 A-11 B) \tan (c+d x) \sec ^3(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]

[Out]

-1/2*(6*A-13*B)*arctanh(sin(d*x+c))/a^3/d+8/15*(9*A-19*B)*tan(d*x+c)/a^3/d-1/2*(6*A-13*B)*sec(d*x+c)*tan(d*x+c
)/a^3/d+1/5*(A-B)*sec(d*x+c)^4*tan(d*x+c)/d/(a+a*sec(d*x+c))^3+1/15*(6*A-11*B)*sec(d*x+c)^3*tan(d*x+c)/a/d/(a+
a*sec(d*x+c))^2+4/15*(9*A-19*B)*sec(d*x+c)^2*tan(d*x+c)/d/(a^3+a^3*sec(d*x+c))

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Rubi [A]  time = 0.47, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ \frac {8 (9 A-19 B) \tan (c+d x)}{15 a^3 d}-\frac {(6 A-13 B) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}+\frac {4 (9 A-19 B) \tan (c+d x) \sec ^2(c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}-\frac {(6 A-13 B) \tan (c+d x) \sec (c+d x)}{2 a^3 d}+\frac {(A-B) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac {(6 A-11 B) \tan (c+d x) \sec ^3(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^5*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^3,x]

[Out]

-((6*A - 13*B)*ArcTanh[Sin[c + d*x]])/(2*a^3*d) + (8*(9*A - 19*B)*Tan[c + d*x])/(15*a^3*d) - ((6*A - 13*B)*Sec
[c + d*x]*Tan[c + d*x])/(2*a^3*d) + ((A - B)*Sec[c + d*x]^4*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) + ((6*A
- 11*B)*Sec[c + d*x]^3*Tan[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) + (4*(9*A - 19*B)*Sec[c + d*x]^2*Tan[c +
d*x])/(15*d*(a^3 + a^3*Sec[c + d*x]))

