\(\int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx\) [539]

Optimal result

Integrand size = 23, antiderivative size = 345 \[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=-\frac {8 a \left (4 a^4-7 a^2 b^2+2 b^4\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^4 \left (a^2-b^2\right )^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 \left (16 a^4-16 a^2 b^2-b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{3 b^4 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {2 a^2 \cos ^2(c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {4 a^3 \left (3 a^2-5 b^2\right ) \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (2 a^2-b^2\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d} \]

[Out]

-2/3*a^2*cos(d*x+c)^2*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))^(3/2)+4/3*a^3*(3*a^2-5*b^2)*sin(d*x+c)/b^3/(a^
2-b^2)^2/d/(a+b*cos(d*x+c))^(1/2)+2/3*(2*a^2-b^2)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/b^3/(a^2-b^2)/d-8/3*a*(4*a
^4-7*a^2*b^2+2*b^4)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2)*(b/(a
+b))^(1/2))*(a+b*cos(d*x+c))^(1/2)/b^4/(a^2-b^2)^2/d/((a+b*cos(d*x+c))/(a+b))^(1/2)+2/3*(16*a^4-16*a^2*b^2-b^4
)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b))^(1/2))*((a+b*
cos(d*x+c))/(a+b))^(1/2)/b^4/(a^2-b^2)/d/(a+b*cos(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 345, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {2871, 3110, 3102, 2831, 2742, 2740, 2734, 2732} \[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=-\frac {2 a^2 \sin (c+d x) \cos ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}+\frac {2 \left (2 a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 b^3 d \left (a^2-b^2\right )}+\frac {2 \left (16 a^4-16 a^2 b^2-b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{3 b^4 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {8 a \left (4 a^4-7 a^2 b^2+2 b^4\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^4 d \left (a^2-b^2\right )^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {4 a^3 \left (3 a^2-5 b^2\right ) \sin (c+d x)}{3 b^3 d \left (a^2-b^2\right )^2 \sqrt {a+b \cos (c+d x)}} \]

[In]

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

[Out]

(-8*a*(4*a^4 - 7*a^2*b^2 + 2*b^4)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(3*b^4*(a^2
- b^2)^2*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + (2*(16*a^4 - 16*a^2*b^2 - b^4)*Sqrt[(a + b*Cos[c + d*x])/(a +
b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(3*b^4*(a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]]) - (2*a^2*Cos[c + d*
x]^2*Sin[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^(3/2)) + (4*a^3*(3*a^2 - 5*b^2)*Sin[c + d*x])/(3*b^
3*(a^2 - b^2)^2*d*Sqrt[a + b*Cos[c + d*x]]) + (2*(2*a^2 - b^2)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(3*b^3*(
a^2 - b^2)*d)

Rule 2732

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2
+ d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2734

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

Rule 2740

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*Sqrt[a + b]))*EllipticF[(1/2)*(c - P
i/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2742

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

Rule 2831

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

Rule 2871

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/
(d*f*(n + 1)*(c^2 - d^2))), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e
+ f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 +
c*d*(a^2 + b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - d^2) - m*(b*c - a*d)^2 +
d*n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3102

