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

Optimal. Leaf size=293 \[ \frac{(79 A-259 B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{192 a^2 d (a \cos (c+d x)+a)^{3/2}}+\frac{(2 A-7 B) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{a^{7/2} d}-\frac{7 (7 A-27 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{64 a^3 d \sqrt{a \cos (c+d x)+a}}-\frac{(177 A-637 B) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{64 \sqrt{2} a^{7/2} d}+\frac{(A-B) \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x)}{6 d (a \cos (c+d x)+a)^{7/2}}+\frac{(3 A-7 B) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{16 a d (a \cos (c+d x)+a)^{5/2}} \]

[Out]

((2*A - 7*B)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(a^(7/2)*d) - ((177*A - 637*B)*ArcTan[(S
qrt[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]^(7/2)*Sin[c + d*x])/(6*d*(a + a*Cos[c + d*x])^(7/2)) + ((3*A - 7*B)*Cos[c + d*x]^(5/2)*Sin[c + d
*x])/(16*a*d*(a + a*Cos[c + d*x])^(5/2)) + ((79*A - 259*B)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(192*a^2*d*(a + a*
Cos[c + d*x])^(3/2)) - (7*(7*A - 27*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(64*a^3*d*Sqrt[a + a*Cos[c + d*x]])

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Rubi [A]  time = 1.04402, antiderivative size = 293, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2977, 2983, 2982, 2782, 205, 2774, 216} \[ \frac{(79 A-259 B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{192 a^2 d (a \cos (c+d x)+a)^{3/2}}+\frac{(2 A-7 B) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{a^{7/2} d}-\frac{7 (7 A-27 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{64 a^3 d \sqrt{a \cos (c+d x)+a}}-\frac{(177 A-637 B) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{64 \sqrt{2} a^{7/2} d}+\frac{(A-B) \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x)}{6 d (a \cos (c+d x)+a)^{7/2}}+\frac{(3 A-7 B) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{16 a d (a \cos (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*A - 7*B)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(a^(7/2)*d) - ((177*A - 637*B)*ArcTan[(S
qrt[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]^(7/2)*Sin[c + d*x])/(6*d*(a + a*Cos[c + d*x])^(7/2)) + ((3*A - 7*B)*Cos[c + d*x]^(5/2)*Sin[c + d
*x])/(16*a*d*(a + a*Cos[c + d*x])^(5/2)) + ((79*A - 259*B)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(192*a^2*d*(a + a*
Cos[c + d*x])^(3/2)) - (7*(7*A - 27*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(64*a^3*d*Sqrt[a + a*Cos[c + d*x]])

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 2983

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*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n)/(f*(
m + n + 1)), x] + Dist[1/(b*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*b*c*(
m + n + 1) + B*(a*c*m + b*d*n) + (A*b*d*(m + n + 1) + B*(a*d*m + b*c*n))*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] && GtQ[n, 0] && (I
ntegerQ[n] || EqQ[m + 1/2, 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{7}{2}}(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{7/2}} \, dx &=\frac{(A-B) \cos ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}+\frac{\int \frac{\cos ^{\frac{5}{2}}(c+d x) \left (\frac{7}{2} a (A-B)-a (A-7 B) \cos (c+d x)\right )}{(a+a \cos (c+d x))^{5/2}} \, dx}{6 a^2}\\ &=\frac{(A-B) \cos ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}+\frac{(3 A-7 B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2}}+\frac{\int \frac{\cos ^{\frac{3}{2}}(c+d x) \left (\frac{15}{4} a^2 (3 A-7 B)-\frac{1}{2} a^2 (17 A-77 B) \cos (c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx}{24 a^4}\\ &=\frac{(A-B) \cos ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}+\frac{(3 A-7 B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2}}+\frac{(79 A-259 B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{\sqrt{\cos (c+d x)} \left (\frac{3}{8} a^3 (79 A-259 B)-\frac{21}{4} a^3 (7 A-27 B) \cos (c+d x)\right )}{\sqrt{a+a \cos (c+d x)}} \, dx}{48 a^6}\\ &=\frac{(A-B) \cos ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}+\frac{(3 A-7 B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2}}+\frac{(79 A-259 B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2}}-\frac{7 (7 A-27 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{64 a^3 d \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{-\frac{21}{8} a^4 (7 A-27 B)+24 a^4 (2 A-7 B) \cos (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{48 a^7}\\ &=\frac{(A-B) \cos ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}+\frac{(3 A-7 B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2}}+\frac{(79 A-259 B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2}}-\frac{7 (7 A-27 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{64 a^3 d \sqrt{a+a \cos (c+d x)}}-\frac{(177 A-637 B) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{128 a^3}+\frac{(2 A-7 B) \int \frac{\sqrt{a+a \cos (c+d x)}}{\sqrt{\cos (c+d x)}} \, dx}{2 a^4}\\ &=\frac{(A-B) \cos ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}+\frac{(3 A-7 B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2}}+\frac{(79 A-259 B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2}}-\frac{7 (7 A-27 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{64 a^3 d \sqrt{a+a \cos (c+d x)}}+\frac{(177 A-637 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 A-7 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 A-7 B) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{a^{7/2} d}-\frac{(177 A-637 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{7}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}+\frac{(3 A-7 B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2}}+\frac{(79 A-259 B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2}}-\frac{7 (7 A-27 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{64 a^3 d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 5.51663, size = 396, normalized size = 1.35 \[ \frac{\cos ^7\left (\frac{1}{2} (c+d x)\right ) \left (\frac{1}{4} \sqrt{\cos (c+d x)} \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^5\left (\frac{1}{2} (c+d x)\right ) ((3172 B-724 A) \cos (c+d x)+(1099 B-247 A) \cos (2 (c+d x))-541 A+96 B \cos (3 (c+d x))+2233 B)+\frac{3 \sqrt{2} e^{\frac{1}{2} i (c+d x)} \sqrt{e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \left (i \sqrt{2} (177 A-637 B) \log \left (1+e^{i (c+d x)}\right )-64 i (2 A-7 B) \sinh ^{-1}\left (e^{i (c+d x)}\right )+128 i A \log \left (1+\sqrt{1+e^{2 i (c+d x)}}\right )-177 i \sqrt{2} A \log \left (\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}-e^{i (c+d x)}+1\right )+128 A d x-448 i B \log \left (1+\sqrt{1+e^{2 i (c+d x)}}\right )+637 i \sqrt{2} B \log \left (\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}-e^{i (c+d x)}+1\right )-448 B d x\right )}{\sqrt{1+e^{2 i (c+d x)}}}\right )}{48 d (a (\cos (c+d x)+1))^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^7*((3*Sqrt[2]*E^((I/2)*(c + d*x))*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))]*(128*A*d*x
- 448*B*d*x - (64*I)*(2*A - 7*B)*ArcSinh[E^(I*(c + d*x))] + I*Sqrt[2]*(177*A - 637*B)*Log[1 + E^(I*(c + d*x))
] + (128*I)*A*Log[1 + Sqrt[1 + E^((2*I)*(c + d*x))]] - (448*I)*B*Log[1 + Sqrt[1 + E^((2*I)*(c + d*x))]] - (177
*I)*Sqrt[2]*A*Log[1 - E^(I*(c + d*x)) + Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))]] + (637*I)*Sqrt[2]*B*Log[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]]*(-
541*A + 2233*B + (-724*A + 3172*B)*Cos[c + d*x] + (-247*A + 1099*B)*Cos[2*(c + d*x)] + 96*B*Cos[3*(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 = 0.665, size = 887, 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)^(7/2)*(A+B*cos(d*x+c))/(a+cos(d*x+c)*a)^(7/2),x)

