3.115 \(\int \frac{\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=261 \[ -\frac{(85 A-157 B) \sin (c+d x) \cos ^2(c+d x)}{80 a^2 d \sqrt{a \cos (c+d x)+a}}+\frac{(475 A-787 B) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{240 a^3 d}-\frac{(985 A-1729 B) \sin (c+d x)}{120 a^2 d \sqrt{a \cos (c+d x)+a}}+\frac{(163 A-283 B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{(A-B) \sin (c+d x) \cos ^4(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}+\frac{(13 A-21 B) \sin (c+d x) \cos ^3(c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}} \]

[Out]

((163*A - 283*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) +
((A - B)*Cos[c + d*x]^4*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) + ((13*A - 21*B)*Cos[c + d*x]^3*Sin[c +
d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2)) - ((985*A - 1729*B)*Sin[c + d*x])/(120*a^2*d*Sqrt[a + a*Cos[c + d*x
]]) - ((85*A - 157*B)*Cos[c + d*x]^2*Sin[c + d*x])/(80*a^2*d*Sqrt[a + a*Cos[c + d*x]]) + ((475*A - 787*B)*Sqrt
[a + a*Cos[c + d*x]]*Sin[c + d*x])/(240*a^3*d)

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Rubi [A]  time = 0.798846, antiderivative size = 261, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {2977, 2983, 2968, 3023, 2751, 2649, 206} \[ -\frac{(85 A-157 B) \sin (c+d x) \cos ^2(c+d x)}{80 a^2 d \sqrt{a \cos (c+d x)+a}}+\frac{(475 A-787 B) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{240 a^3 d}-\frac{(985 A-1729 B) \sin (c+d x)}{120 a^2 d \sqrt{a \cos (c+d x)+a}}+\frac{(163 A-283 B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{(A-B) \sin (c+d x) \cos ^4(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}+\frac{(13 A-21 B) \sin (c+d x) \cos ^3(c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^(5/2),x]

[Out]

((163*A - 283*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) +
((A - B)*Cos[c + d*x]^4*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) + ((13*A - 21*B)*Cos[c + d*x]^3*Sin[c +
d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2)) - ((985*A - 1729*B)*Sin[c + d*x])/(120*a^2*d*Sqrt[a + a*Cos[c + d*x
]]) - ((85*A - 157*B)*Cos[c + d*x]^2*Sin[c + d*x])/(80*a^2*d*Sqrt[a + a*Cos[c + d*x]]) + ((475*A - 787*B)*Sqrt
[a + a*Cos[c + d*x]]*Sin[c + d*x])/(240*a^3*d)

