3.3.53 \(\int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{7/2}} \, dx\) [253]

Optimal. Leaf size=177 \[ \frac {5 \text {ArcTan}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{64 \sqrt {2} a^{7/2} d}-\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}-\frac {13 \sqrt {\cos (c+d x)} \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2}}+\frac {67 \sqrt {\cos (c+d x)} \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2}} \]

[Out]

-1/6*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(7/2)+5/128*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^(7/2)/d*2^(1/2)-13/48*sin(d*x+c)*cos(d*x+c)^(1/2)/a/d/(a+a*cos(d*x+c))^(5/2
)+67/192*sin(d*x+c)*cos(d*x+c)^(1/2)/a^2/d/(a+a*cos(d*x+c))^(3/2)

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Rubi [A]
time = 0.27, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2844, 3056, 3057, 12, 2861, 211} \begin {gather*} \frac {5 \text {ArcTan}\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 {67 \sin (c+d x) \sqrt {\cos (c+d x)}}{192 a^2 d (a \cos (c+d x)+a)^{3/2}}-\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{6 d (a \cos (c+d x)+a)^{7/2}}-\frac {13 \sin (c+d x) \sqrt {\cos (c+d x)}}{48 a d (a \cos (c+d x)+a)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*ArcTan[(Sqrt[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
) - (Cos[c + d*x]^(3/2)*Sin[c + d*x])/(6*d*(a + a*Cos[c + d*x])^(7/2)) - (13*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/
(48*a*d*(a + a*Cos[c + d*x])^(5/2)) + (67*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(192*a^2*d*(a + a*Cos[c + d*x])^(3/
2))

Rule 12

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

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2844

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 2861

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 3056

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 3057

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])

Rubi steps

\begin {align*} \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{7/2}} \, dx &=-\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}-\frac {\int \frac {\sqrt {\cos (c+d x)} \left (\frac {3 a}{2}-5 a \cos (c+d x)\right )}{(a+a \cos (c+d x))^{5/2}} \, dx}{6 a^2}\\ &=-\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}-\frac {13 \sqrt {\cos (c+d x)} \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2}}-\frac {\int \frac {\frac {13 a^2}{4}-\frac {27}{2} a^2 \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2}} \, dx}{24 a^4}\\ &=-\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}-\frac {13 \sqrt {\cos (c+d x)} \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2}}+\frac {67 \sqrt {\cos (c+d x)} \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2}}-\frac {\int -\frac {15 a^3}{8 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{48 a^6}\\ &=-\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}-\frac {13 \sqrt {\cos (c+d x)} \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2}}+\frac {67 \sqrt {\cos (c+d x)} \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2}}+\frac {5 \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{128 a^3}\\ &=-\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}-\frac {13 \sqrt {\cos (c+d x)} \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2}}+\frac {67 \sqrt {\cos (c+d x)} \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2}}-\frac {5 \text {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 {5 \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 {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}-\frac {13 \sqrt {\cos (c+d x)} \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2}}+\frac {67 \sqrt {\cos (c+d x)} \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 2.76, size = 176, normalized size = 0.99 \begin {gather*} \frac {\cos ^7\left (\frac {1}{2} (c+d x)\right ) \sqrt {a (1+\cos (c+d x))} \left (15 \text {ArcSin}\left (\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right )}}\right ) \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right )}+\sqrt {2} \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sin \left (\frac {1}{2} (c+d x)\right ) \left (33-26 \tan ^2\left (\frac {1}{2} (c+d x)\right )+8 \tan ^4\left (\frac {1}{2} (c+d x)\right )\right )\right )}{24 a^4 d \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right )} (1+\cos (c+d x))^4} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Cos[(c + d*x)/2]^7*Sqrt[a*(1 + Cos[c + d*x])]*(15*ArcSin[Sin[(c + d*x)/2]/Sqrt[Cos[(c + d*x)/2]^2]]*Sqrt[Cos[
(c + d*x)/2]^2] + Sqrt[2]*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sin[(c + d*x)/2]*(33 - 26*Tan[(c + d*x)/2]^2 +
8*Tan[(c + d*x)/2]^4)))/(24*a^4*d*Sqrt[Cos[(c + d*x)/2]^2]*(1 + Cos[c + d*x])^4)

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Maple [A]
time = 0.21, size = 280, normalized size = 1.58

method	result	size
default	\(\frac {\left (\cos ^{\frac {5}{2}}\left (d x +c \right )\right ) \left (-1+\cos \left (d x +c \right )\right )^{5} \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (67 \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-17 \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right )+15 \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )-35 \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )+30 \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )-15 \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+15 \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right )\right ) \sqrt {2}}{384 d \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \sin \left (d x +c \right )^{11} a^{4}}\)	\(280\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384/d*cos(d*x+c)^(5/2)*(-1+cos(d*x+c))^5*(a*(1+cos(d*x+c)))^(1/2)*(67*cos(d*x+c)^3*2^(1/2)*(cos(d*x+c)/(1+co
s(d*x+c)))^(1/2)-17*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2+15*arcsin((-1+cos(d*x+c))/sin(d*x+c
))*sin(d*x+c)*cos(d*x+c)^2-35*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)+30*arcsin((-1+cos(d*x+c))/s
in(d*x+c))*cos(d*x+c)*sin(d*x+c)-15*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+15*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)/sin(d*x+c)^11*2^(1/2)/a^4

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.43, size = 214, normalized size = 1.21 \begin {gather*} \frac {15 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right )^{2} + 4 \, \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 {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 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (67 \, \cos \left (d x + c\right )^{2} + 50 \, \cos \left (d x + c\right ) + 15\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(15*sqrt(2)*(cos(d*x + c)^4 + 4*cos(d*x + c)^3 + 6*cos(d*x + c)^2 + 4*cos(d*x + c) + 1)*sqrt(a)*arctan(1
/2*sqrt(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*sqrt(a*cos(d*x + c) + a)*(67*cos(d*x + c)^2 + 50*cos(d*x + c) + 15)*sqrt(cos(d*x + c))*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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 6189 deep

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\cos \left (c+d\,x\right )}^{5/2}}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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