3.655 \(\int \frac{(A+C \cos ^2(c+d x)) \sec ^4(c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx\)

Optimal. Leaf size=370 \[ \frac{\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt{a+b \cos (c+d x)}}{24 a^3 d}+\frac{\left (8 a^2 (2 A+3 C)+5 A b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{24 a^2 d \sqrt{a+b \cos (c+d x)}}-\frac{\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{24 a^3 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{b \left (4 a^2 (A+2 C)+5 A b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{8 a^3 d \sqrt{a+b \cos (c+d x)}}-\frac{5 A b \tan (c+d x) \sec (c+d x) \sqrt{a+b \cos (c+d x)}}{12 a^2 d}+\frac{A \tan (c+d x) \sec ^2(c+d x) \sqrt{a+b \cos (c+d x)}}{3 a d} \]

[Out]

-((15*A*b^2 + 8*a^2*(2*A + 3*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(24*a^3*d*Sqr
t[(a + b*Cos[c + d*x])/(a + b)]) + ((5*A*b^2 + 8*a^2*(2*A + 3*C))*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF
[(c + d*x)/2, (2*b)/(a + b)])/(24*a^2*d*Sqrt[a + b*Cos[c + d*x]]) - (b*(5*A*b^2 + 4*a^2*(A + 2*C))*Sqrt[(a + b
*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/(8*a^3*d*Sqrt[a + b*Cos[c + d*x]]) + ((15*A
*b^2 + 8*a^2*(2*A + 3*C))*Sqrt[a + b*Cos[c + d*x]]*Tan[c + d*x])/(24*a^3*d) - (5*A*b*Sqrt[a + b*Cos[c + d*x]]*
Sec[c + d*x]*Tan[c + d*x])/(12*a^2*d) + (A*Sqrt[a + b*Cos[c + d*x]]*Sec[c + d*x]^2*Tan[c + d*x])/(3*a*d)

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Rubi [A]  time = 1.31926, antiderivative size = 370, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 10, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3056, 3055, 3059, 2655, 2653, 3002, 2663, 2661, 2807, 2805} \[ \frac{\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt{a+b \cos (c+d x)}}{24 a^3 d}+\frac{\left (8 a^2 (2 A+3 C)+5 A b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{24 a^2 d \sqrt{a+b \cos (c+d x)}}-\frac{\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{24 a^3 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{b \left (4 a^2 (A+2 C)+5 A b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{8 a^3 d \sqrt{a+b \cos (c+d x)}}-\frac{5 A b \tan (c+d x) \sec (c+d x) \sqrt{a+b \cos (c+d x)}}{12 a^2 d}+\frac{A \tan (c+d x) \sec ^2(c+d x) \sqrt{a+b \cos (c+d x)}}{3 a d} \]

Antiderivative was successfully verified.

[In]

Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/Sqrt[a + b*Cos[c + d*x]],x]

[Out]

-((15*A*b^2 + 8*a^2*(2*A + 3*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(24*a^3*d*Sqr
t[(a + b*Cos[c + d*x])/(a + b)]) + ((5*A*b^2 + 8*a^2*(2*A + 3*C))*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF
[(c + d*x)/2, (2*b)/(a + b)])/(24*a^2*d*Sqrt[a + b*Cos[c + d*x]]) - (b*(5*A*b^2 + 4*a^2*(A + 2*C))*Sqrt[(a + b
*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/(8*a^3*d*Sqrt[a + b*Cos[c + d*x]]) + ((15*A
*b^2 + 8*a^2*(2*A + 3*C))*Sqrt[a + b*Cos[c + d*x]]*Tan[c + d*x])/(24*a^3*d) - (5*A*b*Sqrt[a + b*Cos[c + d*x]]*
Sec[c + d*x]*Tan[c + d*x])/(12*a^2*d) + (A*Sqrt[a + b*Cos[c + d*x]]*Sec[c + d*x]^2*Tan[c + d*x])/(3*a*d)

Rule 3056

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 + 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] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*
(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 3)
*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ
[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 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]

Rule 2655

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*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 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 2663

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*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

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

Rule 2807

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

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]

