3.213 \(\int \frac{A+C \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=154 \[ \frac{(19 A+3 C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(9 A-7 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac{(A+C) \sin (c+d x) \sqrt{\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}} \]

[Out]

((19*A + 3*C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(16*Sqrt[2
]*a^(5/2)*d) - ((A + C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) - ((9*A - 7*C)*Sqrt[
Cos[c + d*x]]*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2))

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Rubi [A]  time = 0.398808, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.135, Rules used = {3042, 2978, 12, 2782, 205} \[ \frac{(19 A+3 C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(9 A-7 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac{(A+C) \sin (c+d x) \sqrt{\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + C*Cos[c + d*x]^2)/(Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^(5/2)),x]

[Out]

((19*A + 3*C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(16*Sqrt[2
]*a^(5/2)*d) - ((A + C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) - ((9*A - 7*C)*Sqrt[
Cos[c + d*x]]*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2))

Rule 3042

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

Rule 2978

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
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 2782

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> D
ist[(-2*a)/f, Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, (b*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c
+ d*Sin[e + f*x]])], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 -
d^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{A+C \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx &=-\frac{(A+C) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{\int \frac{\frac{1}{2} a (7 A-C)-a (A-3 C) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(A+C) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(9 A-7 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{a^2 (19 A+3 C)}{4 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{(A+C) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(9 A-7 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(19 A+3 C) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{32 a^2}\\ &=-\frac{(A+C) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(9 A-7 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac{(19 A+3 C) \operatorname{Subst}\left (\int \frac{1}{2 a^2+a x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right )}{16 a d}\\ &=\frac{(19 A+3 C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(A+C) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(9 A-7 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}\\ \end{align*}

Mathematica [C]  time = 1.4649, size = 200, normalized size = 1.3 \[ \frac{\cos ^5\left (\frac{1}{2} (c+d x)\right ) \left (-\frac{1}{2} \sqrt{\cos (c+d x)} \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^3\left (\frac{1}{2} (c+d x)\right ) ((9 A-7 C) \cos (c+d x)+13 A-3 C)+\frac{i (19 A+3 C) e^{\frac{1}{2} i (c+d x)} \sqrt{e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \tanh ^{-1}\left (\frac{1-e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )}{\sqrt{1+e^{2 i (c+d x)}}}\right )}{4 d (a (\cos (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + C*Cos[c + d*x]^2)/(Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^(5/2)),x]

[Out]

(Cos[(c + d*x)/2]^5*((I*(19*A + 3*C)*E^((I/2)*(c + d*x))*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))]*ArcTa
nh[(1 - E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])])/Sqrt[1 + E^((2*I)*(c + d*x))] - (Sqrt[Cos[c
+ d*x]]*(13*A - 3*C + (9*A - 7*C)*Cos[c + d*x])*Sec[(c + d*x)/2]^3*Tan[(c + d*x)/2])/2))/(4*d*(a*(1 + Cos[c +
d*x]))^(5/2))

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Maple [B]  time = 0.13, size = 449, normalized size = 2.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(5/2)/cos(d*x+c)^(1/2),x)

[Out]

-1/32/d*(a*(1+cos(d*x+c)))^(1/2)*(-1+cos(d*x+c))^3*(18*A*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+44*A*c
os(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+8*A*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-19*A*2^(1/2)*
arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)^3-44*A*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-3
*C*2^(1/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)^3-19*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*
2^(1/2)*sin(d*x+c)*cos(d*x+c)^2-26*A*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-14*C*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*
x+c)))^(1/2)-3*C*arcsin((-1+cos(d*x+c))/sin(d*x+c))*2^(1/2)*sin(d*x+c)*cos(d*x+c)^2+8*C*cos(d*x+c)^3*(cos(d*x+
c)/(1+cos(d*x+c)))^(1/2)+6*C*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))/(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)
/a^3/cos(d*x+c)^(1/2)/sin(d*x+c)^7

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Maxima [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((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(5/2)/cos(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 2.33245, size = 581, normalized size = 3.77 \begin{align*} \frac{\sqrt{2}{\left ({\left (19 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (19 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (19 \, A + 3 \, C\right )} \cos \left (d x + c\right ) + 19 \, A + 3 \, C\right )} \sqrt{a} \arctan \left (\frac{\sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \,{\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )}}\right ) - 2 \,{\left ({\left (9 \, A - 7 \, C\right )} \cos \left (d x + c\right ) + 13 \, A - 3 \, C\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{32 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(5/2)/cos(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

1/32*(sqrt(2)*((19*A + 3*C)*cos(d*x + c)^3 + 3*(19*A + 3*C)*cos(d*x + c)^2 + 3*(19*A + 3*C)*cos(d*x + c) + 19*
A + 3*C)*sqrt(a)*arctan(1/2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*
x + c)^2 + a*cos(d*x + c))) - 2*((9*A - 7*C)*cos(d*x + c) + 13*A - 3*C)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x
+ c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

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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((A+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))**(5/2)/cos(d*x+c)**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \cos \left (d x + c\right )^{2} + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \sqrt{\cos \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(5/2)/cos(d*x+c)^(1/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)/((a*cos(d*x + c) + a)^(5/2)*sqrt(cos(d*x + c))), x)