3.209 \(\int \frac{\cos ^{\frac{5}{2}}(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=241 \[ \frac{(5 A-49 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{64 a^2 d (a \cos (c+d x)+a)^{3/2}}+\frac{(5 A-177 B) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{64 \sqrt{2} a^{7/2} d}+\frac{2 B \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{a^{7/2} d}+\frac{(A-B) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{6 d (a \cos (c+d x)+a)^{7/2}}+\frac{(5 A-17 B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{48 a d (a \cos (c+d x)+a)^{5/2}} \]

[Out]

(2*B*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(a^(7/2)*d) + ((5*A - 177*B)*ArcTan[(Sqrt[a]*Sin
[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(64*Sqrt[2]*a^(7/2)*d) + ((A - B)*Cos[c + d
*x]^(5/2)*Sin[c + d*x])/(6*d*(a + a*Cos[c + d*x])^(7/2)) + ((5*A - 17*B)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(48*
a*d*(a + a*Cos[c + d*x])^(5/2)) + ((5*A - 49*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(64*a^2*d*(a + a*Cos[c + d*x]
)^(3/2))

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Rubi [A]  time = 0.765032, antiderivative size = 241, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {2977, 2982, 2782, 205, 2774, 216} \[ \frac{(5 A-49 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{64 a^2 d (a \cos (c+d x)+a)^{3/2}}+\frac{(5 A-177 B) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{64 \sqrt{2} a^{7/2} d}+\frac{2 B \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{a^{7/2} d}+\frac{(A-B) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{6 d (a \cos (c+d x)+a)^{7/2}}+\frac{(5 A-17 B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{48 a d (a \cos (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^(5/2)*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^(7/2),x]

[Out]

(2*B*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(a^(7/2)*d) + ((5*A - 177*B)*ArcTan[(Sqrt[a]*Sin
[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(64*Sqrt[2]*a^(7/2)*d) + ((A - B)*Cos[c + d
*x]^(5/2)*Sin[c + d*x])/(6*d*(a + a*Cos[c + d*x])^(7/2)) + ((5*A - 17*B)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(48*
a*d*(a + a*Cos[c + d*x])^(5/2)) + ((5*A - 49*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(64*a^2*d*(a + a*Cos[c + d*x]
)^(3/2))

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{\cos ^{\frac{5}{2}}(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{7/2}} \, dx &=\frac{(A-B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}+\frac{\int \frac{\cos ^{\frac{3}{2}}(c+d x) \left (\frac{5}{2} a (A-B)+6 a B \cos (c+d x)\right )}{(a+a \cos (c+d x))^{5/2}} \, dx}{6 a^2}\\ &=\frac{(A-B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}+\frac{(5 A-17 B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2}}+\frac{\int \frac{\sqrt{\cos (c+d x)} \left (\frac{3}{4} a^2 (5 A-17 B)+24 a^2 B \cos (c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx}{24 a^4}\\ &=\frac{(A-B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}+\frac{(5 A-17 B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2}}+\frac{(5 A-49 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{64 a^2 d (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{\frac{3}{8} a^3 (5 A-49 B)+48 a^3 B \cos (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{48 a^6}\\ &=\frac{(A-B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}+\frac{(5 A-17 B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2}}+\frac{(5 A-49 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{64 a^2 d (a+a \cos (c+d x))^{3/2}}+\frac{(5 A-177 B) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{128 a^3}+\frac{B \int \frac{\sqrt{a+a \cos (c+d x)}}{\sqrt{\cos (c+d x)}} \, dx}{a^4}\\ &=\frac{(A-B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}+\frac{(5 A-17 B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2}}+\frac{(5 A-49 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{64 a^2 d (a+a \cos (c+d x))^{3/2}}-\frac{(5 A-177 B) \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 )}{64 a^2 d}-\frac{(2 B) \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^4 d}\\ &=\frac{2 B \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{a^{7/2} d}+\frac{(5 A-177 B) \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 )}{64 \sqrt{2} a^{7/2} d}+\frac{(A-B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}+\frac{(5 A-17 B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2}}+\frac{(5 A-49 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{64 a^2 d (a+a \cos (c+d x))^{3/2}}\\ \end{align*}

Mathematica [C]  time = 3.03985, size = 266, normalized size = 1.1 \[ \frac{\cos ^7\left (\frac{1}{2} (c+d x)\right ) \left (\frac{1}{4} \sqrt{\cos (c+d x)} \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^5\left (\frac{1}{2} (c+d x)\right ) (4 (25 A-181 B) \cos (c+d x)+(67 A-247 B) \cos (2 (c+d x))+97 A-541 B)-\frac{3 i \sqrt{2} e^{\frac{1}{2} i (c+d x)} \sqrt{e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \left (-\sqrt{2} (5 A-177 B) \tanh ^{-1}\left (\frac{1-e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )+128 B \sinh ^{-1}\left (e^{i (c+d x)}\right )-128 B \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )\right )}{\sqrt{1+e^{2 i (c+d x)}}}\right )}{48 d (a (\cos (c+d x)+1))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^(5/2)*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^(7/2),x]

