3.125 \(\int \frac{\cos ^2(c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx\)

Optimal. Leaf size=147 \[ -\frac{\sin (c+d x)}{4 d \sqrt{a \sec (c+d x)+a}}+\frac{7 \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{4 \sqrt{a} d}-\frac{\sqrt{2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{\sin (c+d x) \cos (c+d x)}{2 d \sqrt{a \sec (c+d x)+a}} \]

[Out]

(7*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*Sqrt[a]*d) - (Sqrt[2]*ArcTan[(Sqrt[a]*Tan[c + d
*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d) - Sin[c + d*x]/(4*d*Sqrt[a + a*Sec[c + d*x]]) + (Cos[c +
d*x]*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.248374, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3823, 4022, 3920, 3774, 203, 3795} \[ -\frac{\sin (c+d x)}{4 d \sqrt{a \sec (c+d x)+a}}+\frac{7 \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{4 \sqrt{a} d}-\frac{\sqrt{2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{\sin (c+d x) \cos (c+d x)}{2 d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2/Sqrt[a + a*Sec[c + d*x]],x]

[Out]

(7*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*Sqrt[a]*d) - (Sqrt[2]*ArcTan[(Sqrt[a]*Tan[c + d
*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d) - Sin[c + d*x]/(4*d*Sqrt[a + a*Sec[c + d*x]]) + (Cos[c +
d*x]*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x]])

Rule 3823

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(Cot[e
+ f*x]*(d*Csc[e + f*x])^n)/(f*n*Sqrt[a + b*Csc[e + f*x]]), x] + Dist[1/(2*b*d*n), Int[((d*Csc[e + f*x])^(n + 1
)*(a + b*(2*n + 1)*Csc[e + f*x]))/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b
^2, 0] && LtQ[n, 0] && IntegerQ[2*n]

Rule 4022

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

Rule 3920

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c/a,
Int[Sqrt[a + b*Csc[e + f*x]], x], x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]

Rule 3774

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(a + x^2), x], x, (b*C
ot[c + d*x])/Sqrt[a + b*Csc[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 3795

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2/f, Subst[Int[1/(2
*a + x^2), x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0
]

Rubi steps

\begin{align*} \int \frac{\cos ^2(c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx &=\frac{\cos (c+d x) \sin (c+d x)}{2 d \sqrt{a+a \sec (c+d x)}}-\frac{\int \frac{\cos (c+d x) (a-3 a \sec (c+d x))}{\sqrt{a+a \sec (c+d x)}} \, dx}{4 a}\\ &=-\frac{\sin (c+d x)}{4 d \sqrt{a+a \sec (c+d x)}}+\frac{\cos (c+d x) \sin (c+d x)}{2 d \sqrt{a+a \sec (c+d x)}}-\frac{\int \frac{-\frac{7 a^2}{2}+\frac{1}{2} a^2 \sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{4 a^2}\\ &=-\frac{\sin (c+d x)}{4 d \sqrt{a+a \sec (c+d x)}}+\frac{\cos (c+d x) \sin (c+d x)}{2 d \sqrt{a+a \sec (c+d x)}}+\frac{7 \int \sqrt{a+a \sec (c+d x)} \, dx}{8 a}-\int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx\\ &=-\frac{\sin (c+d x)}{4 d \sqrt{a+a \sec (c+d x)}}+\frac{\cos (c+d x) \sin (c+d x)}{2 d \sqrt{a+a \sec (c+d x)}}-\frac{7 \operatorname{Subst}\left (\int \frac{1}{a+x^2} \, dx,x,-\frac{a \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{4 d}+\frac{2 \operatorname{Subst}\left (\int \frac{1}{2 a+x^2} \, dx,x,-\frac{a \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{d}\\ &=\frac{7 \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{4 \sqrt{a} d}-\frac{\sqrt{2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right )}{\sqrt{a} d}-\frac{\sin (c+d x)}{4 d \sqrt{a+a \sec (c+d x)}}+\frac{\cos (c+d x) \sin (c+d x)}{2 d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.239557, size = 118, normalized size = 0.8 \[ \frac{\tan (c+d x) \left (\cos (c+d x) (2 \cos (c+d x)-1) \sqrt{1-\sec (c+d x)}+7 \tanh ^{-1}\left (\sqrt{1-\sec (c+d x)}\right )-4 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{1-\sec (c+d x)}}{\sqrt{2}}\right )\right )}{4 d \sqrt{1-\sec (c+d x)} \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2/Sqrt[a + a*Sec[c + d*x]],x]

