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

Optimal. Leaf size=183 \[ \frac{95 \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{48 a^3 d}-\frac{197 \sin (c+d x)}{24 a^2 d \sqrt{a \cos (c+d x)+a}}+\frac{163 \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{\sin (c+d x) \cos ^3(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}-\frac{17 \sin (c+d x) \cos ^2(c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}} \]

[Out]

(163*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - (Cos[c + d*x
]^3*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) - (17*Cos[c + d*x]^2*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d
*x])^(3/2)) - (197*Sin[c + d*x])/(24*a^2*d*Sqrt[a + a*Cos[c + d*x]]) + (95*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*
x])/(48*a^3*d)

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Rubi [A]  time = 0.41154, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {2765, 2977, 2968, 3023, 2751, 2649, 206} \[ \frac{95 \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{48 a^3 d}-\frac{197 \sin (c+d x)}{24 a^2 d \sqrt{a \cos (c+d x)+a}}+\frac{163 \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{\sin (c+d x) \cos ^3(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}-\frac{17 \sin (c+d x) \cos ^2(c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(163*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - (Cos[c + d*x
]^3*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) - (17*Cos[c + d*x]^2*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d
*x])^(3/2)) - (197*Sin[c + d*x])/(24*a^2*d*Sqrt[a + a*Cos[c + d*x]]) + (95*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*
x])/(48*a^3*d)

Rule 2765

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[((b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 1))/(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 - 2)*Simp[b*(c^2*(m + 1) + d^2*(n -
1)) + a*c*d*(m - n + 1) + d*(a*d*(m - n + 1) + b*c*(m + n))*Sin[e + f*x], x], 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] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ
[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))

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 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+a \cos (c+d x))^{5/2}} \, dx &=-\frac{\cos ^3(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{\int \frac{\cos ^2(c+d x) \left (3 a-\frac{11}{2} a \cos (c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{\cos ^3(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{17 \cos ^2(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac{\int \frac{\cos (c+d x) \left (17 a^2-\frac{95}{4} a^2 \cos (c+d x)\right )}{\sqrt{a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{\cos ^3(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{17 \cos ^2(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac{\int \frac{17 a^2 \cos (c+d x)-\frac{95}{4} a^2 \cos ^2(c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{\cos ^3(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{17 \cos ^2(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{95 \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{48 a^3 d}-\frac{\int \frac{-\frac{95 a^3}{8}+\frac{197}{4} a^3 \cos (c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{12 a^5}\\ &=-\frac{\cos ^3(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{17 \cos ^2(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac{197 \sin (c+d x)}{24 a^2 d \sqrt{a+a \cos (c+d x)}}+\frac{95 \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{48 a^3 d}+\frac{163 \int \frac{1}{\sqrt{a+a \cos (c+d x)}} \, dx}{32 a^2}\\ &=-\frac{\cos ^3(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{17 \cos ^2(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac{197 \sin (c+d x)}{24 a^2 d \sqrt{a+a \cos (c+d x)}}+\frac{95 \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{48 a^3 d}-\frac{163 \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 \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{\cos ^3(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{17 \cos ^2(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac{197 \sin (c+d x)}{24 a^2 d \sqrt{a+a \cos (c+d x)}}+\frac{95 \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{48 a^3 d}\\ \end{align*}

