3.148 \(\int \frac{\sec ^3(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^4} \, dx\)

Optimal. Leaf size=400 \[ \frac{8 a^2 \tanh ^{-1}(\sin (c+d x))}{b^6 d}+\frac{2 \left (a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{b^6 d}-\frac{4 a^2}{b^5 d (a \cos (c+d x)+b \sin (c+d x))}-\frac{2 \left (a^2+b^2\right )}{b^5 d (a \cos (c+d x)+b \sin (c+d x))}-\frac{a^2+b^2}{3 b^3 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac{4 a^3 \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{b^6 d \sqrt{a^2+b^2}}+\frac{6 a \sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{b^6 d}+\frac{3 a \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{2 b^4 d \sqrt{a^2+b^2}}-\frac{4 a \sec (c+d x)}{b^5 d}+\frac{3 a (b \cos (c+d x)-a \sin (c+d x))}{2 b^4 d (a \cos (c+d x)+b \sin (c+d x))^2}+\frac{\tanh ^{-1}(\sin (c+d x))}{2 b^4 d}+\frac{\tan (c+d x) \sec (c+d x)}{2 b^4 d} \]

[Out]

(8*a^2*ArcTanh[Sin[c + d*x]])/(b^6*d) + ArcTanh[Sin[c + d*x]]/(2*b^4*d) + (2*(a^2 + b^2)*ArcTanh[Sin[c + d*x]]
)/(b^6*d) + (4*a^3*ArcTanh[(b*Cos[c + d*x] - a*Sin[c + d*x])/Sqrt[a^2 + b^2]])/(b^6*Sqrt[a^2 + b^2]*d) + (3*a*
ArcTanh[(b*Cos[c + d*x] - a*Sin[c + d*x])/Sqrt[a^2 + b^2]])/(2*b^4*Sqrt[a^2 + b^2]*d) + (6*a*Sqrt[a^2 + b^2]*A
rcTanh[(b*Cos[c + d*x] - a*Sin[c + d*x])/Sqrt[a^2 + b^2]])/(b^6*d) - (4*a*Sec[c + d*x])/(b^5*d) - (a^2 + b^2)/
(3*b^3*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) + (3*a*(b*Cos[c + d*x] - a*Sin[c + d*x]))/(2*b^4*d*(a*Cos[c + d*
x] + b*Sin[c + d*x])^2) - (4*a^2)/(b^5*d*(a*Cos[c + d*x] + b*Sin[c + d*x])) - (2*(a^2 + b^2))/(b^5*d*(a*Cos[c
+ d*x] + b*Sin[c + d*x])) + (Sec[c + d*x]*Tan[c + d*x])/(2*b^4*d)

________________________________________________________________________________________

Rubi [A]  time = 0.796426, antiderivative size = 400, normalized size of antiderivative = 1., number of steps used = 32, number of rules used = 8, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3106, 3094, 3770, 3074, 206, 3076, 3768, 3104} \[ \frac{8 a^2 \tanh ^{-1}(\sin (c+d x))}{b^6 d}+\frac{2 \left (a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{b^6 d}-\frac{4 a^2}{b^5 d (a \cos (c+d x)+b \sin (c+d x))}-\frac{2 \left (a^2+b^2\right )}{b^5 d (a \cos (c+d x)+b \sin (c+d x))}-\frac{a^2+b^2}{3 b^3 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac{4 a^3 \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{b^6 d \sqrt{a^2+b^2}}+\frac{6 a \sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{b^6 d}+\frac{3 a \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{2 b^4 d \sqrt{a^2+b^2}}-\frac{4 a \sec (c+d x)}{b^5 d}+\frac{3 a (b \cos (c+d x)-a \sin (c+d x))}{2 b^4 d (a \cos (c+d x)+b \sin (c+d x))^2}+\frac{\tanh ^{-1}(\sin (c+d x))}{2 b^4 d}+\frac{\tan (c+d x) \sec (c+d x)}{2 b^4 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3/(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]

[Out]

