3.250 \(\int \frac{1}{\cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=217 \[ \frac{95 \sin (c+d x)}{48 a^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}-\frac{299 \sin (c+d x)}{48 a^2 d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{163 \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{17 \sin (c+d x)}{16 a d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}-\frac{\sin (c+d x)}{4 d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.546145, antiderivative size = 217, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {2766, 2978, 2984, 12, 2782, 205} \[ \frac{95 \sin (c+d x)}{48 a^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}-\frac{299 \sin (c+d x)}{48 a^2 d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{163 \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{17 \sin (c+d x)}{16 a d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}-\frac{\sin (c+d x)}{4 d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x])^(5/2)),x]

[Out]

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

Rule 2766

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

Rule 2978

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*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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 2984

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*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^(n + 1))/(f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin
[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*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] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 0])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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]

Rubi steps

\begin{align*} \int \frac{1}{\cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx &=-\frac{\sin (c+d x)}{4 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}+\frac{\int \frac{\frac{11 a}{2}-3 a \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{\sin (c+d x)}{4 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac{17 \sin (c+d x)}{16 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{\frac{95 a^2}{4}-17 a^2 \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{\sin (c+d x)}{4 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac{17 \sin (c+d x)}{16 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{95 \sin (c+d x)}{48 a^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{-\frac{299 a^3}{8}+\frac{95}{4} a^3 \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{12 a^5}\\ &=-\frac{\sin (c+d x)}{4 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac{17 \sin (c+d x)}{16 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{95 \sin (c+d x)}{48 a^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{299 \sin (c+d x)}{48 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{489 a^4}{16 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{6 a^6}\\ &=-\frac{\sin (c+d x)}{4 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac{17 \sin (c+d x)}{16 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{95 \sin (c+d x)}{48 a^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{299 \sin (c+d x)}{48 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}+\frac{163 \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{32 a^2}\\ &=-\frac{\sin (c+d x)}{4 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac{17 \sin (c+d x)}{16 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{95 \sin (c+d x)}{48 a^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{299 \sin (c+d x)}{48 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}-\frac{163 \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{163 \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{\sin (c+d x)}{4 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac{17 \sin (c+d x)}{16 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{95 \sin (c+d x)}{48 a^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{299 \sin (c+d x)}{48 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 10.6955, size = 639, normalized size = 2.94 \[ -\frac{\cot ^5\left (\frac{c}{2}+\frac{d x}{2}\right ) \csc ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec ^4\left (\frac{1}{2} (c+d x)\right ) \left (640 \sin ^{12}\left (\frac{c}{2}+\frac{d x}{2}\right ) \cos ^8\left (\frac{1}{2} (c+d x)\right ) \text{HypergeometricPFQ}\left (\left \{2,2,2,2,\frac{7}{2}\right \},\left \{1,1,1,\frac{13}{2}\right \},\frac{\sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )-1}\right )-1280 \left (5 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )-6\right ) \sin ^{12}\left (\frac{c}{2}+\frac{d x}{2}\right ) \cos ^6\left (\frac{1}{2} (c+d x)\right ) \text{HypergeometricPFQ}\left (\left \{2,2,2,\frac{7}{2}\right \},\left \{1,1,\frac{13}{2}\right \},\frac{\sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )-1}\right )+33 \left (1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )\right )^3 \sqrt{\frac{\sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )-1}} \left (\sqrt{\frac{\sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )-1}} \left (4344400 \sin ^{14}\left (\frac{c}{2}+\frac{d x}{2}\right )-26448512 \sin ^{12}\left (\frac{c}{2}+\frac{d x}{2}\right )[phone] \sin ^{10}\left (\frac{c}{2}+\frac{d x}{2}\right )-96281836 \sin ^8\left (\frac{c}{2}+\frac{d x}{2}\right )[phone] \sin ^6\left (\frac{c}{2}+\frac{d x}{2}\right )-38990350 \sin ^4\left (\frac{c}{2}+\frac{d x}{2}\right )[phone] \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )-1148175\right )-105 \left (33208 \sin ^{10}\left (\frac{c}{2}+\frac{d x}{2}\right )-140732 \sin ^8\left (\frac{c}{2}+\frac{d x}{2}\right )+234156 \sin ^6\left (\frac{c}{2}+\frac{d x}{2}\right )-188110 \sin ^4\left (\frac{c}{2}+\frac{d x}{2}\right )+72902 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )-10935\right ) \cos ^4\left (\frac{1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sqrt{\frac{\sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )-1}}\right )\right )\right )}{41580 d \left (1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )\right )^{7/2} (a (\cos (c+d x)+1))^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x])^(5/2)),x]

