3.319 \(\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx\)

Optimal. Leaf size=93 \[ \frac {A \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}+\frac {B \sqrt {\cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{d \sqrt {b \cos (c+d x)}}+\frac {C x \sqrt {\cos (c+d x)}}{\sqrt {b \cos (c+d x)}} \]

[Out]

A*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(1/2)+C*x*cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(1/2)+B*arctanh(sin(d
*x+c))*cos(d*x+c)^(1/2)/d/(b*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {18, 3021, 2735, 3770} \[ \frac {A \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}+\frac {B \sqrt {\cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{d \sqrt {b \cos (c+d x)}}+\frac {C x \sqrt {\cos (c+d x)}}{\sqrt {b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(C*x*Sqrt[Cos[c + d*x]])/Sqrt[b*Cos[c + d*x]] + (B*ArcTanh[Sin[c + d*x]]*Sqrt[Cos[c + d*x]])/(d*Sqrt[b*Cos[c +
d*x]]) + (A*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*Sqrt[b*Cos[c + d*x]])

Rule 18

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(a^(m - 1/2)*b^(n + 1/2)*Sqrt[a*v])/Sqrt[b*v]
, Int[u*v^(m + n), x], x] /; FreeQ[{a, b, m}, x] &&  !IntegerQ[m] && ILtQ[n - 1/2, 0] && IntegerQ[m + n]

Rule 2735

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

Rule 3021

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

Rule 3770

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

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 60, normalized size = 0.65 \[ \frac {A \sin (c+d x)+B \cos (c+d x) \tanh ^{-1}(\sin (c+d x))+C d x \cos (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(C*d*x*Cos[c + d*x] + B*ArcTanh[Sin[c + d*x]]*Cos[c + d*x] + A*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*Sqrt[b*Cos[
c + d*x]])

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fricas [A]  time = 1.21, size = 317, normalized size = 3.41 \[ \left [-\frac {2 \, B \sqrt {-b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{2} + C \sqrt {-b} \cos \left (d x + c\right )^{2} \log \left (2 \, b \cos \left (d x + c\right )^{2} + 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) - 2 \, \sqrt {b \cos \left (d x + c\right )} A \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, b d \cos \left (d x + c\right )^{2}}, \frac {2 \, C \sqrt {b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right )^{2} + B \sqrt {b} \cos \left (d x + c\right )^{2} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )} A \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, b d \cos \left (d x + c\right )^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(b*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{\sqrt {b \cos \left (d x + c\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(b*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/(sqrt(b*cos(d*x + c))*cos(d*x + c)^(3/2)), x)

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maple [A]  time = 0.30, size = 72, normalized size = 0.77 \[ \frac {-2 B \cos \left (d x +c \right ) \arctanh \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )+C \cos \left (d x +c \right ) \left (d x +c \right )+A \sin \left (d x +c \right )}{d \sqrt {b \cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(b*cos(d*x+c))^(1/2),x)

[Out]

1/d*(-2*B*cos(d*x+c)*arctanh((-1+cos(d*x+c))/sin(d*x+c))+C*cos(d*x+c)*(d*x+c)+A*sin(d*x+c))/(b*cos(d*x+c))^(1/
2)/cos(d*x+c)^(1/2)

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maxima [A]  time = 0.64, size = 149, normalized size = 1.60 \[ \frac {\frac {B {\left (\log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )\right )}}{\sqrt {b}} + \frac {4 \, C \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{\sqrt {b}} + \frac {4 \, A \sqrt {b} \sin \left (2 \, d x + 2 \, c\right )}{b \cos \left (2 \, d x + 2 \, c\right )^{2} + b \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, b \cos \left (2 \, d x + 2 \, c\right ) + b}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(b*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

1/2*(B*(log(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) - log(cos(d*x + c)^2 + sin(d*x + c)^2 - 2*si
n(d*x + c) + 1))/sqrt(b) + 4*C*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/sqrt(b) + 4*A*sqrt(b)*sin(2*d*x + 2*c)/
(b*cos(2*d*x + 2*c)^2 + b*sin(2*d*x + 2*c)^2 + 2*b*cos(2*d*x + 2*c) + b))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {b\,\cos \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(3/2)*(b*cos(c + d*x))^(1/2)),x)

[Out]

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(3/2)*(b*cos(c + d*x))^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B \cos {\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}}{\sqrt {b \cos {\left (c + d x \right )}} \cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(3/2)/(b*cos(d*x+c))**(1/2),x)

[Out]

Integral((A + B*cos(c + d*x) + C*cos(c + d*x)**2)/(sqrt(b*cos(c + d*x))*cos(c + d*x)**(3/2)), x)

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