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

Optimal. Leaf size=383 \[ \frac{4 a^2 \sec (c+d x)}{b^5 d}+\frac{2 \left (a^2+b^2\right ) \sec (c+d x)}{b^5 d}-\frac{4 a^3 \tanh ^{-1}(\sin (c+d x))}{b^6 d}-\frac{6 a \left (a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{b^6 d}+\frac{4 a \left (a^2+b^2\right )}{b^5 d (a \cos (c+d x)+b \sin (c+d x))}-\frac{\left (a^2+b^2\right ) (b \cos (c+d x)-a \sin (c+d x))}{2 b^4 d (a \cos (c+d x)+b \sin (c+d x))^2}-\frac{8 a^2 \sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{b^6 d}-\frac{2 \left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{b^6 d}-\frac{\sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{2 b^4 d}-\frac{3 a \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}-\frac{3 a \tan (c+d x) \sec (c+d x)}{2 b^4 d}+\frac{\sec ^3(c+d x)}{3 b^3 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.785488, antiderivative size = 383, normalized size of antiderivative = 1., number of steps used = 31, number of rules used = 8, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3106, 3076, 3074, 206, 3104, 3770, 3094, 3768} \[ \frac{4 a^2 \sec (c+d x)}{b^5 d}+\frac{2 \left (a^2+b^2\right ) \sec (c+d x)}{b^5 d}-\frac{4 a^3 \tanh ^{-1}(\sin (c+d x))}{b^6 d}-\frac{6 a \left (a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{b^6 d}+\frac{4 a \left (a^2+b^2\right )}{b^5 d (a \cos (c+d x)+b \sin (c+d x))}-\frac{\left (a^2+b^2\right ) (b \cos (c+d x)-a \sin (c+d x))}{2 b^4 d (a \cos (c+d x)+b \sin (c+d x))^2}-\frac{8 a^2 \sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{b^6 d}-\frac{2 \left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{b^6 d}-\frac{\sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{2 b^4 d}-\frac{3 a \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}-\frac{3 a \tan (c+d x) \sec (c+d x)}{2 b^4 d}+\frac{\sec ^3(c+d x)}{3 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3770

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

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 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]

Rubi steps

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

Mathematica [C]  time = 2.29544, size = 688, normalized size = 1.8 \[ \frac{\sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \left (\frac{6 b^2 \left (a^2+b^2\right )^2 \sin (c+d x)}{a}+2 b \left (36 a^2+13 b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^2+\frac{2 b \left (36 a^2+13 b^2\right ) \sin \left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}-\frac{2 b \left (36 a^2+13 b^2\right ) \sin \left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}+\frac{6 b (a-i b) (a+i b) \left (8 a^2-b^2\right ) (a \cos (c+d x)+b \sin (c+d x))}{a}+30 a \left (4 a^2+3 b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^2-30 a \left (4 a^2+3 b^2\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^2+60 \sqrt{a^2+b^2} \left (4 a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^2 \tanh ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )-b}{\sqrt{a^2+b^2}}\right )+\frac{2 b^3 \sin \left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}-\frac{2 b^3 \sin \left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{b^2 (b-9 a) (a \cos (c+d x)+b \sin (c+d x))^2}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{b^2 (9 a+b) (a \cos (c+d x)+b \sin (c+d x))^2}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}\right )}{12 b^6 d (a+b \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])*((6*b^2*(a^2 + b^2)^2*Sin[c + d*x])/a + (6*(a - I*b)*(a + I*
b)*b*(8*a^2 - b^2)*(a*Cos[c + d*x] + b*Sin[c + d*x]))/a + 2*b*(36*a^2 + 13*b^2)*(a*Cos[c + d*x] + b*Sin[c + d*
x])^2 + 60*Sqrt[a^2 + b^2]*(4*a^2 + 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])^2 + 30*a*(4*a^2 + 3*b^2)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(a*Cos[c + d*x] + b*Sin[c +
d*x])^2 - 30*a*(4*a^2 + 3*b^2)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*(a*Cos[c + d*x] + b*Sin[c + d*x])^2 +
(b^2*(-9*a + b)*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (2*b^3*Sin[(c +
d*x)/2]*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3 + (2*b*(36*a^2 + 13*b^2)
*Sin[(c + d*x)/2]*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) - (2*b^3*Sin[(c +
d*x)/2]*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 + (b^2*(9*a + b)*(a*Cos[
c + d*x] + b*Sin[c + d*x])^2)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 - (2*b*(36*a^2 + 13*b^2)*Sin[(c + d*x)/2
]*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])))/(12*b^6*d*(a + b*Tan[c + d*x])^
3)

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Maple [B]  time = 0.308, size = 1125, normalized size = 2.9 \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)^4/(a*cos(d*x+c)+b*sin(d*x+c))^3,x)

[Out]

