\(\int \frac {\cos ^{\frac {7}{2}}(c+d x)}{(a+b \cos (c+d x))^2} \, dx\) [588]

Optimal result

Integrand size = 23, antiderivative size = 245 \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {a \left (5 a^2-4 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (15 a^4-16 a^2 b^2-2 b^4\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^4 \left (a^2-b^2\right ) d}-\frac {a^3 \left (5 a^2-7 b^2\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{(a-b) b^4 (a+b)^2 d}+\frac {\left (5 a^2-2 b^2\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac {a^2 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 2.16 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.09 \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {4 \sqrt {\cos (c+d x)} \left (2+\frac {3 a^3}{\left (a^2-b^2\right ) (a+b \cos (c+d x))}\right ) \sin (c+d x)-\frac {\frac {2 \left (5 a^3-8 a b^2\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {8 \left (2 a^2+b^2\right ) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{a+b}+\frac {6 \left (5 a^2-4 b^2\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{b^2 \sqrt {\sin ^2(c+d x)}}}{(a-b) (a+b)}}{12 b^2 d} \] Input:

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

Output:

(4*Sqrt[Cos[c + d*x]]*(2 + (3*a^3)/((a^2 - b^2)*(a + b*Cos[c + d*x])))*Sin 
[c + d*x] - ((2*(5*a^3 - 8*a*b^2)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2 
])/(a + b) + (8*(2*a^2 + b^2)*((a + b)*EllipticF[(c + d*x)/2, 2] - a*Ellip 
ticPi[(2*b)/(a + b), (c + d*x)/2, 2]))/(a + b) + (6*(5*a^2 - 4*b^2)*(-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]))/((a - b)*(a 
 + b)))/(12*b^2*d)
 

Rubi [A] (verified)

Time = 1.57 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.96, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.652, Rules used = {3042, 3271, 27, 3042, 3528, 27, 3042, 3538, 25, 3042, 3119, 3481, 3042, 3120, 3284}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\cos ^{\frac {7}{2}}(c+d x)}{(a+b \cos (c+d x))^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\)

\(\Big \downarrow \) 3271

\(\displaystyle -\frac {\int \frac {\sqrt {\cos (c+d x)} \left (3 a^2-2 b \cos (c+d x) a-\left (5 a^2-2 b^2\right ) \cos ^2(c+d x)\right )}{2 (a+b \cos (c+d x))}dx}{b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {\sqrt {\cos (c+d x)} \left (3 a^2-2 b \cos (c+d x) a-\left (5 a^2-2 b^2\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)}dx}{2 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (3 a^2-2 b \sin \left (c+d x+\frac {\pi }{2}\right ) a+\left (2 b^2-5 a^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3528

\(\displaystyle -\frac {\frac {2 \int -\frac {-3 a \left (5 a^2-4 b^2\right ) \cos ^2(c+d x)-2 b \left (2 a^2+b^2\right ) \cos (c+d x)+a \left (5 a^2-2 b^2\right )}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{3 b}-\frac {2 \left (5 a^2-2 b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {\int \frac {-3 a \left (5 a^2-4 b^2\right ) \cos ^2(c+d x)-2 b \left (2 a^2+b^2\right ) \cos (c+d x)+a \left (5 a^2-2 b^2\right )}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{3 b}-\frac {2 \left (5 a^2-2 b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {\int \frac {-3 a \left (5 a^2-4 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-2 b \left (2 a^2+b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+a \left (5 a^2-2 b^2\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 b}-\frac {2 \left (5 a^2-2 b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3538

\(\displaystyle -\frac {-\frac {-\frac {3 a \left (5 a^2-4 b^2\right ) \int \sqrt {\cos (c+d x)}dx}{b}-\frac {\int -\frac {a b \left (5 a^2-2 b^2\right )+\left (15 a^4-16 b^2 a^2-2 b^4\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{3 b}-\frac {2 \left (5 a^2-2 b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {-\frac {\frac {\int \frac {a b \left (5 a^2-2 b^2\right )+\left (15 a^4-16 b^2 a^2-2 b^4\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}-\frac {3 a \left (5 a^2-4 b^2\right ) \int \sqrt {\cos (c+d x)}dx}{b}}{3 b}-\frac {2 \left (5 a^2-2 b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {\frac {\int \frac {a b \left (5 a^2-2 b^2\right )+\left (15 a^4-16 b^2 a^2-2 b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {3 a \left (5 a^2-4 b^2\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{3 b}-\frac {2 \left (5 a^2-2 b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3119