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 3110

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

Rubi steps \begin{align*} \text {integral}& = -\frac {2 a^2 \cos ^2(c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {2 \int \frac {\cos (c+d x) \left (2 a^2-\frac {3}{2} a b \cos (c+d x)-\frac {3}{2} \left (2 a^2-b^2\right ) \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx}{3 b \left (a^2-b^2\right )} \\ & = -\frac {2 a^2 \cos ^2(c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {4 a^3 \left (3 a^2-5 b^2\right ) \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}-\frac {4 \int \frac {\frac {1}{2} a^2 b \left (3 a^2-5 b^2\right )+\frac {1}{2} a \left (6 a^4-11 a^2 b^2+3 b^4\right ) \cos (c+d x)-\frac {3}{4} b \left (a^2-b^2\right ) \left (2 a^2-b^2\right ) \cos ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{3 b^3 \left (a^2-b^2\right )^2} \\ & = -\frac {2 a^2 \cos ^2(c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {4 a^3 \left (3 a^2-5 b^2\right ) \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (2 a^2-b^2\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}-\frac {8 \int \frac {\frac {3}{8} b^2 \left (4 a^4-7 a^2 b^2-b^4\right )+\frac {3}{2} a b \left (4 a^4-7 a^2 b^2+2 b^4\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{9 b^4 \left (a^2-b^2\right )^2} \\ & = -\frac {2 a^2 \cos ^2(c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {4 a^3 \left (3 a^2-5 b^2\right ) \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (2 a^2-b^2\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}+\frac {\left (16 a^4-16 a^2 b^2-b^4\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{3 b^4 \left (a^2-b^2\right )}-\frac {\left (4 a \left (4 a^4-7 a^2 b^2+2 b^4\right )\right ) \int \sqrt {a+b \cos (c+d x)} \, dx}{3 b^4 \left (a^2-b^2\right )^2} \\ & = -\frac {2 a^2 \cos ^2(c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {4 a^3 \left (3 a^2-5 b^2\right ) \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (2 a^2-b^2\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}-\frac {\left (4 a \left (4 a^4-7 a^2 b^2+2 b^4\right ) \sqrt {a+b \cos (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}} \, dx}{3 b^4 \left (a^2-b^2\right )^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (\left (16 a^4-16 a^2 b^2-b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}\right ) \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}} \, dx}{3 b^4 \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}} \\ & = -\frac {8 a \left (4 a^4-7 a^2 b^2+2 b^4\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^4 \left (a^2-b^2\right )^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 \left (16 a^4-16 a^2 b^2-b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{3 b^4 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {2 a^2 \cos ^2(c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {4 a^3 \left (3 a^2-5 b^2\right ) \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (2 a^2-b^2\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.25 (sec) , antiderivative size = 237, normalized size of antiderivative = 0.69 \[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\frac {2 \left (\frac {\left (\frac {a+b \cos (c+d x)}{a+b}\right )^{3/2} \left (-4 \left (4 a^5-7 a^3 b^2+2 a b^4\right ) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )+\left (16 a^5-16 a^4 b-16 a^3 b^2+16 a^2 b^3-a b^4+b^5\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )\right )}{(a-b)^2}+\frac {b \left (16 a^6-25 a^4 b^2+b^6+4 a b \left (5 a^4-8 a^2 b^2+b^4\right ) \cos (c+d x)+\left (-a^2 b+b^3\right )^2 \cos (2 (c+d x))\right ) \sin (c+d x)}{2 \left (a^2-b^2\right )^2}\right )}{3 b^4 d (a+b \cos (c+d x))^{3/2}} \]

[In]

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

[Out]

(2*((((a + b*Cos[c + d*x])/(a + b))^(3/2)*(-4*(4*a^5 - 7*a^3*b^2 + 2*a*b^4)*EllipticE[(c + d*x)/2, (2*b)/(a +
b)] + (16*a^5 - 16*a^4*b - 16*a^3*b^2 + 16*a^2*b^3 - a*b^4 + b^5)*EllipticF[(c + d*x)/2, (2*b)/(a + b)]))/(a -
b)^2 + (b*(16*a^6 - 25*a^4*b^2 + b^6 + 4*a*b*(5*a^4 - 8*a^2*b^2 + b^4)*Cos[c + d*x] + (-(a^2*b) + b^3)^2*Cos[
2*(c + d*x)])*Sin[c + d*x])/(2*(a^2 - b^2)^2)))/(3*b^4*d*(a + b*Cos[c + d*x])^(3/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1294\) vs. \(2(379)=758\).

Time = 10.29 (sec) , antiderivative size = 1295, normalized size of antiderivative = 3.75

method	result	size
default	\(\text {Expression too large to display}\)	\(1295\)

[In]