9/d/b^3/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^2*a^2*tan(1/2*d*x+1/2*c)^2-2/d*b/(tan(1/2*d*x+1/2*c)
^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^2/a^2*tan(1/2*d*x+1/2*c)^2+23/d/b^2/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c
)*b-a)^2*a*tan(1/2*d*x+1/2*c)+25/d/b^4/(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
))*a^2-3/2/d/b^4/(tan(1/2*d*x+1/2*c)-1)^2*a-6/d/b^5/(tan(1/2*d*x+1/2*c)-1)*a^2-3/2/d/b^4/(tan(1/2*d*x+1/2*c)-1
)*a+10/d*a^3/b^6*ln(tan(1/2*d*x+1/2*c)-1)+1/3/d/b^3/(tan(1/2*d*x+1/2*c)+1)^3-1/2/d/b^3/(tan(1/2*d*x+1/2*c)+1)^
2-1/3/d/b^3/(tan(1/2*d*x+1/2*c)-1)^3-1/2/d/b^3/(tan(1/2*d*x+1/2*c)-1)^2+25/d/b^4/(tan(1/2*d*x+1/2*c)^2*a-2*tan
(1/2*d*x+1/2*c)*b-a)^2*a^3*tan(1/2*d*x+1/2*c)+20/d/b^6/(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))*a^4-8/d/b^5/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^2*a^4*tan(1/2*d*x+1/2*c)^2-7/
d/b^4/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^2*a^3*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)^2/a*tan(1/2*d*x+1/2*c)^3+15/d/b/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^2
*tan(1/2*d*x+1/2*c)^2-2/d/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^2/a*tan(1/2*d*x+1/2*c)+7/d/b^3/(ta
n(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^2*a^2+5/d/b^2/(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))-15/2/d*a/b^4*ln(tan(1/2*d*x+1/2*c)+1)+15/2/d*a/b^4*ln(tan(1/2*d*x+1/2*c)-1)+8/d/b^5/(
tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^2*a^4+3/2/d/b^4/(tan(1/2*d*x+1/2*c)+1)^2*a+6/d/b^5/(tan(1/2*d
*x+1/2*c)+1)*a^2-3/2/d/b^4/(tan(1/2*d*x+1/2*c)+1)*a-10/d*a^3/b^6*ln(tan(1/2*d*x+1/2*c)+1)-5/d/b^2/(tan(1/2*d*x
+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^2*a*tan(1/2*d*x+1/2*c)^3-1/d/b/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2
*c)*b-a)^2+5/2/d/b^3/(tan(1/2*d*x+1/2*c)+1)-5/2/d/b^3/(tan(1/2*d*x+1/2*c)-1)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.856379, size = 1299, normalized size = 3.39 \begin{align*} \frac{4 \, b^{5} + 30 \,{\left (4 \, a^{4} b + a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{4} + 20 \,{\left (2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} + 15 \,{\left ({\left (4 \, a^{4} - 3 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{5} + 2 \,{\left (4 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) +{\left (4 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{3}\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 ) - 15 \,{\left ({\left (4 \, a^{5} - a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} + 2 \,{\left (4 \, a^{4} b + 3 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) +{\left (4 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \,{\left ({\left (4 \, a^{5} - a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} + 2 \,{\left (4 \, a^{4} b + 3 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) +{\left (4 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 10 \,{\left (a b^{4} \cos \left (d x + c\right ) - 6 \,{\left (3 \, a^{3} b^{2} + 2 \, a b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{12 \,{\left (2 \, a b^{7} d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + b^{8} d \cos \left (d x + c\right )^{3} +{\left (a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [A]  time = 1.41592, size = 689, normalized size = 1.8 \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)^4/(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/6*(15*(4*a^3 + 3*a*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(tan(1/2*d*x +
1/2*c) - 1))/b^6 + 15*(4*a^4 + 5*a^2*b^2 + b^4)*log(abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b - 2*sqrt(a^2 + b^2))/a
bs(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) + 2*(9*a*b*tan(1/2*d*x + 1/2*c)^
5 + 36*a^2*tan(1/2*d*x + 1/2*c)^4 + 18*b^2*tan(1/2*d*x + 1/2*c)^4 - 72*a^2*tan(1/2*d*x + 1/2*c)^2 - 24*b^2*tan
(1/2*d*x + 1/2*c)^2 - 9*a*b*tan(1/2*d*x + 1/2*c) + 36*a^2 + 14*b^2)/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*b^5) + 6*(
7*a^5*b*tan(1/2*d*x + 1/2*c)^3 + 5*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 - 2*a*b^5*tan(1/2*d*x + 1/2*c)^3 + 8*a^6*tan
(1/2*d*x + 1/2*c)^2 - 9*a^4*b^2*tan(1/2*d*x + 1/2*c)^2 - 15*a^2*b^4*tan(1/2*d*x + 1/2*c)^2 + 2*b^6*tan(1/2*d*x
+ 1/2*c)^2 - 25*a^5*b*tan(1/2*d*x + 1/2*c) - 23*a^3*b^3*tan(1/2*d*x + 1/2*c) + 2*a*b^5*tan(1/2*d*x + 1/2*c) -
8*a^6 - 7*a^4*b^2 + a^2*b^4)/((a*tan(1/2*d*x + 1/2*c)^2 - 2*b*tan(1/2*d*x + 1/2*c) - a)^2*a^2*b^5))/d