[Out]

((7*ArcTanh[Sqrt[1 - Sec[c + d*x]]] - 4*Sqrt[2]*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]] + Cos[c + d*x]*(-1 + 2
*Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]])*Tan[c + d*x])/(4*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])

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Maple [B]  time = 0.226, size = 380, normalized size = 2.6 \begin{align*}{\frac{1}{16\,ad\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) } \left ( 7\,\sqrt{2} \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \sin \left ( dx+c \right ) \cos \left ( dx+c \right ) +7\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}\sin \left ( dx+c \right ) +8\,\ln \left ({\frac{1}{\sin \left ( dx+c \right ) } \left ( \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sin \left ( dx+c \right ) -\cos \left ( dx+c \right ) +1 \right ) } \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +8\,\ln \left ({\frac{1}{\sin \left ( dx+c \right ) } \left ( \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sin \left ( dx+c \right ) -\cos \left ( dx+c \right ) +1 \right ) } \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}\sin \left ( dx+c \right ) -8\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}+12\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}-4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2/(a+a*sec(d*x+c))^(1/2),x)

[Out]

1/16/d/a*(7*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1
/2)*sin(d*x+c)/cos(d*x+c))*sin(d*x+c)*cos(d*x+c)+7*2^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^
(1/2)*sin(d*x+c)/cos(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*sin(d*x+c)+8*ln(((-2*cos(d*x+c)/(cos(d*x+c)+
1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*cos(d*x+c)*sin(d*x+c)+8*ln
(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/
2)*sin(d*x+c)-8*cos(d*x+c)^4+12*cos(d*x+c)^3-4*cos(d*x+c)^2)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/cos(d*x+c)/si
n(d*x+c)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{2}}{\sqrt{a \sec \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^2/sqrt(a*sec(d*x + c) + a), x)

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Fricas [A]  time = 2.53869, size = 1196, normalized size = 8.14 \begin{align*} \left [\frac{4 \, \sqrt{2}{\left (a \cos \left (d x + c\right ) + a\right )} \sqrt{-\frac{1}{a}} \log \left (\frac{2 \, \sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{-\frac{1}{a}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, \cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 7 \, \sqrt{-a}{\left (\cos \left (d x + c\right ) + 1\right )} \log \left (\frac{2 \, a \cos \left (d x + c\right )^{2} + 2 \, \sqrt{-a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \,{\left (2 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right )\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{8 \,{\left (a d \cos \left (d x + c\right ) + a d\right )}}, -\frac{7 \, \sqrt{a}{\left (\cos \left (d x + c\right ) + 1\right )} \arctan \left (\frac{\sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt{a} \sin \left (d x + c\right )}\right ) -{\left (2 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right )\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right ) - \frac{4 \, \sqrt{2}{\left (a \cos \left (d x + c\right ) + a\right )} \arctan \left (\frac{\sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt{a} \sin \left (d x + c\right )}\right )}{\sqrt{a}}}{4 \,{\left (a d \cos \left (d x + c\right ) + a d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[1/8*(4*sqrt(2)*(a*cos(d*x + c) + a)*sqrt(-1/a)*log((2*sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(-1
/a)*cos(d*x + c)*sin(d*x + c) + 3*cos(d*x + c)^2 + 2*cos(d*x + c) - 1)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1))
- 7*sqrt(-a)*(cos(d*x + c) + 1)*log((2*a*cos(d*x + c)^2 + 2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*c
os(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 2*(2*cos(d*x + c)^2 - cos(d*x + c))*sqrt(
(a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a*d*cos(d*x + c) + a*d), -1/4*(7*sqrt(a)*(cos(d*x + c) + 1)*
arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - (2*cos(d*x + c)^2 - cos(
d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c) - 4*sqrt(2)*(a*cos(d*x + c) + a)*arctan(sqrt(2)
*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)))/sqrt(a))/(a*d*cos(d*x + c) + a*d
)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos ^{2}{\left (c + d x \right )}}{\sqrt{a \left (\sec{\left (c + d x \right )} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2/(a+a*sec(d*x+c))**(1/2),x)

[Out]

Integral(cos(c + d*x)**2/sqrt(a*(sec(c + d*x) + 1)), x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError