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

Optimal. Leaf size=99 \[ \frac{2 b (b B-a C) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d \sqrt{a-b} \sqrt{a+b}}-\frac{(b B-a C) \tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac{B \tan (c+d x)}{a d} \]

[Out]

(2*b*(b*B - a*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^2*Sqrt[a - b]*Sqrt[a + b]*d) - ((b*B -
a*C)*ArcTanh[Sin[c + d*x]])/(a^2*d) + (B*Tan[c + d*x])/(a*d)

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Rubi [A]  time = 0.272975, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.175, Rules used = {3029, 3000, 12, 2747, 3770, 2659, 205} \[ \frac{2 b (b B-a C) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d \sqrt{a-b} \sqrt{a+b}}-\frac{(b B-a C) \tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac{B \tan (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*(b*B - a*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^2*Sqrt[a - b]*Sqrt[a + b]*d) - ((b*B -
a*C)*ArcTanh[Sin[c + d*x]])/(a^2*d) + (B*Tan[c + d*x])/(a*d)

Rule 3029

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[1/b^2, Int[(a + b*Sin[e + f*x])
^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rule 3000

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[((A*b^2 - a*b*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*
Sin[e + f*x])^(1 + n))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int
[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(a*A - b*B)*(b*c - a*d)*(m + 1) + b*d*(A*b - a*B)*(m
+ n + 2) + (A*b - a*B)*(a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(A*b - a*B)*(m + n + 3)*Sin[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^
2, 0] && RationalQ[m] && m < -1 && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n,
-1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2747

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[b/(
b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] - Dist[d/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; Fre
eQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rule 3770

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

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.52719, size = 129, normalized size = 1.3 \[ \frac{-\frac{2 b (b B-a C) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}+(b B-a C) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+a B \tan (c+d x)}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*b*(b*B - a*C)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] + (b*B - a*C)*(Log[C
os[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + a*B*Tan[c + d*x])/(a^2*d)

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Maple [B]  time = 0.061, size = 228, normalized size = 2.3 \begin{align*} -{\frac{B}{ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{bB}{d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{C}{ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+2\,{\frac{{b}^{2}B}{d{a}^{2}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-2\,{\frac{Cb}{ad\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-{\frac{B}{ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{bB}{d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{C}{ad}\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((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+b*cos(d*x+c)),x)

[Out]

-1/a/d*B/(tan(1/2*d*x+1/2*c)-1)+1/d/a^2*ln(tan(1/2*d*x+1/2*c)-1)*b*B-1/a/d*ln(tan(1/2*d*x+1/2*c)-1)*C+2/d*b^2/
a^2/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*B-2/d*b/a/((a+b)*(a-b))^(1/2)*arc
tan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C-1/a/d*B/(tan(1/2*d*x+1/2*c)+1)-1/d/a^2*ln(tan(1/2*d*x+1/2*
c)+1)*b*B+1/a/d*ln(tan(1/2*d*x+1/2*c)+1)*C

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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((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+b*cos(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.39427, size = 1049, normalized size = 10.6 \begin{align*} \left [\frac{{\left (C a b - B b^{2}\right )} \sqrt{-a^{2} + b^{2}} \cos \left (d x + c\right ) \log \left (\frac{2 \, a b \cos \left (d x + c\right ) +{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt{-a^{2} + b^{2}}{\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) +{\left (C a^{3} - B a^{2} b - C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (C a^{3} - B a^{2} b - C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (B a^{3} - B a b^{2}\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{4} - a^{2} b^{2}\right )} d \cos \left (d x + c\right )}, -\frac{2 \,{\left (C a b - B b^{2}\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \cos \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right ) -{\left (C a^{3} - B a^{2} b - C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (C a^{3} - B a^{2} b - C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (B a^{3} - B a b^{2}\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{4} - a^{2} b^{2}\right )} d \cos \left (d x + c\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+b*cos(d*x+c)),x, algorithm="fricas")

[Out]

[1/2*((C*a*b - B*b^2)*sqrt(-a^2 + b^2)*cos(d*x + c)*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d*x + c)^2 + 2
*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x + c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) +
a^2)) + (C*a^3 - B*a^2*b - C*a*b^2 + B*b^3)*cos(d*x + c)*log(sin(d*x + c) + 1) - (C*a^3 - B*a^2*b - C*a*b^2 +
B*b^3)*cos(d*x + c)*log(-sin(d*x + c) + 1) + 2*(B*a^3 - B*a*b^2)*sin(d*x + c))/((a^4 - a^2*b^2)*d*cos(d*x + c)
), -1/2*(2*(C*a*b - B*b^2)*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c)))*cos(d*
x + c) - (C*a^3 - B*a^2*b - C*a*b^2 + B*b^3)*cos(d*x + c)*log(sin(d*x + c) + 1) + (C*a^3 - B*a^2*b - C*a*b^2 +
B*b^3)*cos(d*x + c)*log(-sin(d*x + c) + 1) - 2*(B*a^3 - B*a*b^2)*sin(d*x + c))/((a^4 - a^2*b^2)*d*cos(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((B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**3/(a+b*cos(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.72007, size = 236, normalized size = 2.38 \begin{align*} \frac{\frac{{\left (C a - B b\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac{{\left (C a - B b\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} - \frac{2 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} a} + \frac{2 \,{\left (C a b - B b^{2}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{a^{2} - b^{2}}}\right )\right )}}{\sqrt{a^{2} - b^{2}} a^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+b*cos(d*x+c)),x, algorithm="giac")

[Out]

((C*a - B*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - (C*a - B*b)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^2 - 2*B
*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*a) + 2*(C*a*b - B*b^2)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*s
gn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))/(sqrt(a^2 - b^2)*
a^2))/d