3.2.9 \(\int \frac {\sec ^5(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx\) [109]

Optimal. Leaf size=194 \[ \frac {(A-4 B) \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac {(55 A-244 B) \tan (c+d x)}{105 a^4 d}+\frac {(25 A-88 B) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-4 B) \tan (c+d x)}{a^4 d (1+\sec (c+d x))}+\frac {(A-B) \sec ^4(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(5 A-12 B) \sec ^3(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3} \]

[Out]

(A-4*B)*arctanh(sin(d*x+c))/a^4/d-1/105*(55*A-244*B)*tan(d*x+c)/a^4/d+1/105*(25*A-88*B)*sec(d*x+c)^2*tan(d*x+c
)/a^4/d/(1+sec(d*x+c))^2-(A-4*B)*tan(d*x+c)/a^4/d/(1+sec(d*x+c))+1/7*(A-B)*sec(d*x+c)^4*tan(d*x+c)/d/(a+a*sec(
d*x+c))^4+1/35*(5*A-12*B)*sec(d*x+c)^3*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^3

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Rubi [A]
time = 0.41, antiderivative size = 194, 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 = {4104, 4093, 3872, 3855, 3852, 8} \begin {gather*} -\frac {(55 A-244 B) \tan (c+d x)}{105 a^4 d}+\frac {(A-4 B) \tanh ^{-1}(\sin (c+d x))}{a^4 d}+\frac {(25 A-88 B) \tan (c+d x) \sec ^2(c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}-\frac {(A-4 B) \tan (c+d x)}{a^4 d (\sec (c+d x)+1)}+\frac {(A-B) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}+\frac {(5 A-12 B) \tan (c+d x) \sec ^3(c+d x)}{35 a d (a \sec (c+d x)+a)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^5*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^4,x]

[Out]

((A - 4*B)*ArcTanh[Sin[c + d*x]])/(a^4*d) - ((55*A - 244*B)*Tan[c + d*x])/(105*a^4*d) + ((25*A - 88*B)*Sec[c +
d*x]^2*Tan[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])^2) - ((A - 4*B)*Tan[c + d*x])/(a^4*d*(1 + Sec[c + d*x])) +
((A - B)*Sec[c + d*x]^4*Tan[c + d*x])/(7*d*(a + a*Sec[c + d*x])^4) + ((5*A - 12*B)*Sec[c + d*x]^3*Tan[c + d*x
])/(35*a*d*(a + a*Sec[c + d*x])^3)

