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

Optimal. Leaf size=153 \[ \frac{\left (a^2+b^2\right ) \sec (c+d x)}{b^3 d}-\frac{a \left (a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{b^4 d}-\frac{\left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{b^4 d}-\frac{a \tanh ^{-1}(\sin (c+d x))}{2 b^2 d}-\frac{a \tan (c+d x) \sec (c+d x)}{2 b^2 d}+\frac{\sec ^3(c+d x)}{3 b d} \]

[Out]

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

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Rubi [A]  time = 0.157237, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3104, 3768, 3770, 3074, 206} \[ \frac{\left (a^2+b^2\right ) \sec (c+d x)}{b^3 d}-\frac{a \left (a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{b^4 d}-\frac{\left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{b^4 d}-\frac{a \tanh ^{-1}(\sin (c+d x))}{2 b^2 d}-\frac{a \tan (c+d x) \sec (c+d x)}{2 b^2 d}+\frac{\sec ^3(c+d x)}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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])

Rubi steps

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

Mathematica [B]  time = 1.99009, size = 321, normalized size = 2.1 \[ \frac{48 \left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )-b}{\sqrt{a^2+b^2}}\right )+\sec ^3(c+d x) \left (12 b \left (a^2+b^2\right ) \cos (2 (c+d x))+9 a \left (2 a^2+3 b^2\right ) \cos (c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+12 a^2 b+6 a^3 \cos (3 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-6 a^3 \cos (3 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-6 a b^2 \sin (2 (c+d x))+9 a b^2 \cos (3 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-9 a b^2 \cos (3 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+20 b^3\right )}{24 b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(48*(a^2 + b^2)^(3/2)*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]] + Sec[c + d*x]^3*(12*a^2*b + 20*b^3 +
12*b*(a^2 + b^2)*Cos[2*(c + d*x)] + 6*a^3*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 9*a*b^2
*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 9*a*(2*a^2 + 3*b^2)*Cos[c + d*x]*(Log[Cos[(c + d*
x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - 6*a^3*Cos[3*(c + d*x)]*Log[Cos[(c + d*
x)/2] + Sin[(c + d*x)/2]] - 9*a*b^2*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 6*a*b^2*Sin[2*
(c + d*x)]))/(24*b^4*d)

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Maple [B]  time = 0.202, size = 488, normalized size = 3.2 \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)),x)

[Out]

2/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^4+4/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))*a^2+2/d/(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))+1/3/d/b/(tan(1/2*d*x+1/2*c)+1)^3+1/2/d/b^2/(tan(1/2*d*x+1/2*c)+1)^2*a-1/2/d/b/(t
an(1/2*d*x+1/2*c)+1)^2+1/d/b^3/(tan(1/2*d*x+1/2*c)+1)*a^2-1/2/d/b^2/(tan(1/2*d*x+1/2*c)+1)*a+3/2/d/b/(tan(1/2*
d*x+1/2*c)+1)-1/d*a^3/b^4*ln(tan(1/2*d*x+1/2*c)+1)-3/2/d*a/b^2*ln(tan(1/2*d*x+1/2*c)+1)-1/3/d/b/(tan(1/2*d*x+1
/2*c)-1)^3-1/2/d/b^2/(tan(1/2*d*x+1/2*c)-1)^2*a-1/2/d/b/(tan(1/2*d*x+1/2*c)-1)^2-1/d/b^3/(tan(1/2*d*x+1/2*c)-1
)*a^2-1/2/d/b^2/(tan(1/2*d*x+1/2*c)-1)*a-3/2/d/b/(tan(1/2*d*x+1/2*c)-1)+1/d*a^3/b^4*ln(tan(1/2*d*x+1/2*c)-1)+3
/2/d*a/b^2*ln(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)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.837595, size = 633, normalized size = 4.14 \begin{align*} \frac{6 \,{\left (a^{2} + b^{2}\right )}^{\frac{3}{2}} \cos \left (d x + c\right )^{3} \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 ) - 3 \,{\left (2 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \,{\left (2 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 6 \, a b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 4 \, b^{3} + 12 \,{\left (a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2}}{12 \, b^{4} d \cos \left (d x + c\right )^{3}} \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)),x, algorithm="fricas")

[Out]

1/12*(6*(a^2 + b^2)^(3/2)*cos(d*x + c)^3*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)) - 3*(2*a^3 + 3*a*b^2)*cos(d*x + c)^3*log(sin(d*x + c) + 1) + 3*(2*a^3 + 3*a*b^2)*co
s(d*x + c)^3*log(-sin(d*x + c) + 1) - 6*a*b^2*cos(d*x + c)*sin(d*x + c) + 4*b^3 + 12*(a^2*b + b^3)*cos(d*x + c
)^2)/(b^4*d*cos(d*x + c)^3)

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

[Out]

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

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Giac [A]  time = 1.34997, size = 375, normalized size = 2.45 \begin{align*} -\frac{\frac{3 \,{\left (2 \, a^{3} + 3 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{b^{4}} - \frac{3 \,{\left (2 \, a^{3} + 3 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{b^{4}} + \frac{6 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\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 )}{\sqrt{a^{2} + b^{2}} b^{4}} + \frac{2 \,{\left (3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 12 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 12 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 12 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, a^{2} + 8 \, b^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{3} b^{3}}}{6 \, d} \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)),x, algorithm="giac")

[Out]

-1/6*(3*(2*a^3 + 3*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^4 - 3*(2*a^3 + 3*a*b^2)*log(abs(tan(1/2*d*x + 1
/2*c) - 1))/b^4 + 6*(a^4 + 2*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))/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^4) + 2*(3*a*b*tan(1/2*d*x + 1/2*c)^5 + 6
*a^2*tan(1/2*d*x + 1/2*c)^4 + 12*b^2*tan(1/2*d*x + 1/2*c)^4 - 12*a^2*tan(1/2*d*x + 1/2*c)^2 - 12*b^2*tan(1/2*d
*x + 1/2*c)^2 - 3*a*b*tan(1/2*d*x + 1/2*c) + 6*a^2 + 8*b^2)/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*b^3))/d