3.133 \(\int \frac{\cos ^2(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=185 \[ \frac{19 \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{4 a^{3/2} d}-\frac{13 \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{7 \sin (c+d x)}{4 a d \sqrt{a \sec (c+d x)+a}}+\frac{\sin (c+d x) \cos (c+d x)}{a d \sqrt{a \sec (c+d x)+a}}-\frac{\sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}} \]

[Out]

(19*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*a^(3/2)*d) - (13*ArcTan[(Sqrt[a]*Tan[c + d*x])
/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - (Cos[c + d*x]*Sin[c + d*x])/(2*d*(a + a*Sec[c +
d*x])^(3/2)) - (7*Sin[c + d*x])/(4*a*d*Sqrt[a + a*Sec[c + d*x]]) + (Cos[c + d*x]*Sin[c + d*x])/(a*d*Sqrt[a + a
*Sec[c + d*x]])

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Rubi [A]  time = 0.390697, antiderivative size = 185, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3817, 4022, 3920, 3774, 203, 3795} \[ \frac{19 \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{4 a^{3/2} d}-\frac{13 \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{7 \sin (c+d x)}{4 a d \sqrt{a \sec (c+d x)+a}}+\frac{\sin (c+d x) \cos (c+d x)}{a d \sqrt{a \sec (c+d x)+a}}-\frac{\sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(19*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*a^(3/2)*d) - (13*ArcTan[(Sqrt[a]*Tan[c + d*x])
/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - (Cos[c + d*x]*Sin[c + d*x])/(2*d*(a + a*Sec[c +
d*x])^(3/2)) - (7*Sin[c + d*x])/(4*a*d*Sqrt[a + a*Sec[c + d*x]]) + (Cos[c + d*x]*Sin[c + d*x])/(a*d*Sqrt[a + a
*Sec[c + d*x]])

Rule 3817

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

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)}{(a+a \sec (c+d x))^{3/2}} \, dx &=-\frac{\cos (c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{\int \frac{\cos ^2(c+d x) \left (-4 a+\frac{5}{2} a \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{\cos (c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{\cos (c+d x) \sin (c+d x)}{a d \sqrt{a+a \sec (c+d x)}}-\frac{\int \frac{\cos (c+d x) \left (7 a^2-6 a^2 \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{4 a^3}\\ &=-\frac{\cos (c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{7 \sin (c+d x)}{4 a d \sqrt{a+a \sec (c+d x)}}+\frac{\cos (c+d x) \sin (c+d x)}{a d \sqrt{a+a \sec (c+d x)}}-\frac{\int \frac{-\frac{19 a^3}{2}+\frac{7}{2} a^3 \sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{4 a^4}\\ &=-\frac{\cos (c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{7 \sin (c+d x)}{4 a d \sqrt{a+a \sec (c+d x)}}+\frac{\cos (c+d x) \sin (c+d x)}{a d \sqrt{a+a \sec (c+d x)}}+\frac{19 \int \sqrt{a+a \sec (c+d x)} \, dx}{8 a^2}-\frac{13 \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{4 a}\\ &=-\frac{\cos (c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{7 \sin (c+d x)}{4 a d \sqrt{a+a \sec (c+d x)}}+\frac{\cos (c+d x) \sin (c+d x)}{a d \sqrt{a+a \sec (c+d x)}}-\frac{19 \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 a d}+\frac{13 \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 )}{2 a d}\\ &=\frac{19 \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{4 a^{3/2} d}-\frac{13 \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{\cos (c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{7 \sin (c+d x)}{4 a d \sqrt{a+a \sec (c+d x)}}+\frac{\cos (c+d x) \sin (c+d x)}{a d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 2.83766, size = 197, normalized size = 1.06 \[ -\frac{\sin (2 (c+d x))-\frac{(\cos (c+d x)+1) \tan (c+d x) \sec (c+d x) \left (13 \left (2 \cos ^2(c+d x) \sqrt{1-\sec (c+d x)}-\cos (c+d x) \sqrt{1-\sec (c+d x)}+7 \tanh ^{-1}\left (\sqrt{1-\sec (c+d x)}\right )-4 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{1-\sec (c+d x)}}{\sqrt{2}}\right )\right )-40 \sqrt{1-\sec (c+d x)} \text{Hypergeometric2F1}\left (\frac{1}{2},3,\frac{3}{2},1-\sec (c+d x)\right )\right )}{4 \sqrt{1-\sec (c+d x)}}}{4 d (a (\sec (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sin[2*(c + d*x)] - ((1 + Cos[c + d*x])*(13*(7*ArcTanh[Sqrt[1 - Sec[c + d*x]]] - 4*Sqrt[2]*ArcTanh[Sqrt[1 - S
ec[c + d*x]]/Sqrt[2]] - Cos[c + d*x]*Sqrt[1 - Sec[c + d*x]] + 2*Cos[c + d*x]^2*Sqrt[1 - Sec[c + d*x]]) - 40*Hy
pergeometric2F1[1/2, 3, 3/2, 1 - Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]])*Sec[c + d*x]*Tan[c + d*x])/(4*Sqrt[1 -
Sec[c + d*x]]))/(4*d*(a*(1 + Sec[c + d*x]))^(3/2))