[Out]

-1/384/d*cos(d*x+c)^(7/2)*(a*(1+cos(d*x+c)))^(1/2)*(-1+cos(d*x+c))^7*(494*A*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*
cos(d*x+c)^4+531*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*2^(1/2)*cos(d*x+c)^3*sin(d*x+c)+724*A*(cos(d*x+c)/(1+cos
(d*x+c)))^(3/2)*cos(d*x+c)^3-384*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^5-1911*B*arcsin((-1+cos(d*x+c)
)/sin(d*x+c))*2^(1/2)*cos(d*x+c)^3*sin(d*x+c)+768*A*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*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)-200*A*
(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*cos(d*x+c)^2-1814*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^4-2688*B*ar
ctan(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)-3822*B*arcsin((-1+cos(d*
x+c))/sin(d*x+c))*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)+1536*A*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/c
os(d*x+c))*cos(d*x+c)^2*sin(d*x+c)+531*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*2^(1/2)*cos(d*x+c)-724*
A*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)-686*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^3-5376*B*arc
tan(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)-1911*B*arcsin((-1+cos(d*x
+c))/sin(d*x+c))*sin(d*x+c)*2^(1/2)*cos(d*x+c)+768*A*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*A*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+1750*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*
cos(d*x+c)^2-2688*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)+1134
*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c))/sin(d*x+c)^15/(cos(d*x+c)/(1+cos(d*x+c)))^(9/2)/a^4

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

[Out]

Timed out

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Fricas [A]  time = 141.259, size = 1018, normalized size = 3.47 \begin{align*} \frac{3 \, \sqrt{2}{\left ({\left (177 \, A - 637 \, B\right )} \cos \left (d x + c\right )^{4} + 4 \,{\left (177 \, A - 637 \, B\right )} \cos \left (d x + c\right )^{3} + 6 \,{\left (177 \, A - 637 \, B\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left (177 \, A - 637 \, B\right )} \cos \left (d x + c\right ) + 177 \, A - 637 \, 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 (192 \, B \cos \left (d x + c\right )^{3} -{\left (247 \, A - 1099 \, B\right )} \cos \left (d x + c\right )^{2} - 2 \,{\left (181 \, A - 721 \, B\right )} \cos \left (d x + c\right ) - 147 \, A + 567 \, B\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 384 \,{\left ({\left (2 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{4} + 4 \,{\left (2 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{3} + 6 \,{\left (2 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left (2 \, A - 7 \, B\right )} \cos \left (d x + c\right ) + 2 \, A - 7 \, 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)^(7/2)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/384*(3*sqrt(2)*((177*A - 637*B)*cos(d*x + c)^4 + 4*(177*A - 637*B)*cos(d*x + c)^3 + 6*(177*A - 637*B)*cos(d*
x + c)^2 + 4*(177*A - 637*B)*cos(d*x + c) + 177*A - 637*B)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqr
t(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) + 2*(192*B*cos(d*x + c)^3 - (247*A - 1099*B)*cos(d*x + c)^2 - 2*(181*A
- 721*B)*cos(d*x + c) - 147*A + 567*B)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c) - 384*((2*A -
7*B)*cos(d*x + c)^4 + 4*(2*A - 7*B)*cos(d*x + c)^3 + 6*(2*A - 7*B)*cos(d*x + c)^2 + 4*(2*A - 7*B)*cos(d*x + c
) + 2*A - 7*B)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(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)**(7/2)*(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{7}{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)^(7/2)*(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)^(7/2)/(a*cos(d*x + c) + a)^(7/2), x)