3.214 \(\int \frac{A+C \cos ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=199 \[ \frac{(49 A+C) \sin (c+d x)}{16 a^2 d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}-\frac{5 (15 A-C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(13 A-3 C) \sin (c+d x)}{16 a d \sqrt{\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}-\frac{(A+C) \sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}} \]

[Out]

(-5*(15*A - 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)*Sin[c + d*x])/(4*d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^(5/2)) - ((13*A - 3*C)*Sin
[c + d*x])/(16*a*d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^(3/2)) + ((49*A + C)*Sin[c + d*x])/(16*a^2*d*Sqrt[C
os[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])

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Rubi [A]  time = 0.579113, antiderivative size = 199, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.162, Rules used = {3042, 2978, 2984, 12, 2782, 205} \[ \frac{(49 A+C) \sin (c+d x)}{16 a^2 d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}-\frac{5 (15 A-C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(13 A-3 C) \sin (c+d x)}{16 a d \sqrt{\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}-\frac{(A+C) \sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(5/2)),x]

[Out]

(-5*(15*A - 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)*Sin[c + d*x])/(4*d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^(5/2)) - ((13*A - 3*C)*Sin
[c + d*x])/(16*a*d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^(3/2)) + ((49*A + C)*Sin[c + d*x])/(16*a^2*d*Sqrt[C
os[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])

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 2984

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*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^(n + 1))/(f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin
[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e +
f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2
- d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 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)}{\cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx &=-\frac{(A+C) \sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}+\frac{\int \frac{\frac{1}{2} a (9 A+C)-2 a (A-C) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(A+C) \sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac{(13 A-3 C) \sin (c+d x)}{16 a d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{\frac{1}{4} a^2 (49 A+C)-\frac{1}{2} a^2 (13 A-3 C) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{(A+C) \sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac{(13 A-3 C) \sin (c+d x)}{16 a d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac{(49 A+C) \sin (c+d x)}{16 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}+\frac{\int -\frac{5 a^3 (15 A-C)}{8 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{4 a^5}\\ &=-\frac{(A+C) \sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac{(13 A-3 C) \sin (c+d x)}{16 a d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac{(49 A+C) \sin (c+d x)}{16 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}-\frac{(5 (15 A-C)) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{32 a^2}\\ &=-\frac{(A+C) \sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac{(13 A-3 C) \sin (c+d x)}{16 a d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac{(49 A+C) \sin (c+d x)}{16 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}+\frac{(5 (15 A-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{5 (15 A-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) \sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac{(13 A-3 C) \sin (c+d x)}{16 a d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac{(49 A+C) \sin (c+d x)}{16 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 2.59879, size = 211, normalized size = 1.06 \[ \frac{\cos ^5\left (\frac{1}{2} (c+d x)\right ) \left (\frac{\tan \left (\frac{1}{2} (c+d x)\right ) \sec ^3\left (\frac{1}{2} (c+d x)\right ) (10 (17 A+C) \cos (c+d x)+(49 A+C) \cos (2 (c+d x))+113 A+C)}{4 \sqrt{\cos (c+d x)}}-\frac{5 i (15 A-C) e^{\frac{1}{2} i (c+d x)} \sqrt{e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \tanh ^{-1}\left (\frac{1-e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )}{\sqrt{1+e^{2 i (c+d x)}}}\right )}{4 d (a (\cos (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(5/2)),x]

[Out]

(Cos[(c + d*x)/2]^5*(((-5*I)*(15*A - C)*E^((I/2)*(c + d*x))*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))]*Ar
cTanh[(1 - E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])])/Sqrt[1 + E^((2*I)*(c + d*x))] + ((113*A
+ C + 10*(17*A + C)*Cos[c + d*x] + (49*A + C)*Cos[2*(c + d*x)])*Sec[(c + d*x)/2]^3*Tan[(c + d*x)/2])/(4*Sqrt[C
os[c + d*x]])))/(4*d*(a*(1 + Cos[c + d*x]))^(5/2))

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Maple [B]  time = 0.147, size = 479, normalized size = 2.4 \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)/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(5/2),x)

[Out]

1/32/d*(-1+cos(d*x+c))^2*(-98*A*cos(d*x+c)^5*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-268*A*cos(d*x+c)^4*(cos(d*x+c)/
(1+cos(d*x+c)))^(5/2)-136*A*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+75*A*arcsin((-1+cos(d*x+c))/sin(d*x
+c))*2^(1/2)*sin(d*x+c)*cos(d*x+c)^4+204*A*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-5*C*arcsin((-1+cos(d
*x+c))/sin(d*x+c))*2^(1/2)*sin(d*x+c)*cos(d*x+c)^4+75*A*2^(1/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*
cos(d*x+c)^3+234*A*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-5*C*2^(1/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))
*sin(d*x+c)*cos(d*x+c)^3-2*C*cos(d*x+c)^5*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+64*A*(cos(d*x+c)/(1+cos(d*x+c)))^(
5/2)-8*C*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+10*C*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*(
a*(1+cos(d*x+c)))^(1/2)/sin(d*x+c)^5/cos(d*x+c)^(5/2)/(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/a^3

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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)/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 2.08822, size = 648, normalized size = 3.26 \begin{align*} -\frac{5 \, \sqrt{2}{\left ({\left (15 \, A - C\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (15 \, A - C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (15 \, A - C\right )} \cos \left (d x + c\right )^{2} +{\left (15 \, A - C\right )} \cos \left (d x + c\right )\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 (49 \, A + C\right )} \cos \left (d x + c\right )^{2} + 5 \,{\left (17 \, A + C\right )} \cos \left (d x + c\right ) + 32 \, A\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 )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-1/32*(5*sqrt(2)*((15*A - C)*cos(d*x + c)^4 + 3*(15*A - C)*cos(d*x + c)^3 + 3*(15*A - C)*cos(d*x + c)^2 + (15*
A - C)*cos(d*x + 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*((49*A + C)*cos(d*x + c)^2 + 5*(17*A + C)*cos(d*x + c) + 32*A)*sqr
t(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + 3*a^3*
d*cos(d*x + c)^2 + a^3*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((A+C*cos(d*x+c)**2)/cos(d*x+c)**(3/2)/(a+a*cos(d*x+c))**(5/2),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}} \cos \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)/((a*cos(d*x + c) + a)^(5/2)*cos(d*x + c)^(3/2)), x)