[Out]

(Cos[(c + d*x)/2]^7*(((-3*I)*Sqrt[2]*E^((I/2)*(c + d*x))*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))]*(128*
B*ArcSinh[E^(I*(c + d*x))] - Sqrt[2]*(5*A - 177*B)*ArcTanh[(1 - E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c
+ d*x))])] - 128*B*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]]))/Sqrt[1 + E^((2*I)*(c + d*x))] + (Sqrt[Cos[c + d*x
]]*(97*A - 541*B + 4*(25*A - 181*B)*Cos[c + d*x] + (67*A - 247*B)*Cos[2*(c + d*x)])*Sec[(c + d*x)/2]^5*Tan[(c
+ d*x)/2])/4))/(48*d*(a*(1 + Cos[c + d*x]))^(7/2))

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Maple [B]  time = 0.635, size = 703, 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(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c))/(a+cos(d*x+c)*a)^(7/2),x)

[Out]

-1/384/d*(a*(1+cos(d*x+c)))^(1/2)*(-1+cos(d*x+c))^6*cos(d*x+c)^(5/2)*(134*A*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*
cos(d*x+c)^4+15*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*2^(1/2)*cos(d*x+c)^3*sin(d*x+c)+100*A*(cos(d*x+c)/(1+cos(
d*x+c)))^(3/2)*cos(d*x+c)^3-531*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*2^(1/2)*cos(d*x+c)^3*sin(d*x+c)+30*A*arcs
in((-1+cos(d*x+c))/sin(d*x+c))*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)-104*A*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*cos(d*x
+c)^2-494*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^4-768*B*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))
^(1/2)/cos(d*x+c))*cos(d*x+c)^3*sin(d*x+c)-1062*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*2^(1/2)*cos(d*x+c)^2*sin(
d*x+c)+15*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*2^(1/2)*cos(d*x+c)-100*A*cos(d*x+c)*(cos(d*x+c)/(1+c
os(d*x+c)))^(3/2)-230*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^3-1536*B*arctan(sin(d*x+c)*(cos(d*x+c)/(1
+cos(d*x+c)))^(1/2)/cos(d*x+c))*cos(d*x+c)^2*sin(d*x+c)-531*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*2^
(1/2)*cos(d*x+c)-30*A*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+430*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2-7
68*B*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))*sin(d*x+c)*cos(d*x+c)+294*B*(cos(d*x+c)/(
1+cos(d*x+c)))^(1/2)*cos(d*x+c))/sin(d*x+c)^13/(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)/a^4

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac{5}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)*cos(d*x + c)^(5/2)/(a*cos(d*x + c) + a)^(7/2), x)

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Fricas [A]  time = 98.4538, size = 905, normalized size = 3.76 \begin{align*} -\frac{3 \, \sqrt{2}{\left ({\left (5 \, A - 177 \, B\right )} \cos \left (d x + c\right )^{4} + 4 \,{\left (5 \, A - 177 \, B\right )} \cos \left (d x + c\right )^{3} + 6 \,{\left (5 \, A - 177 \, B\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left (5 \, A - 177 \, B\right )} \cos \left (d x + c\right ) + 5 \, A - 177 \, B\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 (67 \, A - 247 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (25 \, A - 181 \, B\right )} \cos \left (d x + c\right ) + 15 \, A - 147 \, B\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 768 \,{\left (B \cos \left (d x + c\right )^{4} + 4 \, B \cos \left (d x + c\right )^{3} + 6 \, B \cos \left (d x + c\right )^{2} + 4 \, B \cos \left (d x + c\right ) + B\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 )}{384 \,{\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

-1/384*(3*sqrt(2)*((5*A - 177*B)*cos(d*x + c)^4 + 4*(5*A - 177*B)*cos(d*x + c)^3 + 6*(5*A - 177*B)*cos(d*x + c
)^2 + 4*(5*A - 177*B)*cos(d*x + c) + 5*A - 177*B)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x
+ c))/(sqrt(a)*sin(d*x + c))) - 2*((67*A - 247*B)*cos(d*x + c)^2 + 2*(25*A - 181*B)*cos(d*x + c) + 15*A - 147
*B)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c) + 768*(B*cos(d*x + c)^4 + 4*B*cos(d*x + c)^3 + 6*
B*cos(d*x + c)^2 + 4*B*cos(d*x + c) + B)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*s
in(d*x + c))))/(a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c)
+ a^4*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(cos(d*x+c)**(5/2)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac{5}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*cos(d*x + c)^(5/2)/(a*cos(d*x + c) + a)^(7/2), x)