Rule 8

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

Rule 3852

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 3855

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

Rule 3872

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 4093

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 4104

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))^4} \, dx &=\frac {(A-B) \sec ^4(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {\int \frac {\sec ^4(c+d x) (4 a (A-B)-a (A-8 B) \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx}{7 a^2}\\ &=\frac {(A-B) \sec ^4(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(5 A-12 B) \sec ^3(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {\int \frac {\sec ^3(c+d x) \left (3 a^2 (5 A-12 B)-2 a^2 (5 A-26 B) \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{35 a^4}\\ &=\frac {(25 A-88 B) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}+\frac {(A-B) \sec ^4(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(5 A-12 B) \sec ^3(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {\int \frac {\sec ^2(c+d x) \left (2 a^3 (25 A-88 B)-a^3 (55 A-244 B) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{105 a^6}\\ &=\frac {(25 A-88 B) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}+\frac {(A-B) \sec ^4(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(5 A-12 B) \sec ^3(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac {(A-4 B) \tan (c+d x)}{d \left (a^4+a^4 \sec (c+d x)\right )}-\frac {\int \sec (c+d x) \left (-105 a^4 (A-4 B)+a^4 (55 A-244 B) \sec (c+d x)\right ) \, dx}{105 a^8}\\ &=\frac {(25 A-88 B) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}+\frac {(A-B) \sec ^4(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(5 A-12 B) \sec ^3(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac {(A-4 B) \tan (c+d x)}{d \left (a^4+a^4 \sec (c+d x)\right )}-\frac {(55 A-244 B) \int \sec ^2(c+d x) \, dx}{105 a^4}+\frac {(A-4 B) \int \sec (c+d x) \, dx}{a^4}\\ &=\frac {(A-4 B) \tanh ^{-1}(\sin (c+d x))}{a^4 d}+\frac {(25 A-88 B) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}+\frac {(A-B) \sec ^4(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(5 A-12 B) \sec ^3(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac {(A-4 B) \tan (c+d x)}{d \left (a^4+a^4 \sec (c+d x)\right )}+\frac {(55 A-244 B) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{105 a^4 d}\\ &=\frac {(A-4 B) \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac {(55 A-244 B) \tan (c+d x)}{105 a^4 d}+\frac {(25 A-88 B) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}+\frac {(A-B) \sec ^4(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(5 A-12 B) \sec ^3(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac {(A-4 B) \tan (c+d x)}{d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(754\) vs. \(2(194)=388\).
time = 6.42, size = 754, normalized size = 3.89 \begin {gather*} \frac {16 (-A+4 B) \cos ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sec ^3(c+d x) (A+B \sec (c+d x))}{d (B+A \cos (c+d x)) (a+a \sec (c+d x))^4}-\frac {16 (-A+4 B) \cos ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sec ^3(c+d x) (A+B \sec (c+d x))}{d (B+A \cos (c+d x)) (a+a \sec (c+d x))^4}+\frac {\cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec (c) \sec ^4(c+d x) (A+B \sec (c+d x)) \left (4165 A \sin \left (\frac {d x}{2}\right )-10780 B \sin \left (\frac {d x}{2}\right )-4445 A \sin \left (\frac {3 d x}{2}\right )+18788 B \sin \left (\frac {3 d x}{2}\right )+4795 A \sin \left (c-\frac {d x}{2}\right )-20524 B \sin \left (c-\frac {d x}{2}\right )-4795 A \sin \left (c+\frac {d x}{2}\right )+14644 B \sin \left (c+\frac {d x}{2}\right )+4165 A \sin \left (2 c+\frac {d x}{2}\right )-16660 B \sin \left (2 c+\frac {d x}{2}\right )+2275 A \sin \left (c+\frac {3 d x}{2}\right )-4690 B \sin \left (c+\frac {3 d x}{2}\right )-4445 A \sin \left (2 c+\frac {3 d x}{2}\right )+14378 B \sin \left (2 c+\frac {3 d x}{2}\right )+2275 A \sin \left (3 c+\frac {3 d x}{2}\right )-9100 B \sin \left (3 c+\frac {3 d x}{2}\right )-2785 A \sin \left (c+\frac {5 d x}{2}\right )+11668 B \sin \left (c+\frac {5 d x}{2}\right )+735 A \sin \left (2 c+\frac {5 d x}{2}\right )-630 B \sin \left (2 c+\frac {5 d x}{2}\right )-2785 A \sin \left (3 c+\frac {5 d x}{2}\right )+9358 B \sin \left (3 c+\frac {5 d x}{2}\right )+735 A \sin \left (4 c+\frac {5 d x}{2}\right )-2940 B \sin \left (4 c+\frac {5 d x}{2}\right )-1015 A \sin \left (2 c+\frac {7 d x}{2}\right )+4228 B \sin \left (2 c+\frac {7 d x}{2}\right )+105 A \sin \left (3 c+\frac {7 d x}{2}\right )+315 B \sin \left (3 c+\frac {7 d x}{2}\right )-1015 A \sin \left (4 c+\frac {7 d x}{2}\right )+3493 B \sin \left (4 c+\frac {7 d x}{2}\right )+105 A \sin \left (5 c+\frac {7 d x}{2}\right )-420 B \sin \left (5 c+\frac {7 d x}{2}\right )-160 A \sin \left (3 c+\frac {9 d x}{2}\right )+664 B \sin \left (3 c+\frac {9 d x}{2}\right )+105 B \sin \left (4 c+\frac {9 d x}{2}\right )-160 A \sin \left (5 c+\frac {9 d x}{2}\right )+559 B \sin \left (5 c+\frac {9 d x}{2}\right )\right )}{1680 d (B+A \cos (c+d x)) (a+a \sec (c+d x))^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^5*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^4,x]

[Out]

(16*(-A + 4*B)*Cos[c/2 + (d*x)/2]^8*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c + d*x]^3*(A + B*Sec[c +
d*x]))/(d*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^4) - (16*(-A + 4*B)*Cos[c/2 + (d*x)/2]^8*Log[Cos[c/2 + (d
*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c + d*x]^3*(A + B*Sec[c + d*x]))/(d*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])
^4) + (Cos[c/2 + (d*x)/2]*Sec[c/2]*Sec[c]*Sec[c + d*x]^4*(A + B*Sec[c + d*x])*(4165*A*Sin[(d*x)/2] - 10780*B*S
in[(d*x)/2] - 4445*A*Sin[(3*d*x)/2] + 18788*B*Sin[(3*d*x)/2] + 4795*A*Sin[c - (d*x)/2] - 20524*B*Sin[c - (d*x)
/2] - 4795*A*Sin[c + (d*x)/2] + 14644*B*Sin[c + (d*x)/2] + 4165*A*Sin[2*c + (d*x)/2] - 16660*B*Sin[2*c + (d*x)
/2] + 2275*A*Sin[c + (3*d*x)/2] - 4690*B*Sin[c + (3*d*x)/2] - 4445*A*Sin[2*c + (3*d*x)/2] + 14378*B*Sin[2*c +
(3*d*x)/2] + 2275*A*Sin[3*c + (3*d*x)/2] - 9100*B*Sin[3*c + (3*d*x)/2] - 2785*A*Sin[c + (5*d*x)/2] + 11668*B*S
in[c + (5*d*x)/2] + 735*A*Sin[2*c + (5*d*x)/2] - 630*B*Sin[2*c + (5*d*x)/2] - 2785*A*Sin[3*c + (5*d*x)/2] + 93
58*B*Sin[3*c + (5*d*x)/2] + 735*A*Sin[4*c + (5*d*x)/2] - 2940*B*Sin[4*c + (5*d*x)/2] - 1015*A*Sin[2*c + (7*d*x
)/2] + 4228*B*Sin[2*c + (7*d*x)/2] + 105*A*Sin[3*c + (7*d*x)/2] + 315*B*Sin[3*c + (7*d*x)/2] - 1015*A*Sin[4*c
+ (7*d*x)/2] + 3493*B*Sin[4*c + (7*d*x)/2] + 105*A*Sin[5*c + (7*d*x)/2] - 420*B*Sin[5*c + (7*d*x)/2] - 160*A*S
in[3*c + (9*d*x)/2] + 664*B*Sin[3*c + (9*d*x)/2] + 105*B*Sin[4*c + (9*d*x)/2] - 160*A*Sin[5*c + (9*d*x)/2] + 5
59*B*Sin[5*c + (9*d*x)/2]))/(1680*d*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^4)

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Maple [A]
time = 0.22, size = 190, normalized size = 0.98

method	result	size
derivativedivides	\(\frac {-\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{7}+\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{7}-\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A +\frac {7 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{5}-\frac {11 A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {23 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-15 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+49 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (32 B -8 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {8 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-32 B +8 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {8 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{8 d \,a^{4}}\)	\(190\)
default	\(\frac {-\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{7}+\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{7}-\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A +\frac {7 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{5}-\frac {11 A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {23 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-15 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+49 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (32 B -8 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {8 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-32 B +8 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {8 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{8 d \,a^{4}}\)	\(190\)
risch	\(-\frac {2 i \left (105 A \,{\mathrm e}^{8 i \left (d x +c \right )}-420 B \,{\mathrm e}^{8 i \left (d x +c \right )}+735 A \,{\mathrm e}^{7 i \left (d x +c \right )}-2940 B \,{\mathrm e}^{7 i \left (d x +c \right )}+2275 A \,{\mathrm e}^{6 i \left (d x +c \right )}-9100 B \,{\mathrm e}^{6 i \left (d x +c \right )}+4165 A \,{\mathrm e}^{5 i \left (d x +c \right )}-16660 B \,{\mathrm e}^{5 i \left (d x +c \right )}+4795 A \,{\mathrm e}^{4 i \left (d x +c \right )}-20524 B \,{\mathrm e}^{4 i \left (d x +c \right )}+4445 A \,{\mathrm e}^{3 i \left (d x +c \right )}-18788 B \,{\mathrm e}^{3 i \left (d x +c \right )}+2785 A \,{\mathrm e}^{2 i \left (d x +c \right )}-11668 B \,{\mathrm e}^{2 i \left (d x +c \right )}+1015 \,{\mathrm e}^{i \left (d x +c \right )} A -4228 B \,{\mathrm e}^{i \left (d x +c \right )}+160 A -664 B \right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{a^{4} d}+\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{a^{4} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{a^{4} d}-\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a^{4} d}\)	\(323\)

