\(\int \frac {\cos ^{\frac {5}{2}}(c+d x) (a B+b B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx\) [389]

Optimal result

Integrand size = 36, antiderivative size = 116 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (a B+b B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=-\frac {2 a B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d}+\frac {2 \left (3 a^2+b^2\right ) B \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^3 d}-\frac {2 a^3 B \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b^3 (a+b) d}+\frac {2 B \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b d} \]

[Out]

-2*a*B*(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^2/d+2/3*(3*a^2+
b^2)*B*(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^3/d-2*a^3*B*(co
s(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2))/b^3/(a+b)/d+2/3*
B*sin(d*x+c)*cos(d*x+c)^(1/2)/b/d

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {21, 2872, 3138, 2719, 3081, 2720, 2884} \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (a B+b B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=-\frac {2 a^3 B \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b^3 d (a+b)}+\frac {2 B \left (3 a^2+b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^3 d}-\frac {2 a B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d}+\frac {2 B \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d} \]

[In]

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

[Out]

(-2*a*B*EllipticE[(c + d*x)/2, 2])/(b^2*d) + (2*(3*a^2 + b^2)*B*EllipticF[(c + d*x)/2, 2])/(3*b^3*d) - (2*a^3*
B*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(b^3*(a + b)*d) + (2*B*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*b*d)

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2872

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

Rule 2884

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], 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[c + d, 0]

Rule 3081

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

Rule 3138

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

Rubi steps \begin{align*} \text {integral}& = B \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx \\ & = \frac {2 B \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b d}+\frac {(2 B) \int \frac {\frac {a}{2}+\frac {1}{2} b \cos (c+d x)-\frac {3}{2} a \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{3 b} \\ & = \frac {2 B \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b d}-\frac {(2 B) \int \frac {-\frac {a b}{2}-\frac {1}{2} \left (3 a^2+b^2\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{3 b^2}-\frac {(a B) \int \sqrt {\cos (c+d x)} \, dx}{b^2} \\ & = -\frac {2 a B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d}+\frac {2 B \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b d}-\frac {\left (a^3 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{b^3}+\frac {\left (\left (3 a^2+b^2\right ) B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{3 b^3} \\ & = -\frac {2 a B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d}+\frac {2 \left (3 a^2+b^2\right ) B \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^3 d}-\frac {2 a^3 B \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b^3 (a+b) d}+\frac {2 B \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.96 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.37 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (a B+b B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\frac {B \left (4 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {6 a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+4 \sqrt {\cos (c+d x)} \sin (c+d x)-\frac {6 \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{b^2 \sqrt {\sin ^2(c+d x)}}\right )}{6 b d} \]

[In]

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

[Out]

(B*(4*EllipticF[(c + d*x)/2, 2] - (6*a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + 4*Sqrt[Cos[c + d*x
]]*Sin[c + d*x] - (6*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a + b)*EllipticF[ArcSin[Sqrt[Cos
[c + d*x]]], -1] + (-2*a^2 + b^2)*EllipticPi[-(b/a), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(b^2*Sqrt[
Sin[c + d*x]^2])))/(6*b*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(552\) vs. \(2(188)=376\).

Time = 6.43 (sec) , antiderivative size = 553, normalized size of antiderivative = 4.77

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, B \left (4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a \,b^{2}-4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a \,b^{2}+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}+3 a^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 a^{2} b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+a \,b^{2} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-b^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2} b -3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a \,b^{2}-3 a^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \Pi \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), -\frac {2 b}{a -b}, \sqrt {2}\right )\right )}{3 b^{3} \left (a -b \right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(553\)

[In]

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

[Out]

-2/3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*B*(4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*a*b^
2-4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*b^3-2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a*b^2+2*cos(1/2*d*x+
1/2*c)*sin(1/2*d*x+1/2*c)^2*b^3+3*a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(
cos(1/2*d*x+1/2*c),2^(1/2))-3*a^2*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(co
s(1/2*d*x+1/2*c),2^(1/2))+a*b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/
2*d*x+1/2*c),2^(1/2))-b^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+
1/2*c),2^(1/2))+3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2
^(1/2))*a^2*b-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(
1/2))*a*b^2-3*a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),
-2*b/(a-b),2^(1/2)))/b^3/(a-b)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(
1/2*d*x+1/2*c)^2-1)^(1/2)/d

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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