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

Optimal. Leaf size=163 \[ \frac {(12 A-215 B) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)}-\frac {(6 A-55 B) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}+\frac {B \tanh ^{-1}(\sin (c+d x))}{a^4 d}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}+\frac {(3 A-10 B) \tan (c+d x) \sec ^2(c+d x)}{35 a d (a \sec (c+d x)+a)^3} \]

[Out]

B*arctanh(sin(d*x+c))/a^4/d-1/105*(6*A-55*B)*tan(d*x+c)/a^4/d/(1+sec(d*x+c))^2+1/105*(12*A-215*B)*tan(d*x+c)/a
^4/d/(1+sec(d*x+c))+1/7*(A-B)*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^4+1/35*(3*A-10*B)*sec(d*x+c)^2*tan(d*
x+c)/a/d/(a+a*sec(d*x+c))^3

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Rubi [A]  time = 0.48, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {4019, 4008, 3998, 3770, 3794} \[ \frac {(12 A-215 B) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)}-\frac {(6 A-55 B) \tan (c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}+\frac {B \tanh ^{-1}(\sin (c+d x))}{a^4 d}+\frac {(A-B) \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}+\frac {(3 A-10 B) \tan (c+d x) \sec ^2(c+d x)}{35 a d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*ArcTanh[Sin[c + d*x]])/(a^4*d) - ((6*A - 55*B)*Tan[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])^2) + ((12*A - 21
5*B)*Tan[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])) + ((A - B)*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*(a + a*Sec[c +
d*x])^4) + ((3*A - 10*B)*Sec[c + d*x]^2*Tan[c + d*x])/(35*a*d*(a + a*Sec[c + d*x])^3)

Rule 3770

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

Rule 3794

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[Cot[e + f*x]/(f*(b + a*
Csc[e + f*x])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 3998

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

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 ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx &=\frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {\int \frac {\sec ^3(c+d x) (3 a (A-B)+7 a B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx}{7 a^2}\\ &=\frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(3 A-10 B) \sec ^2(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^2 (3 A-10 B)+35 a^2 B \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac {(6 A-55 B) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}+\frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(3 A-10 B) \sec ^2(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac {\int \frac {\sec (c+d x) \left (-2 a^3 (6 A-55 B)-105 a^3 B \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{105 a^6}\\ &=-\frac {(6 A-55 B) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}+\frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(3 A-10 B) \sec ^2(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {(12 A-215 B) \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{105 a^3}+\frac {B \int \sec (c+d x) \, dx}{a^4}\\ &=\frac {B \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac {(6 A-55 B) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}+\frac {(A-B) \sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(3 A-10 B) \sec ^2(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {(12 A-215 B) \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 1.57, size = 239, normalized size = 1.47 \[ \frac {\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (70 (3 A-49 B) \sin \left (\frac {d x}{2}\right )+126 A \sin \left (c+\frac {3 d x}{2}\right )+42 A \sin \left (2 c+\frac {5 d x}{2}\right )+6 A \sin \left (3 c+\frac {7 d x}{2}\right )+2170 B \sin \left (c+\frac {d x}{2}\right )-2625 B \sin \left (c+\frac {3 d x}{2}\right )+735 B \sin \left (2 c+\frac {3 d x}{2}\right )-1015 B \sin \left (2 c+\frac {5 d x}{2}\right )+105 B \sin \left (3 c+\frac {5 d x}{2}\right )-160 B \sin \left (3 c+\frac {7 d x}{2}\right )\right )-6720 B \cos ^8\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{420 a^4 d (\cos (c+d x)+1)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6720*B*Cos[(c + d*x)/2]^8*(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]*(70*(3*A - 49*B)*Sin[(d*x)/2] + 2170*B*Sin[c + (d*x)/2] + 126*A*Sin[c + (3*d*x
)/2] - 2625*B*Sin[c + (3*d*x)/2] + 735*B*Sin[2*c + (3*d*x)/2] + 42*A*Sin[2*c + (5*d*x)/2] - 1015*B*Sin[2*c + (
5*d*x)/2] + 105*B*Sin[3*c + (5*d*x)/2] + 6*A*Sin[3*c + (7*d*x)/2] - 160*B*Sin[3*c + (7*d*x)/2]))/(420*a^4*d*(1
+ Cos[c + d*x])^4)

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fricas [A]  time = 0.45, size = 236, normalized size = 1.45 \[ \frac {105 \, {\left (B \cos \left (d x + c\right )^{4} + 4 \, B \cos \left (d x + c\right )^{3} + 6 \, B \cos \left (d x + c\right )^{2} + 4 \, B \cos \left (d x + c\right ) + B\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left (B \cos \left (d x + c\right )^{4} + 4 \, B \cos \left (d x + c\right )^{3} + 6 \, B \cos \left (d x + c\right )^{2} + 4 \, B \cos \left (d x + c\right ) + B\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (3 \, A - 80 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (24 \, A - 535 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (39 \, A - 620 \, B\right )} \cos \left (d x + c\right ) + 36 \, A - 260 \, B\right )} \sin \left (d x + c\right )}{210 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]