(8*a^2*ArcTanh[Sin[c + d*x]])/(b^6*d) + ArcTanh[Sin[c + d*x]]/(2*b^4*d) + (2*(a^2 + b^2)*ArcTanh[Sin[c + d*x]]
)/(b^6*d) + (4*a^3*ArcTanh[(b*Cos[c + d*x] - a*Sin[c + d*x])/Sqrt[a^2 + b^2]])/(b^6*Sqrt[a^2 + b^2]*d) + (3*a*
ArcTanh[(b*Cos[c + d*x] - a*Sin[c + d*x])/Sqrt[a^2 + b^2]])/(2*b^4*Sqrt[a^2 + b^2]*d) + (6*a*Sqrt[a^2 + b^2]*A
rcTanh[(b*Cos[c + d*x] - a*Sin[c + d*x])/Sqrt[a^2 + b^2]])/(b^6*d) - (4*a*Sec[c + d*x])/(b^5*d) - (a^2 + b^2)/
(3*b^3*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) + (3*a*(b*Cos[c + d*x] - a*Sin[c + d*x]))/(2*b^4*d*(a*Cos[c + d*
x] + b*Sin[c + d*x])^2) - (4*a^2)/(b^5*d*(a*Cos[c + d*x] + b*Sin[c + d*x])) - (2*(a^2 + b^2))/(b^5*d*(a*Cos[c
+ d*x] + b*Sin[c + d*x])) + (Sec[c + d*x]*Tan[c + d*x])/(2*b^4*d)

Rule 3106

Int[cos[(c_.) + (d_.)*(x_)]^(m_)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbo
l] :> Dist[(a^2 + b^2)/b^2, Int[Cos[c + d*x]^(m + 2)*(a*Cos[c + d*x] + b*Sin[c + d*x])^n, x], x] + (Dist[1/b^2
, Int[Cos[c + d*x]^m*(a*Cos[c + d*x] + b*Sin[c + d*x])^(n + 2), x], x] - Dist[(2*a)/b^2, Int[Cos[c + d*x]^(m +
1)*(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] && LtQ[m, -1]

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]

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 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 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 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 3104