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))^4,x,method=_RETURNVERBOSE)

[Out]

1/8/d/a^4*(-1/7*tan(1/2*d*x+1/2*c)^7*A+1/7*tan(1/2*d*x+1/2*c)^7*B-tan(1/2*d*x+1/2*c)^5*A+7/5*tan(1/2*d*x+1/2*c
)^5*B-11/3*A*tan(1/2*d*x+1/2*c)^3+23/3*B*tan(1/2*d*x+1/2*c)^3-15*A*tan(1/2*d*x+1/2*c)+49*B*tan(1/2*d*x+1/2*c)+
(32*B-8*A)*ln(tan(1/2*d*x+1/2*c)-1)-8*B/(tan(1/2*d*x+1/2*c)-1)+(-32*B+8*A)*ln(tan(1/2*d*x+1/2*c)+1)-8*B/(tan(1
/2*d*x+1/2*c)+1))

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Maxima [A]
time = 0.28, size = 326, normalized size = 1.68 \begin {gather*} \frac {B {\left (\frac {1680 \, \sin \left (d x + c\right )}{{\left (a^{4} - \frac {a^{4} \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 {5145 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {805 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {147 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {3360 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {3360 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )} - 5 \, A {\left (\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )}}{840 \, d} \end {gather*}

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))^4,x, algorithm="maxima")

[Out]

1/840*(B*(1680*sin(d*x + c)/((a^4 - a^4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (5145*sin(d
*x + c)/(cos(d*x + c) + 1) + 805*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 147*sin(d*x + c)^5/(cos(d*x + c) + 1)^5
+ 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 3360*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^4 + 3360*log(
sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^4) - 5*A*((315*sin(d*x + c)/(cos(d*x + c) + 1) + 77*sin(d*x + c)^3/(cos
(d*x + c) + 1)^3 + 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 168*l
og(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^4 + 168*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^4))/d

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Fricas [A]
time = 3.61, size = 317, normalized size = 1.63 \begin {gather*} \frac {105 \, {\left ({\left (A - 4 \, B\right )} \cos \left (d x + c\right )^{5} + 4 \, {\left (A - 4 \, B\right )} \cos \left (d x + c\right )^{4} + 6 \, {\left (A - 4 \, B\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (A - 4 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (A - 4 \, B\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left ({\left (A - 4 \, B\right )} \cos \left (d x + c\right )^{5} + 4 \, {\left (A - 4 \, B\right )} \cos \left (d x + c\right )^{4} + 6 \, {\left (A - 4 \, B\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (A - 4 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (A - 4 \, B\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (8 \, {\left (20 \, A - 83 \, B\right )} \cos \left (d x + c\right )^{4} + {\left (535 \, A - 2236 \, B\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (155 \, A - 659 \, B\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (65 \, A - 296 \, B\right )} \cos \left (d x + c\right ) - 105 \, B\right )} \sin \left (d x + c\right )}{210 \, {\left (a^{4} d \cos \left (d x + c\right )^{5} + 4 \, a^{4} d \cos \left (d x + c\right )^{4} + 6 \, a^{4} d \cos \left (d x + c\right )^{3} + 4 \, a^{4} d \cos \left (d x + c\right )^{2} + a^{4} d \cos \left (d x + c\right )\right )}} \end {gather*}

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))^4,x, algorithm="fricas")

[Out]