[Out]

-(Cot[c/2 + (d*x)/2]^5*Csc[c/2 + (d*x)/2]^4*Sec[(c + d*x)/2]^4*(640*Cos[(c + d*x)/2]^8*HypergeometricPFQ[{2, 2
, 2, 2, 7/2}, {1, 1, 1, 13/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^12 - 128
0*Cos[(c + d*x)/2]^6*HypergeometricPFQ[{2, 2, 2, 7/2}, {1, 1, 13/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d
*x)/2]^2)]*Sin[c/2 + (d*x)/2]^12*(-6 + 5*Sin[c/2 + (d*x)/2]^2) + 33*(1 - 2*Sin[c/2 + (d*x)/2]^2)^3*Sqrt[Sin[c/
2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*(-105*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/
2]^2)]]*Cos[(c + d*x)/2]^4*(-10935 + 72902*Sin[c/2 + (d*x)/2]^2 - 188110*Sin[c/2 + (d*x)/2]^4 + 234156*Sin[c/2
+ (d*x)/2]^6 - 140732*Sin[c/2 + (d*x)/2]^8 + 33208*Sin[c/2 + (d*x)/2]^10) + Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2
*Sin[c/2 + (d*x)/2]^2)]*(-1148175 + 10333785*Sin[c/2 + (d*x)/2]^2 - 38990350*Sin[c/2 + (d*x)/2]^4 + 79946462*S
in[c/2 + (d*x)/2]^6 - 96281836*Sin[c/2 + (d*x)/2]^8 + 68243596*Sin[c/2 + (d*x)/2]^10 - 26448512*Sin[c/2 + (d*x
)/2]^12 + 4344400*Sin[c/2 + (d*x)/2]^14))))/(41580*d*(a*(1 + Cos[c + d*x]))^(5/2)*(1 - 2*Sin[c/2 + (d*x)/2]^2)
^(7/2))

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Maple [B]  time = 0.355, size = 377, normalized size = 1.7 \begin{align*}{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{96\,d{a}^{3} \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 1+\cos \left ( dx+c \right ) \right ) ^{3}} \left ( 489\, \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{5/2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}\sin \left ( dx+c \right ) \arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) +1956\,\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{5/2}+2934\,\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{5/2}+1956\,\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{5/2}+489\,\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \sin \left ( dx+c \right ) \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{5/2}-299\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sqrt{2}-204\,\sqrt{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}+343\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sqrt{2}+192\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{2}-32\,\cos \left ( dx+c \right ) \sqrt{2} \right ) \sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) } \left ( \cos \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cos(d*x+c)^(5/2)/(a+cos(d*x+c)*a)^(5/2),x)

[Out]

1/96/d*2^(1/2)/a^3*(489*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*cos(d*x+c)^4*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d
*x+c))+1956*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^3*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+2934*
arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+1956*arcsin((-1+c
os(d*x+c))/sin(d*x+c))*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+489*arcsin((-1+cos(d*x+c))/sin(
d*x+c))*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-299*cos(d*x+c)^5*2^(1/2)-204*2^(1/2)*cos(d*x+c)^4+343*cos
(d*x+c)^3*2^(1/2)+192*cos(d*x+c)^2*2^(1/2)-32*cos(d*x+c)*2^(1/2))*sin(d*x+c)*(a*(1+cos(d*x+c)))^(1/2)/(-1+cos(
d*x+c))/(1+cos(d*x+c))^3/cos(d*x+c)^(5/2)

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

[Out]

Timed out

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Fricas [A]  time = 2.37963, size = 599, normalized size = 2.76 \begin{align*} \frac{489 \, \sqrt{2}{\left (\cos \left (d x + c\right )^{5} + 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \sqrt{a} \arctan \left (\frac{\sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \,{\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )}}\right ) - 2 \,{\left (299 \, \cos \left (d x + c\right )^{3} + 503 \, \cos \left (d x + c\right )^{2} + 160 \, \cos \left (d x + c\right ) - 32\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{96 \,{\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/96*(489*sqrt(2)*(cos(d*x + c)^5 + 3*cos(d*x + c)^4 + 3*cos(d*x + c)^3 + cos(d*x + c)^2)*sqrt(a)*arctan(1/2*s
qrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 + a*cos(d*x + c))) -
2*(299*cos(d*x + c)^3 + 503*cos(d*x + c)^2 + 160*cos(d*x + c) - 32)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c
))*sin(d*x + c))/(a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^
2)

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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(1/cos(d*x+c)**(5/2)/(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{1}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \cos \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(1/((a*cos(d*x + c) + a)^(5/2)*cos(d*x + c)^(5/2)), x)