3.146 \(\int \frac {\sec (c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^4} \, dx\)

Optimal. Leaf size=231 \[ \frac {a (b \cos (c+d x)-a \sin (c+d x))}{2 b^2 d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^2}+\frac {a \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{2 b^2 d \left (a^2+b^2\right )^{3/2}}+\frac {a \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{b^4 d \sqrt {a^2+b^2}}-\frac {1}{b^3 d (a \cos (c+d x)+b \sin (c+d x))}-\frac {1}{3 b d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {\tanh ^{-1}(\sin (c+d x))}{b^4 d} \]

[Out]

arctanh(sin(d*x+c))/b^4/d+1/2*a*arctanh((b*cos(d*x+c)-a*sin(d*x+c))/(a^2+b^2)^(1/2))/b^2/(a^2+b^2)^(3/2)/d-1/3
/b/d/(a*cos(d*x+c)+b*sin(d*x+c))^3+1/2*a*(b*cos(d*x+c)-a*sin(d*x+c))/b^2/(a^2+b^2)/d/(a*cos(d*x+c)+b*sin(d*x+c
))^2-1/b^3/d/(a*cos(d*x+c)+b*sin(d*x+c))+a*arctanh((b*cos(d*x+c)-a*sin(d*x+c))/(a^2+b^2)^(1/2))/b^4/d/(a^2+b^2
)^(1/2)

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Rubi [A]  time = 0.17, antiderivative size = 231, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3094, 3770, 3074, 206, 3076} \[ \frac {a (b \cos (c+d x)-a \sin (c+d x))}{2 b^2 d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^2}+\frac {a \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{b^4 d \sqrt {a^2+b^2}}+\frac {a \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{2 b^2 d \left (a^2+b^2\right )^{3/2}}-\frac {1}{b^3 d (a \cos (c+d x)+b \sin (c+d x))}-\frac {1}{3 b d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {\tanh ^{-1}(\sin (c+d x))}{b^4 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]/(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]

[Out]

ArcTanh[Sin[c + d*x]]/(b^4*d) + (a*ArcTanh[(b*Cos[c + d*x] - a*Sin[c + d*x])/Sqrt[a^2 + b^2]])/(2*b^2*(a^2 + b
^2)^(3/2)*d) + (a*ArcTanh[(b*Cos[c + d*x] - a*Sin[c + d*x])/Sqrt[a^2 + b^2]])/(b^4*Sqrt[a^2 + b^2]*d) - 1/(3*b
*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) + (a*(b*Cos[c + d*x] - a*Sin[c + d*x]))/(2*b^2*(a^2 + b^2)*d*(a*Cos[c
+ d*x] + b*Sin[c + d*x])^2) - 1/(b^3*d*(a*Cos[c + d*x] + b*Sin[c + d*x]))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3074

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Dist[d^(-1), Subst[Int
[1/(a^2 + b^2 - x^2), x], x, b*Cos[c + d*x] - a*Sin[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2,
0]

Rule 3076

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[((b*Cos[c + d*x] -
a*Sin[c + d*x])*(a*Cos[c + d*x] + b*Sin[c + d*x])^(n + 1))/(d*(n + 1)*(a^2 + b^2)), x] + Dist[(n + 2)/((n + 1
)*(a^2 + b^2)), Int[(a*Cos[c + d*x] + b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b
^2, 0] && LtQ[n, -1] && NeQ[n, -2]

Rule 3094

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_)/cos[(c_.) + (d_.)*(x_)], x_Symbol] :>
Simp[(a*Cos[c + d*x] + b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)), x] + (Dist[1/b^2, Int[(a*Cos[c + d*x] + b*Sin[c
+ d*x])^(n + 2)/Cos[c + d*x], x], x] - Dist[a/b^2, Int[(a*Cos[c + d*x] + b*Sin[c + d*x])^(n + 1), x], x]) /;
FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && LtQ[n, -1]