Mathematica [B]  time = 6.35395, size = 587, normalized size = 3.21 \[ -\frac{40 \sin \left (\frac{c}{2}\right ) \cos \left (\frac{d x}{2}\right ) \cos ^5\left (\frac{c}{2}+\frac{d x}{2}\right )}{d (a (\cos (c+d x)+1))^{5/2}}+\frac{8 \sin \left (\frac{3 c}{2}\right ) \cos \left (\frac{3 d x}{2}\right ) \cos ^5\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (a (\cos (c+d x)+1))^{5/2}}-\frac{40 \cos \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \cos ^5\left (\frac{c}{2}+\frac{d x}{2}\right )}{d (a (\cos (c+d x)+1))^{5/2}}+\frac{8 \cos \left (\frac{3 c}{2}\right ) \sin \left (\frac{3 d x}{2}\right ) \cos ^5\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (a (\cos (c+d x)+1))^{5/2}}-\frac{29 \cos ^5\left (\frac{c}{2}+\frac{d x}{2}\right )}{8 d (a (\cos (c+d x)+1))^{5/2} \left (\cos \left (\frac{c}{4}+\frac{d x}{4}\right )-\sin \left (\frac{c}{4}+\frac{d x}{4}\right )\right )^2}+\frac{29 \cos ^5\left (\frac{c}{2}+\frac{d x}{2}\right )}{8 d (a (\cos (c+d x)+1))^{5/2} \left (\sin \left (\frac{c}{4}+\frac{d x}{4}\right )+\cos \left (\frac{c}{4}+\frac{d x}{4}\right )\right )^2}+\frac{\cos ^5\left (\frac{c}{2}+\frac{d x}{2}\right )}{8 d (a (\cos (c+d x)+1))^{5/2} \left (\cos \left (\frac{c}{4}+\frac{d x}{4}\right )-\sin \left (\frac{c}{4}+\frac{d x}{4}\right )\right )^4}-\frac{\cos ^5\left (\frac{c}{2}+\frac{d x}{2}\right )}{8 d (a (\cos (c+d x)+1))^{5/2} \left (\sin \left (\frac{c}{4}+\frac{d x}{4}\right )+\cos \left (\frac{c}{4}+\frac{d x}{4}\right )\right )^4}-\frac{163 \cos ^5\left (\frac{c}{2}+\frac{d x}{2}\right ) \log \left (\cos \left (\frac{c}{4}+\frac{d x}{4}\right )-\sin \left (\frac{c}{4}+\frac{d x}{4}\right )\right )}{4 d (a (\cos (c+d x)+1))^{5/2}}+\frac{163 \cos ^5\left (\frac{c}{2}+\frac{d x}{2}\right ) \log \left (\sin \left (\frac{c}{4}+\frac{d x}{4}\right )+\cos \left (\frac{c}{4}+\frac{d x}{4}\right )\right )}{4 d (a (\cos (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-163*Cos[c/2 + (d*x)/2]^5*Log[Cos[c/4 + (d*x)/4] - Sin[c/4 + (d*x)/4]])/(4*d*(a*(1 + Cos[c + d*x]))^(5/2)) +
(163*Cos[c/2 + (d*x)/2]^5*Log[Cos[c/4 + (d*x)/4] + Sin[c/4 + (d*x)/4]])/(4*d*(a*(1 + Cos[c + d*x]))^(5/2)) - (
40*Cos[(d*x)/2]*Cos[c/2 + (d*x)/2]^5*Sin[c/2])/(d*(a*(1 + Cos[c + d*x]))^(5/2)) + (8*Cos[(3*d*x)/2]*Cos[c/2 +
(d*x)/2]^5*Sin[(3*c)/2])/(3*d*(a*(1 + Cos[c + d*x]))^(5/2)) - (40*Cos[c/2]*Cos[c/2 + (d*x)/2]^5*Sin[(d*x)/2])/
(d*(a*(1 + Cos[c + d*x]))^(5/2)) + (8*Cos[(3*c)/2]*Cos[c/2 + (d*x)/2]^5*Sin[(3*d*x)/2])/(3*d*(a*(1 + Cos[c + d
*x]))^(5/2)) + Cos[c/2 + (d*x)/2]^5/(8*d*(a*(1 + Cos[c + d*x]))^(5/2)*(Cos[c/4 + (d*x)/4] - Sin[c/4 + (d*x)/4]
)^4) - (29*Cos[c/2 + (d*x)/2]^5)/(8*d*(a*(1 + Cos[c + d*x]))^(5/2)*(Cos[c/4 + (d*x)/4] - Sin[c/4 + (d*x)/4])^2
) - Cos[c/2 + (d*x)/2]^5/(8*d*(a*(1 + Cos[c + d*x]))^(5/2)*(Cos[c/4 + (d*x)/4] + Sin[c/4 + (d*x)/4])^4) + (29*
Cos[c/2 + (d*x)/2]^5)/(8*d*(a*(1 + Cos[c + d*x]))^(5/2)*(Cos[c/4 + (d*x)/4] + Sin[c/4 + (d*x)/4])^2)

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Maple [A]  time = 1.463, size = 242, normalized size = 1.3 \begin{align*}{\frac{1}{96\,d}\sqrt{a \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 128\,\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}+489\,\sqrt{2}\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 ) \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}a-512\,\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}-87\,\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}+6\,\sqrt{a}\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}} \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+cos(d*x+c)*a)^(5/2),x)

[Out]

1/96/cos(1/2*d*x+1/2*c)^3*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(128*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*c
os(1/2*d*x+1/2*c)^6+489*2^(1/2)*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))*cos(1/
2*d*x+1/2*c)^4*a-512*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)^4-87*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+6*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2))/a^(7/2)/si
n(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+a*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.72145, size = 566, normalized size = 3.09 \begin{align*} \frac{489 \, \sqrt{2}{\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\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 (32 \, \cos \left (d x + c\right )^{3} - 160 \, \cos \left (d x + c\right )^{2} - 503 \, \cos \left (d x + c\right ) - 299\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{192 \,{\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+a*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/192*(489*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*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*(32*cos(d*x + c)^3 - 160*cos(d*x + c)^2 - 503*cos(d*x + c) - 299)*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+a*cos(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 2.44584, size = 197, normalized size = 1.08 \begin{align*} \frac{\frac{{\left ({\left (3 \,{\left (\frac{2 \, \sqrt{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{a} - \frac{23 \, \sqrt{2}}{a}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - \frac{668 \, \sqrt{2}}{a}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - \frac{465 \, \sqrt{2}}{a}\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{3}{2}}} - \frac{489 \, \sqrt{2} \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}}}}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(((3*(2*sqrt(2)*tan(1/2*d*x + 1/2*c)^2/a - 23*sqrt(2)/a)*tan(1/2*d*x + 1/2*c)^2 - 668*sqrt(2)/a)*tan(1/2*
d*x + 1/2*c)^2 - 465*sqrt(2)/a)*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(3/2) - 489*sqrt(2)*log(ab
s(-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))/d