Rubi steps

\begin{align*} \int \frac{\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx &=\frac{A \sqrt{a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac{\int \frac{\left (-\frac{5 A b}{2}+a (2 A+3 C) \cos (c+d x)+\frac{3}{2} A b \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{3 a}\\ &=-\frac{5 A b \sqrt{a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{12 a^2 d}+\frac{A \sqrt{a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac{\int \frac{\left (\frac{1}{4} \left (15 A b^2+4 a^2 (4 A+6 C)\right )+\frac{1}{2} a A b \cos (c+d x)-\frac{5}{4} A b^2 \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{6 a^2}\\ &=\frac{\left (15 A b^2+8 a^2 (2 A+3 C)\right ) \sqrt{a+b \cos (c+d x)} \tan (c+d x)}{24 a^3 d}-\frac{5 A b \sqrt{a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{12 a^2 d}+\frac{A \sqrt{a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac{\int \frac{\left (-\frac{3}{8} b \left (5 A b^2+4 a^2 (A+2 C)\right )-\frac{5}{4} a A b^2 \cos (c+d x)-\frac{1}{8} b \left (15 A b^2+8 a^2 (2 A+3 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{6 a^3}\\ &=\frac{\left (15 A b^2+8 a^2 (2 A+3 C)\right ) \sqrt{a+b \cos (c+d x)} \tan (c+d x)}{24 a^3 d}-\frac{5 A b \sqrt{a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{12 a^2 d}+\frac{A \sqrt{a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 a d}-\frac{\int \frac{\left (\frac{3}{8} b^2 \left (5 A b^2+4 a^2 (A+2 C)\right )-\frac{1}{8} a b \left (5 A b^2+8 a^2 (2 A+3 C)\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{6 a^3 b}-\frac{\left (15 A b^2+8 a^2 (2 A+3 C)\right ) \int \sqrt{a+b \cos (c+d x)} \, dx}{48 a^3}\\ &=\frac{\left (15 A b^2+8 a^2 (2 A+3 C)\right ) \sqrt{a+b \cos (c+d x)} \tan (c+d x)}{24 a^3 d}-\frac{5 A b \sqrt{a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{12 a^2 d}+\frac{A \sqrt{a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac{1}{48} \left (A \left (16+\frac{5 b^2}{a^2}\right )+24 C\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx-\frac{\left (b \left (5 A b^2+4 a^2 (A+2 C)\right )\right ) \int \frac{\sec (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{16 a^3}-\frac{\left (\left (15 A b^2+8 a^2 (2 A+3 C)\right ) \sqrt{a+b \cos (c+d x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \cos (c+d x)}{a+b}} \, dx}{48 a^3 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}\\ &=-\frac{\left (15 A b^2+8 a^2 (2 A+3 C)\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{24 a^3 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{\left (15 A b^2+8 a^2 (2 A+3 C)\right ) \sqrt{a+b \cos (c+d x)} \tan (c+d x)}{24 a^3 d}-\frac{5 A b \sqrt{a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{12 a^2 d}+\frac{A \sqrt{a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac{\left (\left (A \left (16+\frac{5 b^2}{a^2}\right )+24 C\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}{48 \sqrt{a+b \cos (c+d x)}}-\frac{\left (b \left (5 A b^2+4 a^2 (A+2 C)\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}}\right ) \int \frac{\sec (c+d x)}{\sqrt{\frac{a}{a+b}+\frac{b \cos (c+d x)}{a+b}}} \, dx}{16 a^3 \sqrt{a+b \cos (c+d x)}}\\ &=-\frac{\left (15 A b^2+8 a^2 (2 A+3 C)\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{24 a^3 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{\left (A \left (16+\frac{5 b^2}{a^2}\right )+24 C\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{24 d \sqrt{a+b \cos (c+d x)}}-\frac{b \left (5 A b^2+4 a^2 (A+2 C)\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{8 a^3 d \sqrt{a+b \cos (c+d x)}}+\frac{\left (15 A b^2+8 a^2 (2 A+3 C)\right ) \sqrt{a+b \cos (c+d x)} \tan (c+d x)}{24 a^3 d}-\frac{5 A b \sqrt{a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{12 a^2 d}+\frac{A \sqrt{a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 a d}\\ \end{align*}