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

Rubi steps

\begin{align*} \int \frac{\sec ^3(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^4} \, dx &=\frac{\int \frac{\sec ^3(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx}{b^2}-\frac{(2 a) \int \frac{\sec ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx}{b^2}+\frac{\left (a^2+b^2\right ) \int \frac{\sec (c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^4} \, dx}{b^2}\\ &=-\frac{a^2+b^2}{3 b^3 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac{\int \sec ^3(c+d x) \, dx}{b^4}-2 \frac{(2 a) \int \frac{\sec ^2(c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{b^4}+\frac{\left (4 a^2\right ) \int \frac{\sec (c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx}{b^4}+2 \frac{\left (a^2+b^2\right ) \int \frac{\sec (c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx}{b^4}-\frac{\left (a \left (a^2+b^2\right )\right ) \int \frac{1}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx}{b^4}-\frac{\left (2 a \left (a^2+b^2\right )\right ) \int \frac{1}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx}{b^4}\\ &=-\frac{a^2+b^2}{3 b^3 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac{3 a (b \cos (c+d x)-a \sin (c+d x))}{2 b^4 d (a \cos (c+d x)+b \sin (c+d x))^2}-\frac{4 a^2}{b^5 d (a \cos (c+d x)+b \sin (c+d x))}+\frac{\sec (c+d x) \tan (c+d x)}{2 b^4 d}+\frac{\left (4 a^2\right ) \int \sec (c+d x) \, dx}{b^6}-\frac{\left (4 a^3\right ) \int \frac{1}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{b^6}+\frac{\int \sec (c+d x) \, dx}{2 b^4}-\frac{a \int \frac{1}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{2 b^4}-\frac{a \int \frac{1}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{b^4}+2 \left (-\frac{a^2+b^2}{b^5 d (a \cos (c+d x)+b \sin (c+d x))}+\frac{\left (a^2+b^2\right ) \int \sec (c+d x) \, dx}{b^6}-\frac{\left (a \left (a^2+b^2\right )\right ) \int \frac{1}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{b^6}\right )-2 \left (\frac{2 a \sec (c+d x)}{b^5 d}-\frac{\left (2 a^2\right ) \int \sec (c+d x) \, dx}{b^6}+\frac{\left (2 a \left (a^2+b^2\right )\right ) \int \frac{1}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{b^6}\right )\\ &=\frac{4 a^2 \tanh ^{-1}(\sin (c+d x))}{b^6 d}+\frac{\tanh ^{-1}(\sin (c+d x))}{2 b^4 d}-\frac{a^2+b^2}{3 b^3 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac{3 a (b \cos (c+d x)-a \sin (c+d x))}{2 b^4 d (a \cos (c+d x)+b \sin (c+d x))^2}-\frac{4 a^2}{b^5 d (a \cos (c+d x)+b \sin (c+d x))}+\frac{\sec (c+d x) \tan (c+d x)}{2 b^4 d}+\frac{\left (4 a^3\right ) \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^6 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^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 )}{b^4 d}+2 \left (\frac{\left (a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{b^6 d}-\frac{a^2+b^2}{b^5 d (a \cos (c+d x)+b \sin (c+d x))}+\frac{\left (a \left (a^2+b^2\right )\right ) \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^6 d}\right )-2 \left (-\frac{2 a^2 \tanh ^{-1}(\sin (c+d x))}{b^6 d}+\frac{2 a \sec (c+d x)}{b^5 d}-\frac{\left (2 a \left (a^2+b^2\right )\right ) \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^6 d}\right )\\ &=\frac{4 a^2 \tanh ^{-1}(\sin (c+d x))}{b^6 d}+\frac{\tanh ^{-1}(\sin (c+d x))}{2 b^4 d}+\frac{4 a^3 \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{b^6 \sqrt{a^2+b^2} d}+\frac{3 a \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{2 b^4 \sqrt{a^2+b^2} d}-2 \left (-\frac{2 a^2 \tanh ^{-1}(\sin (c+d x))}{b^6 d}-\frac{2 a \sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{b^6 d}+\frac{2 a \sec (c+d x)}{b^5 d}\right )-\frac{a^2+b^2}{3 b^3 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac{3 a (b \cos (c+d x)-a \sin (c+d x))}{2 b^4 d (a \cos (c+d x)+b \sin (c+d x))^2}-\frac{4 a^2}{b^5 d (a \cos (c+d x)+b \sin (c+d x))}+2 \left (\frac{\left (a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{b^6 d}+\frac{a \sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{b^6 d}-\frac{a^2+b^2}{b^5 d (a \cos (c+d x)+b \sin (c+d x))}\right )+\frac{\sec (c+d x) \tan (c+d x)}{2 b^4 d}\\ \end{align*}

Mathematica [A]  time = 3.20565, size = 538, normalized size = 1.34 \[ -\frac{\sec ^4(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \left (18 b^2 \left (a^2+b^2\right ) \sin (c+d x) (a \cos (c+d x)+b \sin (c+d x))+6 b \left (12 a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^2+30 \left (4 a^2+b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^3-30 \left (4 a^2+b^2\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^3+\frac{60 a \left (4 a^2+3 b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^3 \tanh ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )-b}{\sqrt{a^2+b^2}}\right )}{\sqrt{a^2+b^2}}+4 b^3 \left (a^2+b^2\right )-\frac{3 b^2 (a \cos (c+d x)+b \sin (c+d x))^3}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{3 b^2 (a \cos (c+d x)+b \sin (c+d x))^3}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+48 a b (a \cos (c+d x)+b \sin (c+d x))^3+\frac{48 a b \sin \left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b \sin (c+d x))^3}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}-\frac{48 a b \sin \left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b \sin (c+d x))^3}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}\right )}{12 b^6 d (a+b \tan (c+d x))^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3/(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]

[Out]