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))^4,x, algorithm="fricas")

[Out]

1/210*(105*(B*cos(d*x + c)^4 + 4*B*cos(d*x + c)^3 + 6*B*cos(d*x + c)^2 + 4*B*cos(d*x + c) + B)*log(sin(d*x + c
) + 1) - 105*(B*cos(d*x + c)^4 + 4*B*cos(d*x + c)^3 + 6*B*cos(d*x + c)^2 + 4*B*cos(d*x + c) + B)*log(-sin(d*x
+ c) + 1) + 2*(2*(3*A - 80*B)*cos(d*x + c)^3 + (24*A - 535*B)*cos(d*x + c)^2 + (39*A - 620*B)*cos(d*x + c) + 3
6*A - 260*B)*sin(d*x + c))/(a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*c
os(d*x + c) + a^4*d)

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giac [A]  time = 0.58, size = 181, normalized size = 1.11 \[ \frac {\frac {840 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {840 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 63 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 105 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 105 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 385 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1575 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \]

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))^4,x, algorithm="giac")

[Out]

1/840*(840*B*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 840*B*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^4 + (15*A*a^2
4*tan(1/2*d*x + 1/2*c)^7 - 15*B*a^24*tan(1/2*d*x + 1/2*c)^7 + 63*A*a^24*tan(1/2*d*x + 1/2*c)^5 - 105*B*a^24*ta
n(1/2*d*x + 1/2*c)^5 + 105*A*a^24*tan(1/2*d*x + 1/2*c)^3 - 385*B*a^24*tan(1/2*d*x + 1/2*c)^3 + 105*A*a^24*tan(
1/2*d*x + 1/2*c) - 1575*B*a^24*tan(1/2*d*x + 1/2*c))/a^28)/d

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maple [A]  time = 0.71, size = 199, normalized size = 1.22 \[ \frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}+\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{56 d \,a^{4}}-\frac {B \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 d \,a^{4}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B}{d \,a^{4}}-\frac {15 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B}{d \,a^{4}}+\frac {3 A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d \,a^{4}}-\frac {B \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{4}}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{8 d \,a^{4}}-\frac {11 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d \,a^{4}} \]

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))^4,x)

[Out]

1/8/d/a^4*A*tan(1/2*d*x+1/2*c)+1/56/d/a^4*tan(1/2*d*x+1/2*c)^7*A-1/56/d/a^4*B*tan(1/2*d*x+1/2*c)^7-1/d/a^4*ln(
tan(1/2*d*x+1/2*c)-1)*B-15/8/d/a^4*B*tan(1/2*d*x+1/2*c)+1/d/a^4*ln(tan(1/2*d*x+1/2*c)+1)*B+3/40/d/a^4*A*tan(1/
2*d*x+1/2*c)^5-1/8/d/a^4*B*tan(1/2*d*x+1/2*c)^5+1/8/d/a^4*tan(1/2*d*x+1/2*c)^3*A-11/24/d/a^4*B*tan(1/2*d*x+1/2
*c)^3

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maxima [A]  time = 0.36, size = 228, normalized size = 1.40 \[ -\frac {5 \, B {\left (\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )} - \frac {3 \, A {\left (\frac {35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \]

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))^4,x, algorithm="maxima")

[Out]

-1/840*(5*B*((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*log(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) - 3*A*(35*sin(d*x + c)/(cos(d*x + c) + 1) + 35*sin
(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 5*sin(d*x + c)^7/(cos(d*x + c) + 1
)^7)/a^4)/d

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mupad [B]  time = 2.13, size = 198, normalized size = 1.21 \[ \frac {{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {A\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{8}-\frac {11\,B\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24}\right )+{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {3\,A\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{40}-\frac {B\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{8}\right )+{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}-\frac {15\,B\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}\right )+\frac {A\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{56}-\frac {B\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{56}}{a^4\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}+\frac {2\,B\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{a^4\,d} \]

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))^4),x)

[Out]

(cos(c/2 + (d*x)/2)^4*((A*sin(c/2 + (d*x)/2)^3)/8 - (11*B*sin(c/2 + (d*x)/2)^3)/24) + cos(c/2 + (d*x)/2)^2*((3
*A*sin(c/2 + (d*x)/2)^5)/40 - (B*sin(c/2 + (d*x)/2)^5)/8) + cos(c/2 + (d*x)/2)^6*((A*sin(c/2 + (d*x)/2))/8 - (
15*B*sin(c/2 + (d*x)/2))/8) + (A*sin(c/2 + (d*x)/2)^7)/56 - (B*sin(c/2 + (d*x)/2)^7)/56)/(a^4*d*cos(c/2 + (d*x
)/2)^7) + (2*B*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(a^4*d)

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

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))**4,x)

[Out]

(Integral(A*sec(c + d*x)**4/(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)**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))/a**4

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