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 ^5(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx &=\frac {(A-B) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {\int \frac {\sec ^4(c+d x) (4 a (A-B)-a (2 A-7 B) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=\frac {(A-B) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(6 A-11 B) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {\int \frac {\sec ^3(c+d x) \left (3 a^2 (6 A-11 B)-a^2 (18 A-43 B) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=\frac {(A-B) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(6 A-11 B) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {4 (9 A-19 B) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {\int \sec ^2(c+d x) \left (8 a^3 (9 A-19 B)-15 a^3 (6 A-13 B) \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=\frac {(A-B) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(6 A-11 B) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {4 (9 A-19 B) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {(8 (9 A-19 B)) \int \sec ^2(c+d x) \, dx}{15 a^3}-\frac {(6 A-13 B) \int \sec ^3(c+d x) \, dx}{a^3}\\ &=-\frac {(6 A-13 B) \sec (c+d x) \tan (c+d x)}{2 a^3 d}+\frac {(A-B) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(6 A-11 B) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {4 (9 A-19 B) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {(6 A-13 B) \int \sec (c+d x) \, dx}{2 a^3}-\frac {(8 (9 A-19 B)) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^3 d}\\ &=-\frac {(6 A-13 B) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}+\frac {8 (9 A-19 B) \tan (c+d x)}{15 a^3 d}-\frac {(6 A-13 B) \sec (c+d x) \tan (c+d x)}{2 a^3 d}+\frac {(A-B) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(6 A-11 B) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {4 (9 A-19 B) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [B]  time = 6.40, size = 768, normalized size = 3.80 \[ \frac {\sec \left (\frac {c}{2}\right ) \sec (c) \cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^4(c+d x) \left (-2094 A \sin \left (c-\frac {d x}{2}\right )+1314 A \sin \left (c+\frac {d x}{2}\right )-1650 A \sin \left (2 c+\frac {d x}{2}\right )-450 A \sin \left (c+\frac {3 d x}{2}\right )+1230 A \sin \left (2 c+\frac {3 d x}{2}\right )-1050 A \sin \left (3 c+\frac {3 d x}{2}\right )+1278 A \sin \left (c+\frac {5 d x}{2}\right )-90 A \sin \left (2 c+\frac {5 d x}{2}\right )+918 A \sin \left (3 c+\frac {5 d x}{2}\right )-450 A \sin \left (4 c+\frac {5 d x}{2}\right )+630 A \sin \left (2 c+\frac {7 d x}{2}\right )+60 A \sin \left (3 c+\frac {7 d x}{2}\right )+480 A \sin \left (4 c+\frac {7 d x}{2}\right )-90 A \sin \left (5 c+\frac {7 d x}{2}\right )+144 A \sin \left (3 c+\frac {9 d x}{2}\right )+30 A \sin \left (4 c+\frac {9 d x}{2}\right )+114 A \sin \left (5 c+\frac {9 d x}{2}\right )-870 A \sin \left (\frac {d x}{2}\right )+1830 A \sin \left (\frac {3 d x}{2}\right )+4329 B \sin \left (c-\frac {d x}{2}\right )-1989 B \sin \left (c+\frac {d x}{2}\right )+3575 B \sin \left (2 c+\frac {d x}{2}\right )+475 B \sin \left (c+\frac {3 d x}{2}\right )-2005 B \sin \left (2 c+\frac {3 d x}{2}\right )+2275 B \sin \left (3 c+\frac {3 d x}{2}\right )-2673 B \sin \left (c+\frac {5 d x}{2}\right )-105 B \sin \left (2 c+\frac {5 d x}{2}\right )-1593 B \sin \left (3 c+\frac {5 d x}{2}\right )+975 B \sin \left (4 c+\frac {5 d x}{2}\right )-1325 B \sin \left (2 c+\frac {7 d x}{2}\right )-255 B \sin \left (3 c+\frac {7 d x}{2}\right )-875 B \sin \left (4 c+\frac {7 d x}{2}\right )+195 B \sin \left (5 c+\frac {7 d x}{2}\right )-304 B \sin \left (3 c+\frac {9 d x}{2}\right )-90 B \sin \left (4 c+\frac {9 d x}{2}\right )-214 B \sin \left (5 c+\frac {9 d x}{2}\right )+1235 B \sin \left (\frac {d x}{2}\right )-3805 B \sin \left (\frac {3 d x}{2}\right )\right ) (A+B \sec (c+d x))}{480 d (a \sec (c+d x)+a)^3 (A \cos (c+d x)+B)}-\frac {4 (13 B-6 A) \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^2(c+d x) (A+B \sec (c+d x)) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}{d (a \sec (c+d x)+a)^3 (A \cos (c+d x)+B)}+\frac {4 (13 B-6 A) \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^2(c+d x) (A+B \sec (c+d x)) \log \left (\sin \left (\frac {c}{2}+\frac {d x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}{d (a \sec (c+d x)+a)^3 (A \cos (c+d x)+B)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^5*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^3,x]

[Out]