-(Sec[c + d*x]^4*(a*Cos[c + d*x] + b*Sin[c + d*x])*(4*b^3*(a^2 + b^2) + 18*b^2*(a^2 + b^2)*Sin[c + d*x]*(a*Cos
[c + d*x] + b*Sin[c + d*x]) + 6*b*(12*a^2 + b^2)*(a*Cos[c + d*x] + b*Sin[c + d*x])^2 + 48*a*b*(a*Cos[c + d*x]
+ b*Sin[c + d*x])^3 + (60*a*(4*a^2 + 3*b^2)*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]]*(a*Cos[c + d*x]
+ b*Sin[c + d*x])^3)/Sqrt[a^2 + b^2] + 30*(4*a^2 + b^2)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(a*Cos[c + d
*x] + b*Sin[c + d*x])^3 - 30*(4*a^2 + b^2)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*(a*Cos[c + d*x] + b*Sin[c
+ d*x])^3 - (3*b^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^3)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (48*a*b*Sin[
(c + d*x)/2]*(a*Cos[c + d*x] + b*Sin[c + d*x])^3)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) + (3*b^2*(a*Cos[c + d*
x] + b*Sin[c + d*x])^3)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 - (48*a*b*Sin[(c + d*x)/2]*(a*Cos[c + d*x] + b
*Sin[c + d*x])^3)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])))/(12*b^6*d*(a + b*Tan[c + d*x])^4)

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Maple [B]  time = 0.344, size = 1255, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3/(a*cos(d*x+c)+b*sin(d*x+c))^4,x)

[Out]

9/d/b^4/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^3*a^3*tan(1/2*d*x+1/2*c)^5-5/2/d/b^4*ln(tan(1/2*d*x+
1/2*c)-1)+5/2/d/b^4*ln(tan(1/2*d*x+1/2*c)+1)+18/d/b/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^3*tan(1/
2*d*x+1/2*c)^2+12/d/b^5/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^3*a^4+5/3/d/b^3/(tan(1/2*d*x+1/2*c)^
2*a-2*tan(1/2*d*x+1/2*c)*b-a)^3*a^2+10/d/b^6*ln(tan(1/2*d*x+1/2*c)+1)*a^2+4/d/b^5/(tan(1/2*d*x+1/2*c)-1)*a-10/
d/b^6*ln(tan(1/2*d*x+1/2*c)-1)*a^2+2/d/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^3/a*tan(1/2*d*x+1/2*c
)^5+8/3/d/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^3/a*tan(1/2*d*x+1/2*c)^3+2/d/(tan(1/2*d*x+1/2*c)^2
*a-2*tan(1/2*d*x+1/2*c)*b-a)^3/a*tan(1/2*d*x+1/2*c)-4/d/b^5/(tan(1/2*d*x+1/2*c)+1)*a-1/2/d/b^4/(tan(1/2*d*x+1/
2*c)+1)^2+1/2/d/b^4/(tan(1/2*d*x+1/2*c)+1)+1/2/d/b^4/(tan(1/2*d*x+1/2*c)-1)^2+1/2/d/b^4/(tan(1/2*d*x+1/2*c)-1)
+2/3/d/b/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^3+12/d/b^5/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/
2*c)*b-a)^3*a^4*tan(1/2*d*x+1/2*c)^4-39/d/b^3/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^3*a^2*tan(1/2*
d*x+1/2*c)^4-4/d*b/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^3/a^2*tan(1/2*d*x+1/2*c)^4-72/d/b^4/(tan(
1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^3*a^3*tan(1/2*d*x+1/2*c)^3+38/d/b^2/(tan(1/2*d*x+1/2*c)^2*a-2*tan
(1/2*d*x+1/2*c)*b-a)^3*a*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*tan(1/2*d*x+1/2*c)*b-a)^3/a^
3*tan(1/2*d*x+1/2*c)^3-24/d/b^5/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^3*a^4*tan(1/2*d*x+1/2*c)^2+1
00/d/b^3/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^3*a^2*tan(1/2*d*x+1/2*c)^2+4/d*b/(tan(1/2*d*x+1/2*c
)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^3/a^2*tan(1/2*d*x+1/2*c)^2+63/d/b^4/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*
c)*b-a)^3*a^3*tan(1/2*d*x+1/2*c)+10/d/b^2/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^3*a*tan(1/2*d*x+1/
2*c)-20/d/b^6*a^3/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tan(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2))-15/d/b^4*a/(a^2+b^
2)^(1/2)*arctanh(1/2*(2*a*tan(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.917428, size = 1843, normalized size = 4.61 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