Rule 3770

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

Rubi steps

\begin {align*} \int \frac {\sec (c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^4} \, dx &=-\frac {1}{3 b d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {\int \frac {\sec (c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx}{b^2}-\frac {a \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx}{b^2}\\ &=-\frac {1}{3 b d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {a (b \cos (c+d x)-a \sin (c+d x))}{2 b^2 \left (a^2+b^2\right ) d (a \cos (c+d x)+b \sin (c+d x))^2}-\frac {1}{b^3 d (a \cos (c+d x)+b \sin (c+d x))}+\frac {\int \sec (c+d x) \, dx}{b^4}-\frac {a \int \frac {1}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{b^4}-\frac {a \int \frac {1}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{2 b^2 \left (a^2+b^2\right )}\\ &=\frac {\tanh ^{-1}(\sin (c+d x))}{b^4 d}-\frac {1}{3 b d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {a (b \cos (c+d x)-a \sin (c+d x))}{2 b^2 \left (a^2+b^2\right ) d (a \cos (c+d x)+b \sin (c+d x))^2}-\frac {1}{b^3 d (a \cos (c+d x)+b \sin (c+d x))}+\frac {a \operatorname {Subst}\left (\int \frac {1}{a^2+b^2-x^2} \, dx,x,b \cos (c+d x)-a \sin (c+d x)\right )}{b^4 d}+\frac {a \operatorname {Subst}\left (\int \frac {1}{a^2+b^2-x^2} \, dx,x,b \cos (c+d x)-a \sin (c+d x)\right )}{2 b^2 \left (a^2+b^2\right ) d}\\ &=\frac {\tanh ^{-1}(\sin (c+d x))}{b^4 d}+\frac {a \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{2 b^2 \left (a^2+b^2\right )^{3/2} d}+\frac {a \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{b^4 \sqrt {a^2+b^2} d}-\frac {1}{3 b d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac {a (b \cos (c+d x)-a \sin (c+d x))}{2 b^2 \left (a^2+b^2\right ) d (a \cos (c+d x)+b \sin (c+d x))^2}-\frac {1}{b^3 d (a \cos (c+d x)+b \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 3.28, size = 290, normalized size = 1.26 \[ -\frac {\sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \left (\frac {3 b \left (2 a^2+b^2\right ) \cos (c+d x) (a+b \tan (c+d x))^2}{a^2+b^2}+\frac {6 a \left (2 a^2+3 b^2\right ) \cos ^2(c+d x) (a+b \tan (c+d x))^3 \tanh ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )-b}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}}+3 b^2 \tan (c+d x) (a \cos (c+d x)+b \sin (c+d x))+6 \cos ^2(c+d x) (a+b \tan (c+d x))^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-6 \cos ^2(c+d x) (a+b \tan (c+d x))^3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+2 b^3 \sec (c+d x)\right )}{6 b^4 d (a+b \tan (c+d x))^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]/(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]

[Out]

-1/6*(Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])*(2*b^3*Sec[c + d*x] + 3*b^2*(a*Cos[c + d*x] + b*Sin[c +
d*x])*Tan[c + d*x] + (3*b*(2*a^2 + b^2)*Cos[c + d*x]*(a + b*Tan[c + d*x])^2)/(a^2 + b^2) + (6*a*(2*a^2 + 3*b^
2)*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]]*Cos[c + d*x]^2*(a + b*Tan[c + d*x])^3)/(a^2 + b^2)^(3/2)
+ 6*Cos[c + d*x]^2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(a + b*Tan[c + d*x])^3 - 6*Cos[c + d*x]^2*Log[Cos
[(c + d*x)/2] + Sin[(c + d*x)/2]]*(a + b*Tan[c + d*x])^3))/(b^4*d*(a + b*Tan[c + d*x])^4)