(-4*(-6*A + 13*B)*Cos[c/2 + (d*x)/2]^6*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c + d*x]^2*(A + B*Sec[
c + d*x]))/(d*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^3) + (4*(-6*A + 13*B)*Cos[c/2 + (d*x)/2]^6*Log[Cos[c/2
+ (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c + d*x]^2*(A + B*Sec[c + d*x]))/(d*(B + A*Cos[c + d*x])*(a + a*Sec[c +
d*x])^3) + (Cos[c/2 + (d*x)/2]*Sec[c/2]*Sec[c]*Sec[c + d*x]^4*(A + B*Sec[c + d*x])*(-870*A*Sin[(d*x)/2] + 1235
*B*Sin[(d*x)/2] + 1830*A*Sin[(3*d*x)/2] - 3805*B*Sin[(3*d*x)/2] - 2094*A*Sin[c - (d*x)/2] + 4329*B*Sin[c - (d*
x)/2] + 1314*A*Sin[c + (d*x)/2] - 1989*B*Sin[c + (d*x)/2] - 1650*A*Sin[2*c + (d*x)/2] + 3575*B*Sin[2*c + (d*x)
/2] - 450*A*Sin[c + (3*d*x)/2] + 475*B*Sin[c + (3*d*x)/2] + 1230*A*Sin[2*c + (3*d*x)/2] - 2005*B*Sin[2*c + (3*
d*x)/2] - 1050*A*Sin[3*c + (3*d*x)/2] + 2275*B*Sin[3*c + (3*d*x)/2] + 1278*A*Sin[c + (5*d*x)/2] - 2673*B*Sin[c
+ (5*d*x)/2] - 90*A*Sin[2*c + (5*d*x)/2] - 105*B*Sin[2*c + (5*d*x)/2] + 918*A*Sin[3*c + (5*d*x)/2] - 1593*B*S
in[3*c + (5*d*x)/2] - 450*A*Sin[4*c + (5*d*x)/2] + 975*B*Sin[4*c + (5*d*x)/2] + 630*A*Sin[2*c + (7*d*x)/2] - 1
325*B*Sin[2*c + (7*d*x)/2] + 60*A*Sin[3*c + (7*d*x)/2] - 255*B*Sin[3*c + (7*d*x)/2] + 480*A*Sin[4*c + (7*d*x)/
2] - 875*B*Sin[4*c + (7*d*x)/2] - 90*A*Sin[5*c + (7*d*x)/2] + 195*B*Sin[5*c + (7*d*x)/2] + 144*A*Sin[3*c + (9*
d*x)/2] - 304*B*Sin[3*c + (9*d*x)/2] + 30*A*Sin[4*c + (9*d*x)/2] - 90*B*Sin[4*c + (9*d*x)/2] + 114*A*Sin[5*c +
(9*d*x)/2] - 214*B*Sin[5*c + (9*d*x)/2]))/(480*d*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^3)

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fricas [A]  time = 0.45, size = 295, normalized size = 1.46 \[ -\frac {15 \, {\left ({\left (6 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (6 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (6 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (6 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (6 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (6 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (6 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (6 \, A - 13 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (16 \, {\left (9 \, A - 19 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (114 \, A - 239 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (234 \, A - 479 \, B\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (2 \, A - 3 \, B\right )} \cos \left (d x + c\right ) + 15 \, B\right )} \sin \left (d x + c\right )}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/60*(15*((6*A - 13*B)*cos(d*x + c)^5 + 3*(6*A - 13*B)*cos(d*x + c)^4 + 3*(6*A - 13*B)*cos(d*x + c)^3 + (6*A
- 13*B)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 15*((6*A - 13*B)*cos(d*x + c)^5 + 3*(6*A - 13*B)*cos(d*x + c)^
4 + 3*(6*A - 13*B)*cos(d*x + c)^3 + (6*A - 13*B)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(16*(9*A - 19*B)*c
os(d*x + c)^4 + 3*(114*A - 239*B)*cos(d*x + c)^3 + (234*A - 479*B)*cos(d*x + c)^2 + 15*(2*A - 3*B)*cos(d*x + c
) + 15*B)*sin(d*x + c))/(a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*
x + c)^2)

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giac [A]  time = 0.35, size = 233, normalized size = 1.15 \[ -\frac {\frac {30 \, {\left (6 \, A - 13 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {30 \, {\left (6 \, A - 13 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {60 \, {\left (2 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 30 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 255 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 465 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

-1/60*(30*(6*A - 13*B)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 30*(6*A - 13*B)*log(abs(tan(1/2*d*x + 1/2*c) -
1))/a^3 + 60*(2*A*tan(1/2*d*x + 1/2*c)^3 - 7*B*tan(1/2*d*x + 1/2*c)^3 - 2*A*tan(1/2*d*x + 1/2*c) + 5*B*tan(1/
2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^3) - (3*A*a^12*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^12*tan(1/2*d*x
+ 1/2*c)^5 + 30*A*a^12*tan(1/2*d*x + 1/2*c)^3 - 40*B*a^12*tan(1/2*d*x + 1/2*c)^3 + 255*A*a^12*tan(1/2*d*x + 1
/2*c) - 465*B*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d

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maple [A]  time = 0.70, size = 334, normalized size = 1.65 \[ \frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}-\frac {B \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{2 d \,a^{3}}-\frac {2 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \,a^{3}}+\frac {17 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}-\frac {31 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) A}{d \,a^{3}}-\frac {13 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B}{2 d \,a^{3}}+\frac {7 B}{2 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {A}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {B}{2 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {7 B}{2 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {A}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) A}{d \,a^{3}}+\frac {13 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B}{2 d \,a^{3}}-\frac {B}{2 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x)

