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

Optimal. Leaf size=214 \[ -\frac {5 \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{a^{5/2} d}+\frac {115 \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {35 \sin (c+d x) \sqrt {\cos (c+d x)}}{16 a^2 d \sqrt {a \cos (c+d x)+a}}-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}-\frac {15 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}} \]

[Out]

-5*arcsin(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(5/2)/d-1/4*cos(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*cos(d*x+
c))^(5/2)-15/16*cos(d*x+c)^(3/2)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(3/2)+115/32*arctan(1/2*sin(d*x+c)*a^(1/2)*2^
(1/2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)+35/16*sin(d*x+c)*cos(d*x+c)^(1/2)/a^2/d/(a+a*
cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.58, antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {2765, 2977, 2983, 2982, 2782, 205, 2774, 216} \[ -\frac {5 \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{a^{5/2} d}+\frac {35 \sin (c+d x) \sqrt {\cos (c+d x)}}{16 a^2 d \sqrt {a \cos (c+d x)+a}}+\frac {115 \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {\sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}-\frac {15 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(a^(5/2)*d) + (115*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) - (Cos[c + d*x]^(5/2)*Sin[c + d
*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) - (15*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2
)) + (35*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(16*a^2*d*Sqrt[a + a*Cos[c + d*x]])

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 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]

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 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 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 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 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])

Rubi steps

\begin {align*} \int \frac {\cos ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx &=-\frac {\cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {\int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (\frac {5 a}{2}-5 a \cos (c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {\cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {15 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {\int \frac {\sqrt {\cos (c+d x)} \left (\frac {45 a^2}{4}-\frac {35}{2} a^2 \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac {\cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {15 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {35 \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {\int \frac {-\frac {35 a^3}{4}+20 a^3 \cos (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{8 a^5}\\ &=-\frac {\cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {15 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {35 \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {5 \int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\cos (c+d x)}} \, dx}{2 a^3}+\frac {115 \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{32 a^2}\\ &=-\frac {\cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {15 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {35 \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {5 \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^3 d}-\frac {115 \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 {5 \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^{5/2} d}+\frac {115 \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 {\cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {15 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {35 \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 6.71, size = 385, normalized size = 1.80 \[ \frac {\sqrt {\cos (c+d x)} \cos ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\frac {8 \sin \left (\frac {c}{2}\right ) \cos \left (\frac {d x}{2}\right )}{d}+\frac {8 \cos \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )}{d}-\frac {\sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right )}{2 d}-\frac {\tan \left (\frac {c}{2}\right ) \sec ^3\left (\frac {c}{2}+\frac {d x}{2}\right )}{2 d}+\frac {23 \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{4 d}+\frac {23 \tan \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right )}{4 d}\right )}{(a (\cos (c+d x)+1))^{5/2}}+\frac {5 i e^{\frac {1}{2} i (c+d x)} \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \cos ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (8 \sinh ^{-1}\left (e^{i (c+d x)}\right )+\frac {23 \tanh ^{-1}\left (\frac {1-e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )}{\sqrt {2}}-8 \tanh ^{-1}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right )}{2 \sqrt {2} d \sqrt {1+e^{2 i (c+d x)}} (a (\cos (c+d x)+1))^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(((5*I)/2)*E^((I/2)*(c + d*x))*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))]*(8*ArcSinh[E^(I*(c + d*x))] + (
23*ArcTanh[(1 - E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])])/Sqrt[2] - 8*ArcTanh[Sqrt[1 + E^((2*
I)*(c + d*x))]])*Cos[c/2 + (d*x)/2]^5)/(Sqrt[2]*d*Sqrt[1 + E^((2*I)*(c + d*x))]*(a*(1 + Cos[c + d*x]))^(5/2))
+ (Cos[c/2 + (d*x)/2]^5*Sqrt[Cos[c + d*x]]*((8*Cos[(d*x)/2]*Sin[c/2])/d + (8*Cos[c/2]*Sin[(d*x)/2])/d + (23*Se
c[c/2]*Sec[c/2 + (d*x)/2]^2*Sin[(d*x)/2])/(4*d) - (Sec[c/2]*Sec[c/2 + (d*x)/2]^4*Sin[(d*x)/2])/(2*d) + (23*Sec
[c/2 + (d*x)/2]*Tan[c/2])/(4*d) - (Sec[c/2 + (d*x)/2]^3*Tan[c/2])/(2*d)))/(a*(1 + Cos[c + d*x]))^(5/2)

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fricas [A]  time = 1.78, size = 236, normalized size = 1.10 \[ -\frac {115 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - 2 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (16 \, \cos \left (d x + c\right )^{2} + 55 \, \cos \left (d x + c\right ) + 35\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 160 \, {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(115*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(
d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) - 2*sqrt(a*cos(d*x + c) + a)*(16*cos(d*x + c)^2 + 55*
cos(d*x + c) + 35)*sqrt(cos(d*x + c))*sin(d*x + c) - 160*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) +
1)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x + c\right )^{\frac {7}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.19, size = 344, normalized size = 1.61 \[ \frac {\left (\cos ^{\frac {7}{2}}\left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (-1+\cos \left (d x +c \right )\right )^{5} \left (16 \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+39 \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+80 \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {2}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right )+115 \cos \left (d x +c \right ) \sin \left (d x +c \right ) \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-20 \cos \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+80 \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right ) \sqrt {2}\, \sin \left (d x +c \right )+115 \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right )-35 \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \sqrt {2}}{32 d \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \sin \left (d x +c \right )^{11} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(7/2)/(a+a*cos(d*x+c))^(5/2),x)

[Out]

1/32/d*cos(d*x+c)^(7/2)*(a*(1+cos(d*x+c)))^(1/2)*(-1+cos(d*x+c))^5*(16*cos(d*x+c)^3*2^(1/2)*(cos(d*x+c)/(1+cos
(d*x+c)))^(1/2)+39*cos(d*x+c)^2*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+80*cos(d*x+c)*sin(d*x+c)*2^(1/2)*arc
tan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))+115*cos(d*x+c)*sin(d*x+c)*arcsin((-1+cos(d*x+c))/
sin(d*x+c))-20*cos(d*x+c)*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+80*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*
x+c)))^(1/2)/cos(d*x+c))*2^(1/2)*sin(d*x+c)+115*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)-35*2^(1/2)*(cos(
d*x+c)/(1+cos(d*x+c)))^(1/2))/(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)/sin(d*x+c)^11*2^(1/2)/a^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x + c\right )^{\frac {7}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^{7/2}}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^(7/2)/(a + a*cos(c + d*x))^(5/2),x)

[Out]

int(cos(c + d*x)^(7/2)/(a + a*cos(c + d*x))^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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