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fricas [B]  time = 0.78, size = 745, normalized size = 3.23 \[ -\frac {22 \, a^{4} b^{3} + 38 \, a^{2} b^{5} + 16 \, b^{7} + 12 \, {\left (a^{6} b - 2 \, a^{2} b^{5} - b^{7}\right )} \cos \left (d x + c\right )^{2} + 6 \, {\left (5 \, a^{5} b^{2} + 8 \, a^{3} b^{4} + 3 \, a b^{6}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, {\left ({\left (2 \, a^{6} - 3 \, a^{4} b^{2} - 9 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (2 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \cos \left (d x + c\right ) + {\left (2 \, a^{3} b^{3} + 3 \, a b^{5} + {\left (6 \, a^{5} b + 7 \, a^{3} b^{3} - 3 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) - 6 \, {\left ({\left (a^{7} - a^{5} b^{2} - 5 \, a^{3} b^{4} - 3 \, a b^{6}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (a^{5} b^{2} + 2 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right ) + {\left (a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7} + {\left (3 \, a^{6} b + 5 \, a^{4} b^{3} + a^{2} b^{5} - b^{7}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 6 \, {\left ({\left (a^{7} - a^{5} b^{2} - 5 \, a^{3} b^{4} - 3 \, a b^{6}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (a^{5} b^{2} + 2 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right ) + {\left (a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7} + {\left (3 \, a^{6} b + 5 \, a^{4} b^{3} + a^{2} b^{5} - b^{7}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{12 \, {\left ({\left (a^{7} b^{4} - a^{5} b^{6} - 5 \, a^{3} b^{8} - 3 \, a b^{10}\right )} d \cos \left (d x + c\right )^{3} + 3 \, {\left (a^{5} b^{6} + 2 \, a^{3} b^{8} + a b^{10}\right )} d \cos \left (d x + c\right ) + {\left ({\left (3 \, a^{6} b^{5} + 5 \, a^{4} b^{7} + a^{2} b^{9} - b^{11}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{4} b^{7} + 2 \, a^{2} b^{9} + b^{11}\right )} d\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)/(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

-1/12*(22*a^4*b^3 + 38*a^2*b^5 + 16*b^7 + 12*(a^6*b - 2*a^2*b^5 - b^7)*cos(d*x + c)^2 + 6*(5*a^5*b^2 + 8*a^3*b
^4 + 3*a*b^6)*cos(d*x + c)*sin(d*x + c) - 3*((2*a^6 - 3*a^4*b^2 - 9*a^2*b^4)*cos(d*x + c)^3 + 3*(2*a^4*b^2 + 3
*a^2*b^4)*cos(d*x + c) + (2*a^3*b^3 + 3*a*b^5 + (6*a^5*b + 7*a^3*b^3 - 3*a*b^5)*cos(d*x + c)^2)*sin(d*x + c))*
sqrt(a^2 + b^2)*log((2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 - 2*a^2 - b^2 - 2*sqrt(a^2 +
b^2)*(b*cos(d*x + c) - a*sin(d*x + c)))/(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2))
- 6*((a^7 - a^5*b^2 - 5*a^3*b^4 - 3*a*b^6)*cos(d*x + c)^3 + 3*(a^5*b^2 + 2*a^3*b^4 + a*b^6)*cos(d*x + c) + (a
^4*b^3 + 2*a^2*b^5 + b^7 + (3*a^6*b + 5*a^4*b^3 + a^2*b^5 - b^7)*cos(d*x + c)^2)*sin(d*x + c))*log(sin(d*x + c
) + 1) + 6*((a^7 - a^5*b^2 - 5*a^3*b^4 - 3*a*b^6)*cos(d*x + c)^3 + 3*(a^5*b^2 + 2*a^3*b^4 + a*b^6)*cos(d*x + c
) + (a^4*b^3 + 2*a^2*b^5 + b^7 + (3*a^6*b + 5*a^4*b^3 + a^2*b^5 - b^7)*cos(d*x + c)^2)*sin(d*x + c))*log(-sin(
d*x + c) + 1))/((a^7*b^4 - a^5*b^6 - 5*a^3*b^8 - 3*a*b^10)*d*cos(d*x + c)^3 + 3*(a^5*b^6 + 2*a^3*b^8 + a*b^10)
*d*cos(d*x + c) + ((3*a^6*b^5 + 5*a^4*b^7 + a^2*b^9 - b^11)*d*cos(d*x + c)^2 + (a^4*b^7 + 2*a^2*b^9 + b^11)*d)
*sin(d*x + c))

