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

Optimal. Leaf size=345 \[ \frac {\left (2 a^2 B+3 a b C-5 b^2 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}+\frac {\left (2 a^2 B+3 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a^2 d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (-2 a^3 C+4 a^2 b B+3 a b^2 C-5 b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 d \left (a^2-b^2\right )}+\frac {b \left (-5 a^3 C+7 a^2 b B+3 a b^2 C-5 b^3 B\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 d (a-b) (a+b)^2}-\frac {\left (-2 a^3 C+4 a^2 b B+3 a b^2 C-5 b^3 B\right ) \sin (c+d x)}{a^3 d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)}} \]

[Out]

(4*B*a^2*b-5*B*b^3-2*C*a^3+3*C*a*b^2)*(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))/a^3/(a^2-b^2)/d+1/3*(2*B*a^2-5*B*b^2+3*C*a*b)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*El
lipticF(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/(a^2-b^2)/d+b*(7*B*a^2*b-5*B*b^3-5*C*a^3+3*C*a*b^2)*(cos(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))/a^3/(a-b)/(a+b)^2/d+1/3*(2*B*a^
2-5*B*b^2+3*C*a*b)*sin(d*x+c)/a^2/(a^2-b^2)/d/cos(d*x+c)^(3/2)+b*(B*b-C*a)*sin(d*x+c)/a/(a^2-b^2)/d/cos(d*x+c)
^(3/2)/(a+b*cos(d*x+c))-(4*B*a^2*b-5*B*b^3-2*C*a^3+3*C*a*b^2)*sin(d*x+c)/a^3/(a^2-b^2)/d/cos(d*x+c)^(1/2)

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Rubi [A]  time = 1.38, antiderivative size = 345, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3029, 3000, 3055, 3059, 2639, 3002, 2641, 2805} \[ \frac {\left (2 a^2 B+3 a b C-5 b^2 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d \left (a^2-b^2\right )}+\frac {\left (4 a^2 b B-2 a^3 C+3 a b^2 C-5 b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 d \left (a^2-b^2\right )}+\frac {b \left (7 a^2 b B-5 a^3 C+3 a b^2 C-5 b^3 B\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 d (a-b) (a+b)^2}+\frac {b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}+\frac {\left (2 a^2 B+3 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a^2 d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x)}-\frac {\left (4 a^2 b B-2 a^3 C+3 a b^2 C-5 b^3 B\right ) \sin (c+d x)}{a^3 d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2639

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

Rule 2641

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

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[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 3000

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^2 - a*b*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*
Sin[e + f*x])^(1 + n))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int
[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(a*A - b*B)*(b*c - a*d)*(m + 1) + b*d*(A*b - a*B)*(m
+ n + 2) + (A*b - a*B)*(a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(A*b - a*B)*(m + n + 3)*Sin[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^
2, 0] && RationalQ[m] && m < -1 && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n,
-1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3002

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 3029

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] :> Dist[1/b^2, Int[(a + b*Sin[e + f*x])
^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rule 3055

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*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] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3059