\(\displaystyle -\frac {-\frac {\frac {\int \frac {a b \left (5 a^2-2 b^2\right )+\left (15 a^4-16 b^2 a^2-2 b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {6 a \left (5 a^2-4 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}-\frac {2 \left (5 a^2-2 b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3481

\(\displaystyle -\frac {-\frac {\frac {\frac {\left (15 a^4-16 a^2 b^2-2 b^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{b}-\frac {3 a^3 \left (5 a^2-7 b^2\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{b}-\frac {6 a \left (5 a^2-4 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}-\frac {2 \left (5 a^2-2 b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {\frac {\frac {\left (15 a^4-16 a^2 b^2-2 b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {3 a^3 \left (5 a^2-7 b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{b}-\frac {6 a \left (5 a^2-4 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}-\frac {2 \left (5 a^2-2 b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3120

\(\displaystyle -\frac {-\frac {\frac {\frac {2 \left (15 a^4-16 a^2 b^2-2 b^4\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}-\frac {3 a^3 \left (5 a^2-7 b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{b}-\frac {6 a \left (5 a^2-4 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}-\frac {2 \left (5 a^2-2 b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}}{2 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3284

\(\displaystyle -\frac {a^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {-\frac {2 \left (5 a^2-2 b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}-\frac {\frac {\frac {2 \left (15 a^4-16 a^2 b^2-2 b^4\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}-\frac {6 a^3 \left (5 a^2-7 b^2\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b d (a+b)}}{b}-\frac {6 a \left (5 a^2-4 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}}{2 b \left (a^2-b^2\right )}\)

Input:

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

Output:

-((a^2*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Cos[c + d* 
x]))) - (-1/3*((-6*a*(5*a^2 - 4*b^2)*EllipticE[(c + d*x)/2, 2])/(b*d) + (( 
2*(15*a^4 - 16*a^2*b^2 - 2*b^4)*EllipticF[(c + d*x)/2, 2])/(b*d) - (6*a^3* 
(5*a^2 - 7*b^2)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(b*(a + b)*d))/ 
b)/b - (2*(5*a^2 - 2*b^2)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*b*d))/(2*b*( 
a^2 - b^2))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3119

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 3120

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 3271

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2))*Co 
s[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] + Simp[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 3284

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 3481

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)]))/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[ 
B/d   Int[(a + b*Sin[e + f*x])^m, x], x] - Simp[(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 3528

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

rule 3538

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] :> Simp[C/(b*d)   Int[Sqrt[a + b*Sin[e + f*x]], x] 
, x] - Simp[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] /; Fre 
eQ[{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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1069\) vs. \(2(242)=484\).

Time = 18.15 (sec) , antiderivative size = 1070, normalized size of antiderivative = 4.37

method	result	size
default	\(\text {Expression too large to display}\)	\(1070\)

Input:

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

Output:

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

Fricas [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\cos ^{\frac {7}{2}}(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\int { \frac {\cos \left (d x + c\right )^{\frac {7}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {\cos ^{\frac {7}{2}}(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\int { \frac {\cos \left (d x + c\right )^{\frac {7}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{7/2}}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\cos ^{\frac {7}{2}}(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\int \frac {\sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{3}}{\cos \left (d x +c \right )^{2} b^{2}+2 \cos \left (d x +c \right ) a b +a^{2}}d x \] Input:

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

Output:

int((sqrt(cos(c + d*x))*cos(c + d*x)**3)/(cos(c + d*x)**2*b**2 + 2*cos(c + 
 d*x)*a*b + a**2),x)