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giac [B]  time = 4.07, size = 527, normalized size = 2.28 \[ \frac {\frac {3 \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{2} b^{4} + b^{6}\right )} \sqrt {a^{2} + b^{2}}} + \frac {2 \, {\left (3 \, a^{6} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, a^{4} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, a^{2} b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, a^{7} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 9 \, a^{5} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 12 \, a^{3} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 12 \, a b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 36 \, a^{6} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a^{4} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, a^{2} b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, b^{7} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, a^{7} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 48 \, a^{5} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 42 \, a^{3} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, a b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 33 \, a^{6} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, a^{4} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a^{2} b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a^{7} + 5 \, a^{5} b^{2} + 2 \, a^{3} b^{4}\right )}}{{\left (a^{5} b^{3} + a^{3} b^{5}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}^{3}} + \frac {6 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{4}} - \frac {6 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{4}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)/(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="giac")

[Out]

1/6*(3*(2*a^3 + 3*a*b^2)*log(abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b - 2*sqrt(a^2 + b^2))/abs(2*a*tan(1/2*d*x + 1/2
*c) - 2*b + 2*sqrt(a^2 + b^2)))/((a^2*b^4 + b^6)*sqrt(a^2 + b^2)) + 2*(3*a^6*b*tan(1/2*d*x + 1/2*c)^5 + 6*a^4*
b^3*tan(1/2*d*x + 1/2*c)^5 + 6*a^2*b^5*tan(1/2*d*x + 1/2*c)^5 + 6*a^7*tan(1/2*d*x + 1/2*c)^4 - 9*a^5*b^2*tan(1
/2*d*x + 1/2*c)^4 - 12*a^3*b^4*tan(1/2*d*x + 1/2*c)^4 - 12*a*b^6*tan(1/2*d*x + 1/2*c)^4 - 36*a^6*b*tan(1/2*d*x
+ 1/2*c)^3 - 6*a^4*b^3*tan(1/2*d*x + 1/2*c)^3 + 8*a^2*b^5*tan(1/2*d*x + 1/2*c)^3 + 8*b^7*tan(1/2*d*x + 1/2*c)
^3 - 12*a^7*tan(1/2*d*x + 1/2*c)^2 + 48*a^5*b^2*tan(1/2*d*x + 1/2*c)^2 + 42*a^3*b^4*tan(1/2*d*x + 1/2*c)^2 + 1
2*a*b^6*tan(1/2*d*x + 1/2*c)^2 + 33*a^6*b*tan(1/2*d*x + 1/2*c) + 24*a^4*b^3*tan(1/2*d*x + 1/2*c) + 6*a^2*b^5*t
an(1/2*d*x + 1/2*c) + 6*a^7 + 5*a^5*b^2 + 2*a^3*b^4)/((a^5*b^3 + a^3*b^5)*(a*tan(1/2*d*x + 1/2*c)^2 - 2*b*tan(
1/2*d*x + 1/2*c) - a)^3) + 6*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^4 - 6*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^4
)/d

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maple [B]  time = 0.42, size = 1367, normalized size = 5.92 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)/(a*cos(d*x+c)+b*sin(d*x+c))^4,x)

[Out]