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Maple [B]  time = 0.216, size = 560, normalized size = 3. \begin{align*} \text{result too large to display} \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))^(3/2),x)

[Out]

-1/16/d/a^2*(-1+cos(d*x+c))*(19*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))*cos(d*x+c)^2*sin(d*x+c)+38*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)
+26*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)*cos(d*x+c)^2+19*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)+52*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*cos(d*x+c)^5+26*
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)+20*cos(d*x+c)^4+16*cos(d*x+c)^3-28*cos(d*x+c)^2)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/cos(d*x+c
)/sin(d*x+c)^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{2}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{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))^(3/2),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^2/(a*sec(d*x + c) + a)^(3/2), x)

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Fricas [A]  time = 2.82666, size = 1435, normalized size = 7.76 \begin{align*} \left [-\frac{13 \, \sqrt{2}{\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt{-a} \log \left (-\frac{2 \, \sqrt{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 ) - 3 \, a \cos \left (d x + c\right )^{2} - 2 \, a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 19 \,{\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt{-a} \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 )^{3} - 3 \, \cos \left (d x + c\right )^{2} - 7 \, \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^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}, \frac{13 \, \sqrt{2}{\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt{a} \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 ) - 19 \,{\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt{a} \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 )^{3} - 3 \, \cos \left (d x + c\right )^{2} - 7 \, \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 )}{4 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} 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))^(3/2),x, algorithm="fricas")

[Out]

[-1/8*(13*sqrt(2)*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*sqrt(-a)*log(-(2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c)
+ a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) - 3*a*cos(d*x + c)^2 - 2*a*cos(d*x + c) + a)/(cos(d*x + c)^2 + 2
*cos(d*x + c) + 1)) + 19*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*sqrt(-a)*log((2*a*cos(d*x + c)^2 + 2*sqrt(-a)*s
qrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) - 2
*(2*cos(d*x + c)^3 - 3*cos(d*x + c)^2 - 7*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/
(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d), 1/4*(13*sqrt(2)*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*s
qrt(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))) - 19*(cos(d
*x + c)^2 + 2*cos(d*x + c) + 1)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*s
in(d*x + c))) + (2*cos(d*x + c)^3 - 3*cos(d*x + c)^2 - 7*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))
*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*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 )}}{\left (a \left (\sec{\left (c + d x \right )} + 1\right )\right )^{\frac{3}{2}}}\, 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))**(3/2),x)

[Out]

Integral(cos(c + d*x)**2/(a*(sec(c + d*x) + 1))**(3/2), 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))^(3/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError