3.129 \(\int \frac{\sec ^3(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=179 \[ \frac{2 a^2 \tanh ^{-1}(\sin (c+d x))}{b^4 d}+\frac{\left (a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{b^4 d}-\frac{a^2+b^2}{b^3 d (a \cos (c+d x)+b \sin (c+d x))}+\frac{3 a \sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{b^4 d}-\frac{2 a \sec (c+d x)}{b^3 d}+\frac{\tanh ^{-1}(\sin (c+d x))}{2 b^2 d}+\frac{\tan (c+d x) \sec (c+d x)}{2 b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.23961, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3106, 3094, 3770, 3074, 206, 3768, 3104} \[ \frac{2 a^2 \tanh ^{-1}(\sin (c+d x))}{b^4 d}+\frac{\left (a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{b^4 d}-\frac{a^2+b^2}{b^3 d (a \cos (c+d x)+b \sin (c+d x))}+\frac{3 a \sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{b^4 d}-\frac{2 a \sec (c+d x)}{b^3 d}+\frac{\tanh ^{-1}(\sin (c+d x))}{2 b^2 d}+\frac{\tan (c+d x) \sec (c+d x)}{2 b^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3/(a*Cos[c + d*x] + b*Sin[c + d*x])^2,x]

[Out]

(2*a^2*ArcTanh[Sin[c + d*x]])/(b^4*d) + ArcTanh[Sin[c + d*x]]/(2*b^2*d) + ((a^2 + b^2)*ArcTanh[Sin[c + d*x]])/
(b^4*d) + (3*a*Sqrt[a^2 + b^2]*ArcTanh[(b*Cos[c + d*x] - a*Sin[c + d*x])/Sqrt[a^2 + b^2]])/(b^4*d) - (2*a*Sec[
c + d*x])/(b^3*d) - (a^2 + b^2)/(b^3*d*(a*Cos[c + d*x] + b*Sin[c + d*x])) + (Sec[c + d*x]*Tan[c + d*x])/(2*b^2
*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 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))^2} \, dx &=\frac{\int \sec ^3(c+d x) \, dx}{b^2}-\frac{(2 a) \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 ) \int \frac{\sec (c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx}{b^2}\\ &=-\frac{2 a \sec (c+d x)}{b^3 d}-\frac{a^2+b^2}{b^3 d (a \cos (c+d x)+b \sin (c+d x))}+\frac{\sec (c+d x) \tan (c+d x)}{2 b^2 d}+\frac{\left (2 a^2\right ) \int \sec (c+d x) \, dx}{b^4}+\frac{\int \sec (c+d x) \, dx}{2 b^2}+\frac{\left (a^2+b^2\right ) \int \sec (c+d x) \, 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)} \, 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)} \, dx}{b^4}\\ &=\frac{2 a^2 \tanh ^{-1}(\sin (c+d x))}{b^4 d}+\frac{\tanh ^{-1}(\sin (c+d x))}{2 b^2 d}+\frac{\left (a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{b^4 d}-\frac{2 a \sec (c+d x)}{b^3 d}-\frac{a^2+b^2}{b^3 d (a \cos (c+d x)+b \sin (c+d x))}+\frac{\sec (c+d x) \tan (c+d x)}{2 b^2 d}+\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^4 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^4 d}\\ &=\frac{2 a^2 \tanh ^{-1}(\sin (c+d x))}{b^4 d}+\frac{\tanh ^{-1}(\sin (c+d x))}{2 b^2 d}+\frac{\left (a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{b^4 d}+\frac{3 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^4 d}-\frac{2 a \sec (c+d x)}{b^3 d}-\frac{a^2+b^2}{b^3 d (a \cos (c+d x)+b \sin (c+d x))}+\frac{\sec (c+d x) \tan (c+d x)}{2 b^2 d}\\ \end{align*}

