3.634 \(\int (a+b \cos (c+d x))^{3/2} (A+C \cos ^2(c+d x)) \sec ^2(c+d x) \, dx\)

Optimal. Leaf size=270 \[ \frac {\left (a^2 (3 A-2 C)+2 b^2 (3 A+C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 d \sqrt {a+b \cos (c+d x)}}-\frac {b (3 A-2 C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}-\frac {a (3 A-8 C) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}+\frac {3 a A b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}} \]

[Out]

-1/3*b*(3*A-2*C)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/d-1/3*a*(3*A-8*C)*(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)/d/((a+b*cos(d*x+c))/(a+b))
^(1/2)+1/3*(a^2*(3*A-2*C)+2*b^2*(3*A+C))*(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)/d/(a+b*cos(d*x+c))^(1/2)+3*a*A*b*(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),2,2^(1/2)*(b/(a+b))^(1/2))*((a+b*cos(d*x+c))/
(a+b))^(1/2)/d/(a+b*cos(d*x+c))^(1/2)+A*(a+b*cos(d*x+c))^(3/2)*tan(d*x+c)/d

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Rubi [A]  time = 0.93, antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3048, 3049, 3059, 2655, 2653, 3002, 2663, 2661, 2807, 2805} \[ \frac {\left (a^2 (3 A-2 C)+2 b^2 (3 A+C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 d \sqrt {a+b \cos (c+d x)}}-\frac {b (3 A-2 C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}-\frac {a (3 A-8 C) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {A \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}+\frac {3 a A b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^(3/2)*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^2,x]

[Out]

-(a*(3*A - 8*C)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(3*d*Sqrt[(a + b*Cos[c + d*x])
/(a + b)]) + ((a^2*(3*A - 2*C) + 2*b^2*(3*A + C))*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2
*b)/(a + b)])/(3*d*Sqrt[a + b*Cos[c + d*x]]) + (3*a*A*b*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c +
d*x)/2, (2*b)/(a + b)])/(d*Sqrt[a + b*Cos[c + d*x]]) - (b*(3*A - 2*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(
3*d) + (A*(a + b*Cos[c + d*x])^(3/2)*Tan[c + d*x])/d

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 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 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 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 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 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 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 3048

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[((c^2*C + A*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(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
- 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n +
2) - b*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*(A*d^2*(m + n + 2) + C*(c^2*(
m + 1) + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3049

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] + Dist[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 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 (a+b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx &=\frac {A (a+b \cos (c+d x))^{3/2} \tan (c+d x)}{d}+\int \sqrt {a+b \cos (c+d x)} \left (\frac {3 A b}{2}+a C \cos (c+d x)-\frac {1}{2} b (3 A-2 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac {b (3 A-2 C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 d}+\frac {A (a+b \cos (c+d x))^{3/2} \tan (c+d x)}{d}+\frac {2}{3} \int \frac {\left (\frac {9 a A b}{4}+\frac {1}{2} \left (3 A b^2+\left (3 a^2+b^2\right ) C\right ) \cos (c+d x)-\frac {1}{4} a b (3 A-8 C) \cos ^2(c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx\\ &=-\frac {b (3 A-2 C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 d}+\frac {A (a+b \cos (c+d x))^{3/2} \tan (c+d x)}{d}-\frac {2 \int \frac {\left (-\frac {9}{4} a A b^2-\frac {1}{4} b \left (a^2 (3 A-2 C)+2 b^2 (3 A+C)\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{3 b}-\frac {1}{6} (a (3 A-8 C)) \int \sqrt {a+b \cos (c+d x)} \, dx\\ &=-\frac {b (3 A-2 C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 d}+\frac {A (a+b \cos (c+d x))^{3/2} \tan (c+d x)}{d}+\frac {1}{2} (3 a A b) \int \frac {\sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx-\frac {1}{6} \left (-a^2 (3 A-2 C)-2 b^2 (3 A+C)\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx-\frac {\left (a (3 A-8 C) \sqrt {a+b \cos (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}} \, dx}{6 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\\ &=-\frac {a (3 A-8 C) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {b (3 A-2 C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 d}+\frac {A (a+b \cos (c+d x))^{3/2} \tan (c+d x)}{d}+\frac {\left (3 a A b \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}{2 \sqrt {a+b \cos (c+d x)}}-\frac {\left (\left (-a^2 (3 A-2 C)-2 b^2 (3 A+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}{6 \sqrt {a+b \cos (c+d x)}}\\ &=-\frac {a (3 A-8 C) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (a^2 (3 A-2 C)+2 b^2 (3 A+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 )}{3 d \sqrt {a+b \cos (c+d x)}}+\frac {3 a A b \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 )}{d \sqrt {a+b \cos (c+d x)}}-\frac {b (3 A-2 C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 d}+\frac {A (a+b \cos (c+d x))^{3/2} \tan (c+d x)}{d}\\ \end {align*}