Mathematica [C]  time = 6.74647, size = 604, normalized size = 1.63 \[ \frac{\sqrt{a+b \cos (c+d x)} \left (\frac{\sec (c+d x) \left (16 a^2 A \sin (c+d x)+24 a^2 C \sin (c+d x)+15 A b^2 \sin (c+d x)\right )}{24 a^3}-\frac{5 A b \tan (c+d x) \sec (c+d x)}{12 a^2}+\frac{A \tan (c+d x) \sec ^2(c+d x)}{3 a}\right )}{d}-\frac{b \left (\frac{2 \left (40 a^2 A+72 a^2 C+45 A b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{\sqrt{a+b \cos (c+d x)}}-\frac{2 i \left (16 a^2 A+24 a^2 C+15 A b^2\right ) \sin (c+d x) \cos (2 (c+d x)) \sqrt{\frac{b-b \cos (c+d x)}{a+b}} \sqrt{-\frac{b \cos (c+d x)+b}{a-b}} \left (2 a (a-b) E\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \cos (c+d x)}\right )|\frac{a+b}{a-b}\right )+b \left (2 a F\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \cos (c+d x)}\right )|\frac{a+b}{a-b}\right )-b \Pi \left (\frac{a+b}{a};i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \cos (c+d x)}\right )|\frac{a+b}{a-b}\right )\right )\right )}{a \sqrt{-\frac{1}{a+b}} \sqrt{1-\cos ^2(c+d x)} \sqrt{-\frac{a^2-2 a (a+b \cos (c+d x))+(a+b \cos (c+d x))^2-b^2}{b^2}} \left (2 a^2-4 a (a+b \cos (c+d x))+2 (a+b \cos (c+d x))^2-b^2\right )}+\frac{40 a A b \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{\sqrt{a+b \cos (c+d x)}}\right )}{96 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/Sqrt[a + b*Cos[c + d*x]],x]

[Out]

-(b*((40*a*A*b*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/Sqrt[a + b*Cos[c + d*
x]] + (2*(40*a^2*A + 45*A*b^2 + 72*a^2*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/
(a + b)])/Sqrt[a + b*Cos[c + d*x]] - ((2*I)*(16*a^2*A + 15*A*b^2 + 24*a^2*C)*Sqrt[(b - b*Cos[c + d*x])/(a + b)
]*Sqrt[-((b + b*Cos[c + d*x])/(a - b))]*Cos[2*(c + d*x)]*(2*a*(a - b)*EllipticE[I*ArcSinh[Sqrt[-(a + b)^(-1)]*
Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)] + b*(2*a*EllipticF[I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Cos[c
+ d*x]]], (a + b)/(a - b)] - b*EllipticPi[(a + b)/a, I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]],
(a + b)/(a - b)]))*Sin[c + d*x])/(a*Sqrt[-(a + b)^(-1)]*Sqrt[1 - Cos[c + d*x]^2]*Sqrt[-((a^2 - b^2 - 2*a*(a +
b*Cos[c + d*x]) + (a + b*Cos[c + d*x])^2)/b^2)]*(2*a^2 - b^2 - 4*a*(a + b*Cos[c + d*x]) + 2*(a + b*Cos[c + d*x
])^2))))/(96*a^3*d) + (Sqrt[a + b*Cos[c + d*x]]*((Sec[c + d*x]*(16*a^2*A*Sin[c + d*x] + 15*A*b^2*Sin[c + d*x]
+ 24*a^2*C*Sin[c + d*x]))/(24*a^3) - (5*A*b*Sec[c + d*x]*Tan[c + d*x])/(12*a^2) + (A*Sec[c + d*x]^2*Tan[c + d*
x])/(3*a)))/d

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Maple [B]  time = 1.063, size = 1562, normalized size = 4.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c))^(1/2),x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2*b-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*C*(-1/a*cos(1/2*d*x+1/2*c)*(-2*b*sin(1/2*d*x
+1/2*c)^4+(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*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*co
s(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(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))-1/2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))
^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^
(1/2))+1/2/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c
)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*b*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+1/2/a*b*(sin(1/2*d*x+
1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*
c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2)))+2*A*(-1/3/a*cos(1/2*d*x+1/2*c)*(-2*b*sin(1/2*
d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)^3+5/12*b/a^2*cos(1/2*d*x+1/2*c)*(-2*
b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)^2-1/24*(16*a^2+15*b^2)/a^3
*cos(1/2*d*x+1/2*c)*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)+5/
48*b^2/a^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^
4+(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))+1/3*(sin(1/2*d*x+1/2*c)^2
)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(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))-1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2
*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c)
,(-2*b/(a-b))^(1/2))+1/3/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin
(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*b*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-5/16*b^
2/a^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+
b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+5/16/a^3*(sin(1/2*d*x+1/2*c)^2
)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1
/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^3+1/4/a*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+
1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d
*x+1/2*c),2,(-2*b/(a-b))^(1/2))+5/16*b^3/a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b
))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a
-b))^(1/2))))/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*b+a+b)^(1/2)/d

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)**4/(a+b*cos(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{4}}{\sqrt{b \cos \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*sec(d*x + c)^4/sqrt(b*cos(d*x + c) + a), x)