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

Optimal. Leaf size=192 \[ \frac{(5 A-43 C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{2 C \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{a^{5/2} d}-\frac{(A+C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}+\frac{(5 A-11 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{16 a d (a \cos (c+d x)+a)^{3/2}} \]

[Out]

(2*C*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(a^(5/2)*d) + ((5*A - 43*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)*Cos[c + d*
x]^(3/2)*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) + ((5*A - 11*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 = 0.579869, antiderivative size = 192, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.189, Rules used = {3042, 2977, 2982, 2782, 205, 2774, 216} \[ \frac{(5 A-43 C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{2 C \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{a^{5/2} d}-\frac{(A+C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}+\frac{(5 A-11 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{16 a d (a \cos (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*C*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(a^(5/2)*d) + ((5*A - 43*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)*Cos[c + d*
x]^(3/2)*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) + ((5*A - 11*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 2977

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^n)/(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n -
1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B}, 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 2982

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin
[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[(A*b - a*B)/b, Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*
x]]), x], x] + Dist[B/b, Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*x]], x], x] /; FreeQ[{a, b, c, d, e
, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

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]

Rule 2774

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2/f, Su
bst[Int[1/Sqrt[1 - x^2/a], x], x, (b*Cos[e + f*x])/Sqrt[a + b*Sin[e + f*x]]], x] /; FreeQ[{a, b, d, e, f}, x]
&& EqQ[a^2 - b^2, 0] && EqQ[d, a/b]

Rule 216

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

Rubi steps

\begin{align*} \int \frac{\sqrt{\cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{5/2}} \, dx &=-\frac{(A+C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{\int \frac{\sqrt{\cos (c+d x)} \left (\frac{1}{2} a (5 A-3 C)+4 a C \cos (c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(A+C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{(5 A-11 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{\frac{1}{4} a^2 (5 A-11 C)+8 a^2 C \cos (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{(A+C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{(5 A-11 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(5 A-43 C) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{32 a^2}+\frac{C \int \frac{\sqrt{a+a \cos (c+d x)}}{\sqrt{\cos (c+d x)}} \, dx}{a^3}\\ &=-\frac{(A+C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{(5 A-11 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac{(5 A-43 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{(2 C) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{a}}} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{a^3 d}\\ &=\frac{2 C \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{a^{5/2} d}+\frac{(5 A-43 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) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{(5 A-11 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 = 1.91051, size = 244, normalized size = 1.27 \[ \frac{\cos ^5\left (\frac{1}{2} (c+d x)\right ) \left (\sqrt{\cos (c+d x)} \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^3\left (\frac{1}{2} (c+d x)\right ) ((A-15 C) \cos (c+d x)+5 A-11 C)-\frac{i \sqrt{2} e^{\frac{1}{2} i (c+d x)} \sqrt{e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \left (-\sqrt{2} (5 A-43 C) \tanh ^{-1}\left (\frac{1-e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )+32 C \sinh ^{-1}\left (e^{i (c+d x)}\right )-32 C \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )\right )}{\sqrt{1+e^{2 i (c+d x)}}}\right )}{8 d (a (\cos (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^5*(((-I)*Sqrt[2]*E^((I/2)*(c + d*x))*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))]*(32*C*A
rcSinh[E^(I*(c + d*x))] - Sqrt[2]*(5*A - 43*C)*ArcTanh[(1 - E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d
*x))])] - 32*C*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]]))/Sqrt[1 + E^((2*I)*(c + d*x))] + Sqrt[Cos[c + d*x]]*(5*
A - 11*C + (A - 15*C)*Cos[c + d*x])*Sec[(c + d*x)/2]^3*Tan[(c + d*x)/2]))/(8*d*(a*(1 + Cos[c + d*x]))^(5/2))

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Maple [B]  time = 0.195, size = 553, 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)*cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(5/2),x)

[Out]

-1/32/d*cos(d*x+c)^(1/2)*(a*(1+cos(d*x+c)))^(1/2)*(-1+cos(d*x+c))^4*(2*A*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c
)))^(5/2)+12*A*cos(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)+5*A*2^(1/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)^3-12*A*cos(d*x+c)*(cos(d*x+c)/(1+cos(d
*x+c)))^(5/2)-43*C*2^(1/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)^3+5*A*arcsin((-1+cos(d*x+c
))/sin(d*x+c))*2^(1/2)*sin(d*x+c)*cos(d*x+c)^2-10*A*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-43*C*arcsin((-1+cos(d*x+
c))/sin(d*x+c))*2^(1/2)*sin(d*x+c)*cos(d*x+c)^2-64*C*sin(d*x+c)*cos(d*x+c)^3*arctan(sin(d*x+c)*(cos(d*x+c)/(1+
cos(d*x+c)))^(1/2)/cos(d*x+c))-30*C*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-64*C*sin(d*x+c)*cos(d*x+c)^
2*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))+8*C*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))
^(1/2)+22*C*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))/a^3/(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)/sin(d*x+c)^9

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sqrt{\cos \left (d x + c\right )}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{5}{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)^(1/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 97.3025, size = 733, normalized size = 3.82 \begin{align*} -\frac{\sqrt{2}{\left ({\left (5 \, A - 43 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (5 \, A - 43 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (5 \, A - 43 \, C\right )} \cos \left (d x + c\right ) + 5 \, A - 43 \, C\right )} \sqrt{a} \arctan \left (\frac{\sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )}}{\sqrt{a} \sin \left (d x + c\right )}\right ) - 2 \,{\left ({\left (A - 15 \, C\right )} \cos \left (d x + c\right ) + 5 \, A - 11 \, C\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 64 \,{\left (C \cos \left (d x + c\right )^{3} + 3 \, C \cos \left (d x + c\right )^{2} + 3 \, C \cos \left (d x + c\right ) + C\right )} \sqrt{a} \arctan \left (\frac{\sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )}}{\sqrt{a} \sin \left (d x + c\right )}\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)*cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-1/32*(sqrt(2)*((5*A - 43*C)*cos(d*x + c)^3 + 3*(5*A - 43*C)*cos(d*x + c)^2 + 3*(5*A - 43*C)*cos(d*x + c) + 5*
A - 43*C)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) - 2*((A -
15*C)*cos(d*x + c) + 5*A - 11*C)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c) + 64*(C*cos(d*x + c
)^3 + 3*C*cos(d*x + c)^2 + 3*C*cos(d*x + c) + C)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(s
qrt(a)*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)*cos(d*x+c)**(1/2)/(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{{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sqrt{\cos \left (d x + c\right )}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{5}{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)^(1/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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