2/d/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^3*a/(a^2+b^2)*tan(1/2*d*x+1/2*c)^5+16/d/b/(tan(1/2*d*x+1
/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^3*a^2/(a^2+b^2)*tan(1/2*d*x+1/2*c)^2+4/d*b^3/(tan(1/2*d*x+1/2*c)^2*a-2*b*t
an(1/2*d*x+1/2*c)-a)^3/a^2/(a^2+b^2)*tan(1/2*d*x+1/2*c)^2+11/d/b^2/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2
*c)-a)^3*a^3/(a^2+b^2)*tan(1/2*d*x+1/2*c)+2/d*b^2/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^3/a/(a^2+b
^2)*tan(1/2*d*x+1/2*c)+1/d/b^2/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^3*a^3/(a^2+b^2)*tan(1/2*d*x+1
/2*c)^5+2/d*b^2/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^3/a/(a^2+b^2)*tan(1/2*d*x+1/2*c)^5+2/d/b^3/(
tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^3/(a^2+b^2)*a^4*tan(1/2*d*x+1/2*c)^4-3/d/b/(tan(1/2*d*x+1/2*c
)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^3/(a^2+b^2)*a^2*tan(1/2*d*x+1/2*c)^4-4/d*b^3/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1
/2*d*x+1/2*c)-a)^3/(a^2+b^2)/a^2*tan(1/2*d*x+1/2*c)^4-12/d/b^2/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-
a)^3*a^3/(a^2+b^2)*tan(1/2*d*x+1/2*c)^3+8/3/d*b^2/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^3/a/(a^2+b
^2)*tan(1/2*d*x+1/2*c)^3+8/3/d*b^4/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^3/a^3/(a^2+b^2)*tan(1/2*d
*x+1/2*c)^3-4/d/b^3/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^3*a^4/(a^2+b^2)*tan(1/2*d*x+1/2*c)^2+2/3
/d*b/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^3/(a^2+b^2)-2/d/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x
+1/2*c)-a)^3*a/(a^2+b^2)*tan(1/2*d*x+1/2*c)^3+8/d/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^3*a/(a^2+b
^2)*tan(1/2*d*x+1/2*c)-4/d*b/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^3/(a^2+b^2)*tan(1/2*d*x+1/2*c)^
4+14/d*b/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^3/(a^2+b^2)*tan(1/2*d*x+1/2*c)^2+2/d/b^3/(tan(1/2*d
*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)^3/(a^2+b^2)*a^4+5/3/d/b/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c)
-a)^3/(a^2+b^2)*a^2-2/d/b^4*a^3/(a^2+b^2)^(3/2)*arctanh(1/2*(2*a*tan(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2))-3/d/
b^2*a/(a^2+b^2)^(3/2)*arctanh(1/2*(2*a*tan(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2))-1/d/b^4*ln(tan(1/2*d*x+1/2*c)-
1)+1/d/b^4*ln(tan(1/2*d*x+1/2*c)+1)

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maxima [B]  time = 0.47, size = 661, normalized size = 2.86 \[ \frac {\frac {3 \, {\left (2 \, a^{2} + 3 \, b^{2}\right )} a \log \left (\frac {b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{2} b^{4} + b^{6}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (6 \, a^{7} + 5 \, a^{5} b^{2} + 2 \, a^{3} b^{4} + \frac {3 \, {\left (11 \, a^{6} b + 8 \, a^{4} b^{3} + 2 \, a^{2} b^{5}\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {6 \, {\left (2 \, a^{7} - 8 \, a^{5} b^{2} - 7 \, a^{3} b^{4} - 2 \, a b^{6}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {2 \, {\left (18 \, a^{6} b + 3 \, a^{4} b^{3} - 4 \, a^{2} b^{5} - 4 \, b^{7}\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, {\left (2 \, a^{7} - 3 \, a^{5} b^{2} - 4 \, a^{3} b^{4} - 4 \, a b^{6}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {3 \, {\left (a^{6} b + 2 \, a^{4} b^{3} + 2 \, a^{2} b^{5}\right )} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{8} b^{3} + a^{6} b^{5} + \frac {6 \, {\left (a^{7} b^{4} + a^{5} b^{6}\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {3 \, {\left (a^{8} b^{3} - 3 \, a^{6} b^{5} - 4 \, a^{4} b^{7}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {4 \, {\left (3 \, a^{7} b^{4} + a^{5} b^{6} - 2 \, a^{3} b^{8}\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, {\left (a^{8} b^{3} - 3 \, a^{6} b^{5} - 4 \, a^{4} b^{7}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {6 \, {\left (a^{7} b^{4} + a^{5} b^{6}\right )} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {{\left (a^{8} b^{3} + a^{6} b^{5}\right )} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{b^{4}} - \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{b^{4}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)/(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

1/6*(3*(2*a^2 + 3*b^2)*a*log((b - a*sin(d*x + c)/(cos(d*x + c) + 1) + sqrt(a^2 + b^2))/(b - a*sin(d*x + c)/(co
s(d*x + c) + 1) - sqrt(a^2 + b^2)))/((a^2*b^4 + b^6)*sqrt(a^2 + b^2)) - 2*(6*a^7 + 5*a^5*b^2 + 2*a^3*b^4 + 3*(
11*a^6*b + 8*a^4*b^3 + 2*a^2*b^5)*sin(d*x + c)/(cos(d*x + c) + 1) - 6*(2*a^7 - 8*a^5*b^2 - 7*a^3*b^4 - 2*a*b^6
)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 2*(18*a^6*b + 3*a^4*b^3 - 4*a^2*b^5 - 4*b^7)*sin(d*x + c)^3/(cos(d*x +
c) + 1)^3 + 3*(2*a^7 - 3*a^5*b^2 - 4*a^3*b^4 - 4*a*b^6)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 3*(a^6*b + 2*a^
4*b^3 + 2*a^2*b^5)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/(a^8*b^3 + a^6*b^5 + 6*(a^7*b^4 + a^5*b^6)*sin(d*x + c
)/(cos(d*x + c) + 1) - 3*(a^8*b^3 - 3*a^6*b^5 - 4*a^4*b^7)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 4*(3*a^7*b^4
+ a^5*b^6 - 2*a^3*b^8)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*(a^8*b^3 - 3*a^6*b^5 - 4*a^4*b^7)*sin(d*x + c)^
4/(cos(d*x + c) + 1)^4 + 6*(a^7*b^4 + a^5*b^6)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - (a^8*b^3 + a^6*b^5)*sin(d
*x + c)^6/(cos(d*x + c) + 1)^6) + 6*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/b^4 - 6*log(sin(d*x + c)/(cos(d*x
+ c) + 1) - 1)/b^4)/d

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mupad [B]  time = 4.83, size = 2848, normalized size = 12.33 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)*(a*cos(c + d*x) + b*sin(c + d*x))^4),x)

[Out]