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 2968

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

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2751

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

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 206

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

Rubi steps

\begin{align*} \int \frac{\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx &=\frac{(A-B) \cos ^4(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{\int \frac{\cos ^3(c+d x) \left (4 a (A-B)-\frac{1}{2} a (5 A-13 B) \cos (c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=\frac{(A-B) \cos ^4(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{(13 A-21 B) \cos ^3(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{\cos ^2(c+d x) \left (\frac{3}{2} a^2 (13 A-21 B)-\frac{1}{4} a^2 (85 A-157 B) \cos (c+d x)\right )}{\sqrt{a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=\frac{(A-B) \cos ^4(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{(13 A-21 B) \cos ^3(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac{(85 A-157 B) \cos ^2(c+d x) \sin (c+d x)}{80 a^2 d \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{\cos (c+d x) \left (-\frac{1}{2} a^3 (85 A-157 B)+\frac{1}{8} a^3 (475 A-787 B) \cos (c+d x)\right )}{\sqrt{a+a \cos (c+d x)}} \, dx}{20 a^5}\\ &=\frac{(A-B) \cos ^4(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{(13 A-21 B) \cos ^3(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac{(85 A-157 B) \cos ^2(c+d x) \sin (c+d x)}{80 a^2 d \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{-\frac{1}{2} a^3 (85 A-157 B) \cos (c+d x)+\frac{1}{8} a^3 (475 A-787 B) \cos ^2(c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{20 a^5}\\ &=\frac{(A-B) \cos ^4(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{(13 A-21 B) \cos ^3(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac{(85 A-157 B) \cos ^2(c+d x) \sin (c+d x)}{80 a^2 d \sqrt{a+a \cos (c+d x)}}+\frac{(475 A-787 B) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{240 a^3 d}+\frac{\int \frac{\frac{1}{16} a^4 (475 A-787 B)-\frac{1}{8} a^4 (985 A-1729 B) \cos (c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{30 a^6}\\ &=\frac{(A-B) \cos ^4(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{(13 A-21 B) \cos ^3(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac{(985 A-1729 B) \sin (c+d x)}{120 a^2 d \sqrt{a+a \cos (c+d x)}}-\frac{(85 A-157 B) \cos ^2(c+d x) \sin (c+d x)}{80 a^2 d \sqrt{a+a \cos (c+d x)}}+\frac{(475 A-787 B) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{240 a^3 d}+\frac{(163 A-283 B) \int \frac{1}{\sqrt{a+a \cos (c+d x)}} \, dx}{32 a^2}\\ &=\frac{(A-B) \cos ^4(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{(13 A-21 B) \cos ^3(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac{(985 A-1729 B) \sin (c+d x)}{120 a^2 d \sqrt{a+a \cos (c+d x)}}-\frac{(85 A-157 B) \cos ^2(c+d x) \sin (c+d x)}{80 a^2 d \sqrt{a+a \cos (c+d x)}}+\frac{(475 A-787 B) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{240 a^3 d}-\frac{(163 A-283 B) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{16 a^2 d}\\ &=\frac{(163 A-283 B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a+a \cos (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{(A-B) \cos ^4(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{(13 A-21 B) \cos ^3(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac{(985 A-1729 B) \sin (c+d x)}{120 a^2 d \sqrt{a+a \cos (c+d x)}}-\frac{(85 A-157 B) \cos ^2(c+d x) \sin (c+d x)}{80 a^2 d \sqrt{a+a \cos (c+d x)}}+\frac{(475 A-787 B) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{240 a^3 d}\\ \end{align*}

Mathematica [A]  time = 1.46063, size = 139, normalized size = 0.53 \[ \frac{\tan \left (\frac{1}{2} (c+d x)\right ) (-5 (479 A-887 B) \cos (c+d x)+(832 B-400 A) \cos (2 (c+d x))+40 A \cos (3 (c+d x))-1895 A-40 B \cos (3 (c+d x))+12 B \cos (4 (c+d x))+3491 B)+30 (163 A-283 B) \cos ^3\left (\frac{1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{240 a d (a (\cos (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^(5/2),x]

[Out]

(30*(163*A - 283*B)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^3 + (-1895*A + 3491*B - 5*(479*A - 887*B)*Cos[c
+ d*x] + (-400*A + 832*B)*Cos[2*(c + d*x)] + 40*A*Cos[3*(c + d*x)] - 40*B*Cos[3*(c + d*x)] + 12*B*Cos[4*(c +
d*x)])*Tan[(c + d*x)/2])/(240*a*d*(a*(1 + Cos[c + d*x]))^(3/2))

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Maple [B]  time = 2.654, size = 467, normalized size = 1.8 \begin{align*}{\frac{1}{480\,d}\sqrt{a \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 768\,B\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}+640\,A\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}-2176\,B\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}+2445\,A\ln \left ( 2\,{\frac{2\,\sqrt{a}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+2\,a}{\cos \left ( 1/2\,dx+c/2 \right ) }} \right ) \sqrt{2} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}a-4245\,B\ln \left ( 2\,{\frac{2\,\sqrt{a}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+2\,a}{\cos \left ( 1/2\,dx+c/2 \right ) }} \right ) \sqrt{2} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}a-2560\,A\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+5248\,B\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-435\,A\sqrt{a}\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+555\,B\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+30\,A\sqrt{a}\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}-30\,B\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a} \right ) \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}{a}^{-{\frac{7}{2}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{ \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(A+B*cos(d*x+c))/(a+cos(d*x+c)*a)^(5/2),x)

[Out]