Mathematica [C]  time = 6.11163, size = 709, normalized size = 3.96 \[ -\frac{3 \left (2 a^2+b^2\right ) \sec ^2(c+d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{2 b^4 d (a+b \tan (c+d x))^2}+\frac{3 \left (2 a^2+b^2\right ) \sec ^2(c+d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{2 b^4 d (a+b \tan (c+d x))^2}-\frac{6 a \sqrt{a^2+b^2} \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \tanh ^{-1}\left (\frac{\sqrt{a^2+b^2} \left (a \sin \left (\frac{1}{2} (c+d x)\right )-b \cos \left (\frac{1}{2} (c+d x)\right )\right )}{a^2 \cos \left (\frac{1}{2} (c+d x)\right )+b^2 \cos \left (\frac{1}{2} (c+d x)\right )}\right )}{b^4 d (a+b \tan (c+d x))^2}-\frac{2 a \sin \left (\frac{1}{2} (c+d x)\right ) \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2}{b^3 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^2}-\frac{2 a \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2}{b^3 d (a+b \tan (c+d x))^2}+\frac{2 a \sin \left (\frac{1}{2} (c+d x)\right ) \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2}{b^3 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^2}+\frac{\sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2}{4 b^2 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2 (a+b \tan (c+d x))^2}-\frac{\sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2}{4 b^2 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2 (a+b \tan (c+d x))^2}-\frac{(a-i b) (a+i b) \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))}{b^3 d (a+b \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3/(a*Cos[c + d*x] + b*Sin[c + d*x])^2,x]

[Out]

-(((a - I*b)*(a + I*b)*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x]))/(b^3*d*(a + b*Tan[c + d*x])^2)) - (2*
a*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(b^3*d*(a + b*Tan[c + d*x])^2) - (6*a*Sqrt[a^2 + b^2]*Ar
cTanh[(Sqrt[a^2 + b^2]*(-(b*Cos[(c + d*x)/2]) + a*Sin[(c + d*x)/2]))/(a^2*Cos[(c + d*x)/2] + b^2*Cos[(c + d*x)
/2])]*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(b^4*d*(a + b*Tan[c + d*x])^2) - (3*(2*a^2 + b^2)*Lo
g[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(2*b^4*d*(a + b*Tan
[c + d*x])^2) + (3*(2*a^2 + b^2)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sec[c + d*x]^2*(a*Cos[c + d*x] + b*S
in[c + d*x])^2)/(2*b^4*d*(a + b*Tan[c + d*x])^2) + (Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(4*b^2
*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2*(a + b*Tan[c + d*x])^2) - (2*a*Sec[c + d*x]^2*Sin[(c + d*x)/2]*(a*C
os[c + d*x] + b*Sin[c + d*x])^2)/(b^3*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(a + b*Tan[c + d*x])^2) - (Sec[c
+ d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(4*b^2*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2*(a + b*Tan[c +
d*x])^2) + (2*a*Sec[c + d*x]^2*Sin[(c + d*x)/2]*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)/(b^3*d*(Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2])*(a + b*Tan[c + d*x])^2)

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Maple [B]  time = 0.247, size = 440, normalized size = 2.5 \begin{align*} 2\,{\frac{a\tan \left ( 1/2\,dx+c/2 \right ) }{{b}^{2}d \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\,\tan \left ( 1/2\,dx+c/2 \right ) b-a \right ) }}+2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) }{d \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\,\tan \left ( 1/2\,dx+c/2 \right ) b-a \right ) a}}+2\,{\frac{{a}^{2}}{{b}^{3}d \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\,\tan \left ( 1/2\,dx+c/2 \right ) b-a \right ) }}+2\,{\frac{1}{db \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\,\tan \left ( 1/2\,dx+c/2 \right ) b-a \right ) }}-6\,{\frac{\sqrt{{a}^{2}+{b}^{2}}a}{d{b}^{4}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }-{\frac{1}{2\,{b}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-2\,{\frac{a}{{b}^{3}d \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }}+{\frac{1}{2\,{b}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+3\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ){a}^{2}}{d{b}^{4}}}+{\frac{3}{2\,{b}^{2}d}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{1}{2\,{b}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+2\,{\frac{a}{{b}^{3}d \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }}+{\frac{1}{2\,{b}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-3\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ){a}^{2}}{d{b}^{4}}}-{\frac{3}{2\,{b}^{2}d}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) } \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))^2,x)

[Out]

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)*a*tan(1/2*d*x+1/2*c)+2/d/(tan(1/2*d*x+1/2*c)^2*a-2*t
an(1/2*d*x+1/2*c)*b-a)/a*tan(1/2*d*x+1/2*c)+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)*a^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)-6/d/b^4*(a^2+b^2)^(1/2)*a*arctanh(1/2*(2*a*tan(1/2*d*x+1/2
*c)-2*b)/(a^2+b^2)^(1/2))-1/2/d/b^2/(tan(1/2*d*x+1/2*c)+1)^2-2/d/b^3/(tan(1/2*d*x+1/2*c)+1)*a+1/2/d/b^2/(tan(1
/2*d*x+1/2*c)+1)+3/d/b^4*ln(tan(1/2*d*x+1/2*c)+1)*a^2+3/2/d/b^2*ln(tan(1/2*d*x+1/2*c)+1)+1/2/d/b^2/(tan(1/2*d*
x+1/2*c)-1)^2+2/d/b^3/(tan(1/2*d*x+1/2*c)-1)*a+1/2/d/b^2/(tan(1/2*d*x+1/2*c)-1)-3/d/b^4*ln(tan(1/2*d*x+1/2*c)-
1)*a^2-3/2/d/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)^3/(a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.697012, size = 860, normalized size = 4.8 \begin{align*} -\frac{6 \, a b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, b^{3} + 6 \,{\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2} - 6 \,{\left (a^{2} \cos \left (d x + c\right )^{3} + a b \cos \left (d x + c\right )^{2} \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 ) - 3 \,{\left ({\left (2 \, a^{3} + a b^{2}\right )} \cos \left (d x + c\right )^{3} +{\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \,{\left ({\left (2 \, a^{3} + a b^{2}\right )} \cos \left (d x + c\right )^{3} +{\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{4 \,{\left (a b^{4} d \cos \left (d x + c\right )^{3} + b^{5} d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right )\right )}} \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))^2,x, algorithm="fricas")

[Out]

-1/4*(6*a*b^2*cos(d*x + c)*sin(d*x + c) - 2*b^3 + 6*(2*a^2*b + b^3)*cos(d*x + c)^2 - 6*(a^2*cos(d*x + c)^3 + a
*b*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)) - 3*((2*a^3 + a*b^2)*cos(d*x + c)^3 + (2*a^2*b + b^3)*cos(d*x + c)^2*sin(d*x
+ c))*log(sin(d*x + c) + 1) + 3*((2*a^3 + a*b^2)*cos(d*x + c)^3 + (2*a^2*b + b^3)*cos(d*x + c)^2*sin(d*x + c))
*log(-sin(d*x + c) + 1))/(a*b^4*d*cos(d*x + c)^3 + b^5*d*cos(d*x + c)^2*sin(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{3}{\left (c + d x \right )}}{\left (a \cos{\left (c + d x \right )} + b \sin{\left (c + d x \right )}\right )^{2}}\, dx \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))**2,x)

[Out]

Integral(sec(c + d*x)**3/(a*cos(c + d*x) + b*sin(c + d*x))**2, x)

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Giac [A]  time = 1.32164, size = 378, normalized size = 2.11 \begin{align*} \frac{\frac{3 \,{\left (2 \, a^{2} + 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^{2} + 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^{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 )}{\sqrt{a^{2} + b^{2}} b^{4}} + \frac{2 \,{\left (b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, a\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{2} b^{3}} + \frac{4 \,{\left (a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{3} + a b^{2}\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 )} a b^{3}}}{2 \, d} \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))^2,x, algorithm="giac")

[Out]

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