int(cos(d*x+c)^4/(a+cos(d*x+c)*b)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-(-(-2*b*cos(1/2*d*x+1/2*c)^2-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(8/b^2*(-1/6/b*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d
*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)+1/6*(a-b)/b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2
*c)^2+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2
*c),(-2*b/(a-b))^(1/2))-1/12/b^2*(-2*a+6*b)*(a-b)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)
/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),(-2*b
/(a-b))^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))))+4*(a+b)/b^4*(a-b)*(sin(1/2*d*x+1/2*c)^2)^(1/
2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(
EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2)))+2/b^4*a^4*(
1/6/b/(a-b)/(a+b)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x
+1/2*c)^2+1/2*(a-b)/b)^2+8/3*sin(1/2*d*x+1/2*c)^2*b/(a-b)^2/(a+b)^2*cos(1/2*d*x+1/2*c)*a/(-(-2*b*cos(1/2*d*x+1
/2*c)^2-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)+(3*a-b)/(3*a^3+3*a^2*b-3*a*b^2-3*b^3)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((
2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellipt
icF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-4/3*a/(a-b)/(a+b)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+
1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x
+1/2*c),(-2*b/(a-b))^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))))+2*(3*a^2+2*a*b+b^2)/b^4*(sin(1/
2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*
x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+8*a^3/b^4/sin(1/2*d*x+1/2*c)^2/(2*b*sin(1/2
*d*x+1/2*c)^2-a-b)/(a^2-b^2)*(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*cos(1/2*d*x+1/2*c
)*sin(1/2*d*x+1/2*c)^2*b+EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+
b)/(a-b))^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a-EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*(-2*b/(a-b)*si
n(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*b))/sin(1/2*d*x+1/2*c)/(-2*b*sin(1/2*d*x+1/
2*c)^2+a+b)^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.18 (sec) , antiderivative size = 954, normalized size of antiderivative = 2.77 \[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/9*(6*(8*a^6*b^2 - 13*a^4*b^4 + a^2*b^6 + (a^4*b^4 - 2*a^2*b^6 + b^8)*cos(d*x + c)^2 + 2*(5*a^5*b^3 - 8*a^3*b
^5 + a*b^7)*cos(d*x + c))*sqrt(b*cos(d*x + c) + a)*sin(d*x + c) + (sqrt(2)*(-32*I*a^6*b^2 + 68*I*a^4*b^4 - 37*
I*a^2*b^6 - 3*I*b^8)*cos(d*x + c)^2 - 2*sqrt(2)*(32*I*a^7*b - 68*I*a^5*b^3 + 37*I*a^3*b^5 + 3*I*a*b^7)*cos(d*x
+ c) + sqrt(2)*(-32*I*a^8 + 68*I*a^6*b^2 - 37*I*a^4*b^4 - 3*I*a^2*b^6))*sqrt(b)*weierstrassPInverse(4/3*(4*a^
2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*a)/b) + (sqrt(2)*(
32*I*a^6*b^2 - 68*I*a^4*b^4 + 37*I*a^2*b^6 + 3*I*b^8)*cos(d*x + c)^2 - 2*sqrt(2)*(-32*I*a^7*b + 68*I*a^5*b^3 -
37*I*a^3*b^5 - 3*I*a*b^7)*cos(d*x + c) + sqrt(2)*(32*I*a^8 - 68*I*a^6*b^2 + 37*I*a^4*b^4 + 3*I*a^2*b^6))*sqrt
(b)*weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*si
n(d*x + c) + 2*a)/b) - 12*(sqrt(2)*(4*I*a^5*b^3 - 7*I*a^3*b^5 + 2*I*a*b^7)*cos(d*x + c)^2 + 2*sqrt(2)*(4*I*a^6
*b^2 - 7*I*a^4*b^4 + 2*I*a^2*b^6)*cos(d*x + c) + sqrt(2)*(4*I*a^7*b - 7*I*a^5*b^3 + 2*I*a^3*b^5))*sqrt(b)*weie
rstrassZeta(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2,
-8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*a)/b)) - 12*(sqrt(2)*(-4*I*a^5*b^
3 + 7*I*a^3*b^5 - 2*I*a*b^7)*cos(d*x + c)^2 + 2*sqrt(2)*(-4*I*a^6*b^2 + 7*I*a^4*b^4 - 2*I*a^2*b^6)*cos(d*x + c
) + sqrt(2)*(-4*I*a^7*b + 7*I*a^5*b^3 - 2*I*a^3*b^5))*sqrt(b)*weierstrassZeta(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(
8*a^3 - 9*a*b^2)/b^3, weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d
*x + c) - 3*I*b*sin(d*x + c) + 2*a)/b)))/((a^4*b^7 - 2*a^2*b^9 + b^11)*d*cos(d*x + c)^2 + 2*(a^5*b^6 - 2*a^3*b
^8 + a*b^10)*d*cos(d*x + c) + (a^6*b^5 - 2*a^4*b^7 + a^2*b^9)*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**4/(a+b*cos(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {\cos \left (d x + c\right )^{4}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

integrate(cos(d*x + c)^4/(b*cos(d*x + c) + a)^(5/2), x)

Giac [F]

\[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {\cos \left (d x + c\right )^{4}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

integrate(cos(d*x + c)^4/(b*cos(d*x + c) + a)^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^4(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^4}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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