[Out]

1/20/d/a^3*A*tan(1/2*d*x+1/2*c)^5-1/20/d/a^3*B*tan(1/2*d*x+1/2*c)^5+1/2/d/a^3*tan(1/2*d*x+1/2*c)^3*A-2/3/d/a^3
*B*tan(1/2*d*x+1/2*c)^3+17/4/d/a^3*A*tan(1/2*d*x+1/2*c)-31/4/d/a^3*B*tan(1/2*d*x+1/2*c)+3/d/a^3*ln(tan(1/2*d*x
+1/2*c)-1)*A-13/2/d/a^3*ln(tan(1/2*d*x+1/2*c)-1)*B+7/2/d/a^3/(tan(1/2*d*x+1/2*c)-1)*B-1/d/a^3/(tan(1/2*d*x+1/2
*c)-1)*A+1/2/d/a^3*B/(tan(1/2*d*x+1/2*c)-1)^2+7/2/d/a^3/(tan(1/2*d*x+1/2*c)+1)*B-1/d/a^3/(tan(1/2*d*x+1/2*c)+1
)*A-3/d/a^3*ln(tan(1/2*d*x+1/2*c)+1)*A+13/2/d/a^3*ln(tan(1/2*d*x+1/2*c)+1)*B-1/2/d/a^3*B/(tan(1/2*d*x+1/2*c)+1
)^2

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maxima [A]  time = 0.36, size = 377, normalized size = 1.87 \[ -\frac {B {\left (\frac {60 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} - \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - 3 \, A {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/60*(B*(60*(5*sin(d*x + c)/(cos(d*x + c) + 1) - 7*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^3 - 2*a^3*sin(d*x
+ c)^2/(cos(d*x + c) + 1)^2 + a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (465*sin(d*x + c)/(cos(d*x + c) + 1)
+ 40*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 390*log(sin(d*x + c)/(
cos(d*x + c) + 1) + 1)/a^3 + 390*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3) - 3*A*(40*sin(d*x + c)/((a^3 -
a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (85*sin(d*x + c)/(cos(d*x + c) + 1) + 10*sin(d*
x + c)^3/(cos(d*x + c) + 1)^3 + sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 60*log(sin(d*x + c)/(cos(d*x + c) +
1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3))/d

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mupad [B]  time = 2.03, size = 216, normalized size = 1.07 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A-B\right )}{2\,a^3}+\frac {3\,\left (3\,A-5\,B\right )}{4\,a^3}+\frac {2\,A-10\,B}{4\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,A-7\,B\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A-5\,B\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B}{4\,a^3}+\frac {3\,A-5\,B}{12\,a^3}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B\right )}{20\,a^3\,d}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (6\,A-13\,B\right )}{a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B/cos(c + d*x))/(cos(c + d*x)^5*(a + a/cos(c + d*x))^3),x)

[Out]

(tan(c/2 + (d*x)/2)*((3*(A - B))/(2*a^3) + (3*(3*A - 5*B))/(4*a^3) + (2*A - 10*B)/(4*a^3)))/d - (tan(c/2 + (d*
x)/2)^3*(2*A - 7*B) - tan(c/2 + (d*x)/2)*(2*A - 5*B))/(d*(a^3*tan(c/2 + (d*x)/2)^4 - 2*a^3*tan(c/2 + (d*x)/2)^
2 + a^3)) + (tan(c/2 + (d*x)/2)^3*((A - B)/(4*a^3) + (3*A - 5*B)/(12*a^3)))/d + (tan(c/2 + (d*x)/2)^5*(A - B))
/(20*a^3*d) - (atanh(tan(c/2 + (d*x)/2))*(6*A - 13*B))/(a^3*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {A \sec ^{5}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{6}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**3,x)

[Out]

(Integral(A*sec(c + d*x)**5/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(B*sec(c
+ d*x)**6/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x))/a**3

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