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

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

[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]))

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Rubi [A]  time = 0.169594, antiderivative size = 231, normalized size of antiderivative = 1., 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 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]

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*}

Mathematica [A]  time = 3.2569, 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]

-(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)*A
rcTanh[(-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))/(6*b^4*d*(a + b*Tan[c + d*x])^4)

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Maple [B]  time = 0.316, size = 1367, normalized size = 5.9 \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)/(a*cos(d*x+c)+b*sin(d*x+c))^4,x)

[Out]

-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+b^2)*tan(1/2*d*x+1/2*c)^4+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/(a^2+b^2)+14/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+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*tan(1/2*d*x+1/2*c)*b-a)^3/(a^2+b^2)*a^4+5/3/d/
b/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^3/(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))+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^2+b^2)*a*tan(1/2*d*x+1/2*c)^5-2/d
/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^3*a/(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*tan(1/2*d*x+1/2*c)*b-a)^3*a/(a^2+b^2)*tan(1/2*d*x+1/2*c)-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)+4/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/(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*tan(1/2*d*x+1/2*c)*b-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*tan(1/2*d*x+1/2*c)*b-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*tan(1/2*d*x+1/2*c)*b-a)^3/(a^2+b^2)*a^3*tan(1/2*d*x+1/2*c)^5+2/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^2+b^2)/a*tan(1/2*d*x+1/2*c)^5+2/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+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*tan(1/2*d*x+1/2*c)*b-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*tan(1/2*d*x+1/2*c)*b-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*tan(1/2*d*x+1/2*c)*b-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*tan(1/2*d*x+1/2*c)*b-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*tan(1/2*d*x+1/2*c)*b-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*tan(1/2*
d*x+1/2*c)*b-a)^3*a^4/(a^2+b^2)*tan(1/2*d*x+1/2*c)^2+16/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/(a^2+b^2)*tan(1/2*d*x+1/2*c)^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)/(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 = 1.03503, size = 1655, normalized size = 7.16 \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)/(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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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)/(a*cos(d*x+c)+b*sin(d*x+c))**4,x)

[Out]

Timed out

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Giac [B]  time = 1.3403, size = 711, normalized size = 3.08 \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)/(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