(2*atanh((64*a*b^5*tan(c/2 + (d*x)/2))/((176*a^3*b^15)/(b^12 + 2*a^2*b^10 + a^4*b^8) + (160*a^5*b^13)/(b^12 +
2*a^2*b^10 + a^4*b^8) + (48*a^7*b^11)/(b^12 + 2*a^2*b^10 + a^4*b^8) + (64*a*b^17)/(b^12 + 2*a^2*b^10 + a^4*b^8
)) + (48*a^3*b^3*tan(c/2 + (d*x)/2))/((176*a^3*b^15)/(b^12 + 2*a^2*b^10 + a^4*b^8) + (160*a^5*b^13)/(b^12 + 2*
a^2*b^10 + a^4*b^8) + (48*a^7*b^11)/(b^12 + 2*a^2*b^10 + a^4*b^8) + (64*a*b^17)/(b^12 + 2*a^2*b^10 + a^4*b^8))
))/(b^4*d) - ((6*a^4 + 2*b^4 + 5*a^2*b^2)/(3*b^3*(a^2 + b^2)) + (tan(c/2 + (d*x)/2)*(11*a^4 + 2*b^4 + 8*a^2*b^
2))/(a*b^2*(a^2 + b^2)) + (tan(c/2 + (d*x)/2)^5*(a^4 + 2*b^4 + 2*a^2*b^2))/(a*b^2*(a^2 + b^2)) - (tan(c/2 + (d
*x)/2)^4*(4*b^6 - 2*a^6 + 4*a^2*b^4 + 3*a^4*b^2))/(a^2*b^3*(a^2 + b^2)) + (2*tan(c/2 + (d*x)/2)^2*(2*b^6 - 2*a
^6 + 7*a^2*b^4 + 8*a^4*b^2))/(a^2*b^3*(a^2 + b^2)) - (2*tan(c/2 + (d*x)/2)^3*(3*a^2 - 2*b^2)*(6*a^4 + 2*b^4 +
5*a^2*b^2))/(3*a^3*b^2*(a^2 + b^2)))/(d*(tan(c/2 + (d*x)/2)^2*(12*a*b^2 - 3*a^3) - a^3*tan(c/2 + (d*x)/2)^6 -
tan(c/2 + (d*x)/2)^4*(12*a*b^2 - 3*a^3) - tan(c/2 + (d*x)/2)^3*(12*a^2*b - 8*b^3) + a^3 + 6*a^2*b*tan(c/2 + (d
*x)/2) + 6*a^2*b*tan(c/2 + (d*x)/2)^5)) - (a*atan(((a*((a^2 + b^2)^3)^(1/2)*(2*a^2 + 3*b^2)*((8*(4*a^2*b^7 + 8
*a^4*b^5 + 4*a^6*b^3))/(b^12 + 2*a^2*b^10 + a^4*b^8) + (8*tan(c/2 + (d*x)/2)*(8*a*b^9 + 29*a^3*b^7 + 28*a^5*b^
5 + 8*a^7*b^3))/(b^13 + 2*a^2*b^11 + a^4*b^9) - (a*((a^2 + b^2)^3)^(1/2)*(2*a^2 + 3*b^2)*((8*tan(c/2 + (d*x)/2
)*(12*a^2*b^12 + 20*a^4*b^10 + 8*a^6*b^8))/(b^13 + 2*a^2*b^11 + a^4*b^9) - (8*(4*a*b^12 + 6*a^3*b^10 + 2*a^5*b
^8))/(b^12 + 2*a^2*b^10 + a^4*b^8) + (a*((a^2 + b^2)^3)^(1/2)*(2*a^2 + 3*b^2)*((8*(4*a^2*b^15 + 8*a^4*b^13 + 4
*a^6*b^11))/(b^12 + 2*a^2*b^10 + a^4*b^8) + (8*tan(c/2 + (d*x)/2)*(12*a*b^17 + 32*a^3*b^15 + 28*a^5*b^13 + 8*a
^7*b^11))/(b^13 + 2*a^2*b^11 + a^4*b^9)))/(2*(b^10 + 3*a^2*b^8 + 3*a^4*b^6 + a^6*b^4))))/(2*(b^10 + 3*a^2*b^8
+ 3*a^4*b^6 + a^6*b^4)))*1i)/(2*(b^10 + 3*a^2*b^8 + 3*a^4*b^6 + a^6*b^4)) + (a*((a^2 + b^2)^3)^(1/2)*(2*a^2 +
3*b^2)*((8*(4*a^2*b^7 + 8*a^4*b^5 + 4*a^6*b^3))/(b^12 + 2*a^2*b^10 + a^4*b^8) + (8*tan(c/2 + (d*x)/2)*(8*a*b^9
+ 29*a^3*b^7 + 28*a^5*b^5 + 8*a^7*b^3))/(b^13 + 2*a^2*b^11 + a^4*b^9) - (a*((a^2 + b^2)^3)^(1/2)*(2*a^2 + 3*b
^2)*((8*(4*a*b^12 + 6*a^3*b^10 + 2*a^5*b^8))/(b^12 + 2*a^2*b^10 + a^4*b^8) - (8*tan(c/2 + (d*x)/2)*(12*a^2*b^1
2 + 20*a^4*b^10 + 8*a^6*b^8))/(b^13 + 2*a^2*b^11 + a^4*b^9) + (a*((a^2 + b^2)^3)^(1/2)*(2*a^2 + 3*b^2)*((8*(4*