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Mathematica [C]  time = 3.52, size = 406, normalized size = 1.50 \[ \frac {\frac {8 \left (C \left (3 a^2+b^2\right )+3 A b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+4 \tan (c+d x) \sqrt {a+b \cos (c+d x)} (3 a A+2 b C \cos (c+d x))+\frac {2 a b (15 A+8 C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}-\frac {2 i (3 A-8 C) \csc (c+d x) \sqrt {-\frac {b (\cos (c+d x)-1)}{a+b}} \sqrt {\frac {b (\cos (c+d x)+1)}{b-a}} \left (b \left (b \Pi \left (\frac {a+b}{a};i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )-2 a F\left (i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )\right )-2 a (a-b) E\left (i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )\right )}{b \sqrt {-\frac {1}{a+b}}}}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^(3/2)*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^2,x]

[Out]

((8*(3*A*b^2 + (3*a^2 + b^2)*C)*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*a*b*(15*A + 8*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)*(3*A - 8*C)*Sqrt[-((b*(-1 + Cos[c + d*x]))/(a + b))]*Sqrt[(b*(1 +
Cos[c + d*x]))/(-a + b)]*Csc[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)]
)))/(b*Sqrt[-(a + b)^(-1)]) + 4*Sqrt[a + b*Cos[c + d*x]]*(3*a*A + 2*b*C*Cos[c + d*x])*Tan[c + d*x])/(12*d)

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fricas [F]  time = 4.35, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b \cos \left (d x + c\right )^{3} + C a \cos \left (d x + c\right )^{2} + A b \cos \left (d x + c\right ) + A a\right )} \sqrt {b \cos \left (d x + c\right ) + a} \sec \left (d x + c\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, algorithm="fricas")

[Out]

integral((C*b*cos(d*x + c)^3 + C*a*cos(d*x + c)^2 + A*b*cos(d*x + c) + A*a)*sqrt(b*cos(d*x + c) + a)*sec(d*x +
c)^2, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^(3/2)*sec(d*x + c)^2, x)

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maple [B]  time = 2.76, size = 1221, normalized size = 4.52 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^2,x)

[Out]

-1/3*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-16*C*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2
*c)^6+(12*A*a*b+8*C*a*b+16*C*b^2)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-6*A*a^2-6*A*a*b-4*C*a*b-4*C*b^2)*s
in(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*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(
a-b))^(1/2)*(3*A*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2+6*A*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(
a-b))^(1/2))*b^2-3*A*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2+3*A*EllipticE(cos(1/2*d*x+1/2*c),(-2
*b/(a-b))^(1/2))*a*b-9*A*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))*a*b-2*C*EllipticF(cos(1/2*d*x+1/2
*c),(-2*b/(a-b))^(1/2))*a^2+2*C*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^2+8*C*EllipticE(cos(1/2*d*x
+1/2*c),(-2*b/(a-b))^(1/2))*a^2-8*C*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b)*sin(1/2*d*x+1/2*c)^2
+3*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/
2*c),(-2*b/(a-b))^(1/2))*a^2+6*A*b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b)
)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-3*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*
d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2+3*A*(sin(1/2*d*x+1/2*c)^2
)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a
*b-9*A*a*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticPi(cos(1/2
*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))-2*a^2*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/
(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+2*b^2*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)
*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+8*C*(sin(1/2*d*x+1/2
*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/
2))*a^2-8*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2
*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b)/(-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)^2-1)/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]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^(3/2)*sec(d*x + c)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^(3/2))/cos(c + d*x)^2,x)

[Out]

int(((A + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^(3/2))/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((a+b*cos(d*x+c))**(3/2)*(A+C*cos(d*x+c)**2)*sec(d*x+c)**2,x)

[Out]

Timed out

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