1/480*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(768*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^
8+640*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^6-2176*B*2^(1/2)*(a*sin(1/2*d*x+1/2*
c)^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^6+2445*A*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*
x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^4*a-4245*B*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*
x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^4*a-2560*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2
*c)^4+5248*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^4-435*A*a^(1/2)*2^(1/2)*(a*sin(
1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)^2+555*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*cos(1/2*d*x+
1/2*c)^2+30*A*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-30*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/
2))/cos(1/2*d*x+1/2*c)^3/a^(7/2)/sin(1/2*d*x+1/2*c)/(cos(1/2*d*x+1/2*c)^2*a)^(1/2)/d

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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)^4*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.72255, size = 741, normalized size = 2.84 \begin{align*} -\frac{15 \, \sqrt{2}{\left ({\left (163 \, A - 283 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (163 \, A - 283 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (163 \, A - 283 \, B\right )} \cos \left (d x + c\right ) + 163 \, A - 283 \, B\right )} \sqrt{a} \log \left (-\frac{a \cos \left (d x + c\right )^{2} + 2 \, \sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \,{\left (96 \, B \cos \left (d x + c\right )^{4} + 160 \,{\left (A - B\right )} \cos \left (d x + c\right )^{3} - 32 \,{\left (25 \, A - 49 \, B\right )} \cos \left (d x + c\right )^{2} - 5 \,{\left (503 \, A - 911 \, B\right )} \cos \left (d x + c\right ) - 1495 \, A + 2671 \, B\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{960 \,{\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(cos(d*x+c)^4*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-1/960*(15*sqrt(2)*((163*A - 283*B)*cos(d*x + c)^3 + 3*(163*A - 283*B)*cos(d*x + c)^2 + 3*(163*A - 283*B)*cos(
d*x + c) + 163*A - 283*B)*sqrt(a)*log(-(a*cos(d*x + c)^2 + 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sin(d*x
+ c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - 4*(96*B*cos(d*x + c)^4 + 160*(A - B)*c
os(d*x + c)^3 - 32*(25*A - 49*B)*cos(d*x + c)^2 - 5*(503*A - 911*B)*cos(d*x + c) - 1495*A + 2671*B)*sqrt(a*cos
(d*x + c) + 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(cos(d*x+c)**4*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 2.28102, size = 347, normalized size = 1.33 \begin{align*} -\frac{\frac{15 \,{\left (163 \, \sqrt{2} A - 283 \, \sqrt{2} B\right )} \log \left ({\left | -\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} \right |}\right )}{a^{\frac{5}{2}}} - \frac{{\left ({\left ({\left (15 \,{\left (\frac{2 \,{\left (\sqrt{2} A a^{2} - \sqrt{2} B a^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{a^{2}} - \frac{21 \, \sqrt{2} A a^{2} - 29 \, \sqrt{2} B a^{2}}{a^{2}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - \frac{3685 \, \sqrt{2} A a^{2} - 6733 \, \sqrt{2} B a^{2}}{a^{2}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - \frac{5 \,{\left (1133 \, \sqrt{2} A a^{2} - 1973 \, \sqrt{2} B a^{2}\right )}}{a^{2}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - \frac{15 \,{\left (155 \, \sqrt{2} A a^{2} - 291 \, \sqrt{2} B a^{2}\right )}}{a^{2}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{\frac{5}{2}}}}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

-1/480*(15*(163*sqrt(2)*A - 283*sqrt(2)*B)*log(abs(-sqrt(a)*tan(1/2*d*x + 1/2*c) + sqrt(a*tan(1/2*d*x + 1/2*c)
^2 + a)))/a^(5/2) - (((15*(2*(sqrt(2)*A*a^2 - sqrt(2)*B*a^2)*tan(1/2*d*x + 1/2*c)^2/a^2 - (21*sqrt(2)*A*a^2 -
29*sqrt(2)*B*a^2)/a^2)*tan(1/2*d*x + 1/2*c)^2 - (3685*sqrt(2)*A*a^2 - 6733*sqrt(2)*B*a^2)/a^2)*tan(1/2*d*x + 1
/2*c)^2 - 5*(1133*sqrt(2)*A*a^2 - 1973*sqrt(2)*B*a^2)/a^2)*tan(1/2*d*x + 1/2*c)^2 - 15*(155*sqrt(2)*A*a^2 - 29
1*sqrt(2)*B*a^2)/a^2)*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(5/2))/d