a^2*b^15 + 8*a^4*b^13 + 4*a^6*b^11))/(b^12 + 2*a^2*b^10 + a^4*b^8) + (8*tan(c/2 + (d*x)/2)*(12*a*b^17 + 32*a^3
*b^15 + 28*a^5*b^13 + 8*a^7*b^11))/(b^13 + 2*a^2*b^11 + a^4*b^9)))/(2*(b^10 + 3*a^2*b^8 + 3*a^4*b^6 + a^6*b^4)
)))/(2*(b^10 + 3*a^2*b^8 + 3*a^4*b^6 + a^6*b^4)))*1i)/(2*(b^10 + 3*a^2*b^8 + 3*a^4*b^6 + a^6*b^4)))/((16*(2*a^
5 + 3*a^3*b^2))/(b^12 + 2*a^2*b^10 + a^4*b^8) - (16*tan(c/2 + (d*x)/2)*(8*a^6 + 12*a^2*b^4 + 20*a^4*b^2))/(b^1
3 + 2*a^2*b^11 + a^4*b^9) - (a*((a^2 + b^2)^3)^(1/2)*(2*a^2 + 3*b^2)*((8*(4*a^2*b^7 + 8*a^4*b^5 + 4*a^6*b^3))/
(b^12 + 2*a^2*b^10 + a^4*b^8) + (8*tan(c/2 + (d*x)/2)*(8*a*b^9 + 29*a^3*b^7 + 28*a^5*b^5 + 8*a^7*b^3))/(b^13 +
2*a^2*b^11 + a^4*b^9) - (a*((a^2 + b^2)^3)^(1/2)*(2*a^2 + 3*b^2)*((8*tan(c/2 + (d*x)/2)*(12*a^2*b^12 + 20*a^4
*b^10 + 8*a^6*b^8))/(b^13 + 2*a^2*b^11 + a^4*b^9) - (8*(4*a*b^12 + 6*a^3*b^10 + 2*a^5*b^8))/(b^12 + 2*a^2*b^10
+ a^4*b^8) + (a*((a^2 + b^2)^3)^(1/2)*(2*a^2 + 3*b^2)*((8*(4*a^2*b^15 + 8*a^4*b^13 + 4*a^6*b^11))/(b^12 + 2*a
^2*b^10 + a^4*b^8) + (8*tan(c/2 + (d*x)/2)*(12*a*b^17 + 32*a^3*b^15 + 28*a^5*b^13 + 8*a^7*b^11))/(b^13 + 2*a^2
*b^11 + a^4*b^9)))/(2*(b^10 + 3*a^2*b^8 + 3*a^4*b^6 + a^6*b^4))))/(2*(b^10 + 3*a^2*b^8 + 3*a^4*b^6 + a^6*b^4))
))/(2*(b^10 + 3*a^2*b^8 + 3*a^4*b^6 + a^6*b^4)) + (a*((a^2 + b^2)^3)^(1/2)*(2*a^2 + 3*b^2)*((8*(4*a^2*b^7 + 8*
a^4*b^5 + 4*a^6*b^3))/(b^12 + 2*a^2*b^10 + a^4*b^8) + (8*tan(c/2 + (d*x)/2)*(8*a*b^9 + 29*a^3*b^7 + 28*a^5*b^5
+ 8*a^7*b^3))/(b^13 + 2*a^2*b^11 + a^4*b^9) - (a*((a^2 + b^2)^3)^(1/2)*(2*a^2 + 3*b^2)*((8*(4*a*b^12 + 6*a^3*
b^10 + 2*a^5*b^8))/(b^12 + 2*a^2*b^10 + a^4*b^8) - (8*tan(c/2 + (d*x)/2)*(12*a^2*b^12 + 20*a^4*b^10 + 8*a^6*b^
8))/(b^13 + 2*a^2*b^11 + a^4*b^9) + (a*((a^2 + b^2)^3)^(1/2)*(2*a^2 + 3*b^2)*((8*(4*a^2*b^15 + 8*a^4*b^13 + 4*
a^6*b^11))/(b^12 + 2*a^2*b^10 + a^4*b^8) + (8*tan(c/2 + (d*x)/2)*(12*a*b^17 + 32*a^3*b^15 + 28*a^5*b^13 + 8*a^
7*b^11))/(b^13 + 2*a^2*b^11 + a^4*b^9)))/(2*(b^10 + 3*a^2*b^8 + 3*a^4*b^6 + a^6*b^4))))/(2*(b^10 + 3*a^2*b^8 +
3*a^4*b^6 + a^6*b^4))))/(2*(b^10 + 3*a^2*b^8 + 3*a^4*b^6 + a^6*b^4))))*((a^2 + b^2)^3)^(1/2)*(2*a^2 + 3*b^2)*
1i)/(d*(b^10 + 3*a^2*b^8 + 3*a^4*b^6 + a^6*b^4))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec {\left (c + d x \right )}}{\left (a \cos {\left (c + d x \right )} + b \sin {\left (c + d x \right )}\right )^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)/(a*cos(d*x+c)+b*sin(d*x+c))**4,x)

[Out]

Integral(sec(c + d*x)/(a*cos(c + d*x) + b*sin(c + d*x))**4, x)

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