1/12*(6*a^2*b^5 + 6*b^7 - 30*(4*a^6*b - 3*a^4*b^3 - 8*a^2*b^5 - b^7)*cos(d*x + c)^4 - 20*(11*a^4*b^3 + 13*a^2*
b^5 + 2*b^7)*cos(d*x + c)^2 + 15*((4*a^6 - 9*a^4*b^2 - 9*a^2*b^4)*cos(d*x + c)^5 + 3*(4*a^4*b^2 + 3*a^2*b^4)*c
os(d*x + c)^3 + ((12*a^5*b + 5*a^3*b^3 - 3*a*b^5)*cos(d*x + c)^4 + (4*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*s
qrt(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)) + 15*((4*a^7 - 7*a^5*b^2 - 14*a^3*b^4 - 3*a*b^6)*cos(d*x + c)^5 + 3*(4*a^5*b^2 + 5*a^3*b^4 + a*b^6)*
cos(d*x + c)^3 + ((12*a^6*b + 11*a^4*b^3 - 2*a^2*b^5 - b^7)*cos(d*x + c)^4 + (4*a^4*b^3 + 5*a^2*b^5 + b^7)*cos
(d*x + c)^2)*sin(d*x + c))*log(sin(d*x + c) + 1) - 15*((4*a^7 - 7*a^5*b^2 - 14*a^3*b^4 - 3*a*b^6)*cos(d*x + c)
^5 + 3*(4*a^5*b^2 + 5*a^3*b^4 + a*b^6)*cos(d*x + c)^3 + ((12*a^6*b + 11*a^4*b^3 - 2*a^2*b^5 - b^7)*cos(d*x + c
)^4 + (4*a^4*b^3 + 5*a^2*b^5 + b^7)*cos(d*x + c)^2)*sin(d*x + c))*log(-sin(d*x + c) + 1) - 30*(10*(a^5*b^2 + a
^3*b^4)*cos(d*x + c)^3 + (a^3*b^4 + a*b^6)*cos(d*x + c))*sin(d*x + c))/((a^5*b^6 - 2*a^3*b^8 - 3*a*b^10)*d*cos
(d*x + c)^5 + 3*(a^3*b^8 + a*b^10)*d*cos(d*x + c)^3 + ((3*a^4*b^7 + 2*a^2*b^9 - b^11)*d*cos(d*x + c)^4 + (a^2*
b^9 + b^11)*d*cos(d*x + c)^2)*sin(d*x + c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3/(a*cos(d*x+c)+b*sin(d*x+c))**4,x)

[Out]

Timed out

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Giac [A]  time = 1.35717, size = 740, normalized size = 1.85 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="giac")

[Out]

1/6*(15*(4*a^2 + b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^6 - 15*(4*a^2 + b^2)*log(abs(tan(1/2*d*x + 1/2*c) -
1))/b^6 + 15*(4*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)))/(sqrt(a^2 + b^2)*b^6) + 6*(b*tan(1/2*d*x + 1/2*c)^3 + 8*a*tan(1/2*d*x +
1/2*c)^2 + b*tan(1/2*d*x + 1/2*c) - 8*a)/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*b^5) + 2*(27*a^6*b*tan(1/2*d*x + 1/2
*c)^5 + 6*a^2*b^5*tan(1/2*d*x + 1/2*c)^5 + 36*a^7*tan(1/2*d*x + 1/2*c)^4 - 117*a^5*b^2*tan(1/2*d*x + 1/2*c)^4
- 12*a*b^6*tan(1/2*d*x + 1/2*c)^4 - 216*a^6*b*tan(1/2*d*x + 1/2*c)^3 + 114*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 - 72*a^7*tan(1/2*d*x + 1/2*c)^2 + 300*a^5*b^2*ta
n(1/2*d*x + 1/2*c)^2 + 54*a^3*b^4*tan(1/2*d*x + 1/2*c)^2 + 12*a*b^6*tan(1/2*d*x + 1/2*c)^2 + 189*a^6*b*tan(1/2
*d*x + 1/2*c) + 30*a^4*b^3*tan(1/2*d*x + 1/2*c) + 6*a^2*b^5*tan(1/2*d*x + 1/2*c) + 36*a^7 + 5*a^5*b^2 + 2*a^3*
b^4)/((a*tan(1/2*d*x + 1/2*c)^2 - 2*b*tan(1/2*d*x + 1/2*c) - a)^3*a^3*b^5))/d