1/210*(105*((A - 4*B)*cos(d*x + c)^5 + 4*(A - 4*B)*cos(d*x + c)^4 + 6*(A - 4*B)*cos(d*x + c)^3 + 4*(A - 4*B)*c
os(d*x + c)^2 + (A - 4*B)*cos(d*x + c))*log(sin(d*x + c) + 1) - 105*((A - 4*B)*cos(d*x + c)^5 + 4*(A - 4*B)*co
s(d*x + c)^4 + 6*(A - 4*B)*cos(d*x + c)^3 + 4*(A - 4*B)*cos(d*x + c)^2 + (A - 4*B)*cos(d*x + c))*log(-sin(d*x
+ c) + 1) - 2*(8*(20*A - 83*B)*cos(d*x + c)^4 + (535*A - 2236*B)*cos(d*x + c)^3 + 4*(155*A - 659*B)*cos(d*x +
c)^2 + 4*(65*A - 296*B)*cos(d*x + c) - 105*B)*sin(d*x + c))/(a^4*d*cos(d*x + c)^5 + 4*a^4*d*cos(d*x + c)^4 + 6
*a^4*d*cos(d*x + c)^3 + 4*a^4*d*cos(d*x + c)^2 + a^4*d*cos(d*x + c))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {A \sec ^{5}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{6}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx}{a^{4}} \end {gather*}

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))**4,x)

[Out]

(Integral(A*sec(c + d*x)**5/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1), x)
+ Integral(B*sec(c + d*x)**6/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1),
x))/a**4

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Giac [A]
time = 0.50, size = 220, normalized size = 1.13 \begin {gather*} \frac {\frac {840 \, {\left (A - 4 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {840 \, {\left (A - 4 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} - \frac {1680 \, 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^{4}} - \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 105 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 147 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 385 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 805 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1575 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5145 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \end {gather*}

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))^4,x, algorithm="giac")

[Out]

1/840*(840*(A - 4*B)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 840*(A - 4*B)*log(abs(tan(1/2*d*x + 1/2*c) - 1))
/a^4 - 1680*B*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*a^4) - (15*A*a^24*tan(1/2*d*x + 1/2*c)^7 - 15
*B*a^24*tan(1/2*d*x + 1/2*c)^7 + 105*A*a^24*tan(1/2*d*x + 1/2*c)^5 - 147*B*a^24*tan(1/2*d*x + 1/2*c)^5 + 385*A
*a^24*tan(1/2*d*x + 1/2*c)^3 - 805*B*a^24*tan(1/2*d*x + 1/2*c)^3 + 1575*A*a^24*tan(1/2*d*x + 1/2*c) - 5145*B*a
^24*tan(1/2*d*x + 1/2*c))/a^28)/d

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Mupad [B]
time = 2.03, size = 237, normalized size = 1.22 \begin {gather*} \frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A-4\,B\right )}{a^4\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {A-B}{20\,a^4}+\frac {3\,A-5\,B}{40\,a^4}\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A-B}{2\,a^4}+\frac {3\,\left (3\,A-5\,B\right )}{8\,a^4}+\frac {2\,A-10\,B}{4\,a^4}-\frac {2\,A+10\,B}{8\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A-B\right )}{56\,a^4\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B}{8\,a^4}+\frac {3\,A-5\,B}{12\,a^4}+\frac {2\,A-10\,B}{24\,a^4}\right )}{d}-\frac {2\,B\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^4\right )} \end {gather*}

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))^4),x)

[Out]

(2*atanh(tan(c/2 + (d*x)/2))*(A - 4*B))/(a^4*d) - (tan(c/2 + (d*x)/2)^5*((A - B)/(20*a^4) + (3*A - 5*B)/(40*a^
4)))/d - (tan(c/2 + (d*x)/2)*((A - B)/(2*a^4) + (3*(3*A - 5*B))/(8*a^4) + (2*A - 10*B)/(4*a^4) - (2*A + 10*B)/
(8*a^4)))/d - (tan(c/2 + (d*x)/2)^7*(A - B))/(56*a^4*d) - (tan(c/2 + (d*x)/2)^3*((A - B)/(8*a^4) + (3*A - 5*B)
/(12*a^4) + (2*A - 10*B)/(24*a^4)))/d - (2*B*tan(c/2 + (d*x)/2))/(d*(a^4*tan(c/2 + (d*x)/2)^2 - a^4))

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