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*} \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx &=\int \frac {B+C \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx\\ &=\frac {b (b B-a C) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}+\frac {\int \frac {\frac {1}{2} \left (2 a^2 B-5 b^2 B+3 a b C\right )-a (b B-a C) \cos (c+d x)+\frac {3}{2} b (b B-a C) \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac {\left (2 a^2 B-5 b^2 B+3 a b C\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}+\frac {b (b B-a C) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}+\frac {2 \int \frac {-\frac {3}{4} \left (4 a^2 b B-5 b^3 B-2 a^3 C+3 a b^2 C\right )+\frac {1}{2} a \left (a^2 B+2 b^2 B-3 a b C\right ) \cos (c+d x)+\frac {1}{4} b \left (2 a^2 B-5 b^2 B+3 a b C\right ) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{3 a^2 \left (a^2-b^2\right )}\\ &=\frac {\left (2 a^2 B-5 b^2 B+3 a b C\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\left (4 a^2 b B-5 b^3 B-2 a^3 C+3 a b^2 C\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {b (b B-a C) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}+\frac {4 \int \frac {\frac {1}{8} \left (2 a^4 B+16 a^2 b^2 B-15 b^4 B-12 a^3 b C+9 a b^3 C\right )+\frac {1}{4} a \left (7 a^2 b B-10 b^3 B-3 a^3 C+6 a b^2 C\right ) \cos (c+d x)+\frac {3}{8} b \left (4 a^2 b B-5 b^3 B-2 a^3 C+3 a b^2 C\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{3 a^3 \left (a^2-b^2\right )}\\ &=\frac {\left (2 a^2 B-5 b^2 B+3 a b C\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\left (4 a^2 b B-5 b^3 B-2 a^3 C+3 a b^2 C\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {b (b B-a C) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}-\frac {4 \int \frac {-\frac {1}{8} b \left (2 a^4 B+16 a^2 b^2 B-15 b^4 B-12 a^3 b C+9 a b^3 C\right )-\frac {1}{8} a b^2 \left (2 a^2 B-5 b^2 B+3 a b C\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{3 a^3 b \left (a^2-b^2\right )}+\frac {\left (4 a^2 b B-5 b^3 B-2 a^3 C+3 a b^2 C\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=\frac {\left (4 a^2 b B-5 b^3 B-2 a^3 C+3 a b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (2 a^2 B-5 b^2 B+3 a b C\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\left (4 a^2 b B-5 b^3 B-2 a^3 C+3 a b^2 C\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {b (b B-a C) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}+\frac {\left (2 a^2 B-5 b^2 B+3 a b C\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 a^2 \left (a^2-b^2\right )}+\frac {\left (b \left (7 a^2 b B-5 b^3 B-5 a^3 C+3 a b^2 C\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=\frac {\left (4 a^2 b B-5 b^3 B-2 a^3 C+3 a b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (2 a^2 B-5 b^2 B+3 a b C\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 \left (a^2-b^2\right ) d}+\frac {b \left (7 a^2 b B-5 b^3 B-5 a^3 C+3 a b^2 C\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 (a-b) (a+b)^2 d}+\frac {\left (2 a^2 B-5 b^2 B+3 a b C\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\left (4 a^2 b B-5 b^3 B-2 a^3 C+3 a b^2 C\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {b (b B-a C) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 6.87, size = 427, normalized size = 1.24 \[ \frac {\sqrt {\cos (c+d x)} \left (\frac {2 \sec (c+d x) (a C \sin (c+d x)-2 b B \sin (c+d x))}{a^3}+\frac {2 B \tan (c+d x) \sec (c+d x)}{3 a^2}+\frac {b^4 B \sin (c+d x)-a b^3 C \sin (c+d x)}{a^3 \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )}{d}+\frac {\frac {2 \left (-6 a^3 b C+12 a^2 b^2 B+9 a b^3 C-15 b^4 B\right ) \sin (c+d x) \cos (2 (c+d x)) \left (\left (b^2-2 a^2\right ) \Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )\right )}{a b^2 \sqrt {1-\cos ^2(c+d x)} \left (2 \cos ^2(c+d x)-1\right )}+\frac {\left (-12 a^4 C+28 a^3 b B+24 a^2 b^2 C-40 a b^3 B\right ) \left (2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-\frac {2 a \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}\right )}{b}+\frac {2 \left (4 a^4 B-30 a^3 b C+44 a^2 b^2 B+27 a b^3 C-45 b^4 B\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}}{12 a^3 d (a-b) (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*(4*a^4*B + 44*a^2*b^2*B - 45*b^4*B - 30*a^3*b*C + 27*a*b^3*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(
a + b) + ((28*a^3*b*B - 40*a*b^3*B - 12*a^4*C + 24*a^2*b^2*C)*(2*EllipticF[(c + d*x)/2, 2] - (2*a*EllipticPi[(
2*b)/(a + b), (c + d*x)/2, 2])/(a + b)))/b + (2*(12*a^2*b^2*B - 15*b^4*B - 6*a^3*b*C + 9*a*b^3*C)*Cos[2*(c + d
*x)]*(-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])/(a*b^2*Sqrt[1 - Cos[c + d*
x]^2]*(-1 + 2*Cos[c + d*x]^2)))/(12*a^3*(a - b)*(a + b)*d) + (Sqrt[Cos[c + d*x]]*((2*Sec[c + d*x]*(-2*b*B*Sin[
c + d*x] + a*C*Sin[c + d*x]))/a^3 + (b^4*B*Sin[c + d*x] - a*b^3*C*Sin[c + d*x])/(a^3*(a^2 - b^2)*(a + b*Cos[c
+ d*x])) + (2*B*Sec[c + d*x]*Tan[c + d*x])/(3*a^2)))/d

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 10.87, size = 1031, normalized size = 2.99 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-4*b^2*(2*B*b-C*a)/a^3/(-2*a*b+2*b^2)*(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)*Ellipt
icPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))+2*(B*b-C*a)*b/a^2*(-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*cos(1/2*d*x+1/2*c)^2*b+a-b)-1/2/(a+b)/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/(a^2-b^2)*(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/(a^2-b^2)*(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)*Ellip
ticE(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)))+2*(-2*B*b+
C*a)/a^3*(-(-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)^(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))+2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*c
os(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)+2*B/a^2*(-1/6*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)/(-1/2+cos(1/2*d*x+1/2*c)^2)^2+1/3*(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)*Ellipt
icF(cos(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

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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