3.667 \(\int \frac{(A+C \cos ^2(c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx\)
Optimal. Leaf size=375 \[ -\frac{2 \left (-a^2 b^2 (7 A+3 C)+a^4 (-C)+3 A b^4\right ) \sin (c+d x)}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}+\frac{2 \left (a^2 C+A b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 a b d \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (-a^2 b^2 (7 A+3 C)+a^4 (-C)+3 A b^4\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 a^2 b d \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 A \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{a^2 d \sqrt{a+b \cos (c+d x)}} \]
[Out]
(2*(3*A*b^4 - a^4*C - a^2*b^2*(7*A + 3*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(3*
a^2*b*(a^2 - b^2)^2*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + (2*(A*b^2 + a^2*C)*Sqrt[(a + b*Cos[c + d*x])/(a +
b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(3*a*b*(a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]]) + (2*A*Sqrt[(a + b*C
os[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/(a^2*d*Sqrt[a + b*Cos[c + d*x]]) + (2*(A*b^2
+ a^2*C)*Sin[c + d*x])/(3*a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^(3/2)) - (2*(3*A*b^4 - a^4*C - a^2*b^2*(7*A + 3
*C))*Sin[c + d*x])/(3*a^2*(a^2 - b^2)^2*d*Sqrt[a + b*Cos[c + d*x]])
________________________________________________________________________________________
Rubi [A] time = 1.08117, antiderivative size = 375, normalized size of antiderivative
= 1., number of steps used = 10, number of rules used = 10, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\)
= 0.303, Rules used = {3056, 3055, 3059, 2655, 2653, 3002, 2663, 2661, 2807, 2805}
\[ -\frac{2 \left (-a^2 b^2 (7 A+3 C)+a^4 (-C)+3 A b^4\right ) \sin (c+d x)}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}+\frac{2 \left (a^2 C+A b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 a b d \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (-a^2 b^2 (7 A+3 C)+a^4 (-C)+3 A b^4\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 a^2 b d \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 A \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{a^2 d \sqrt{a+b \cos (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x])/(a + b*Cos[c + d*x])^(5/2),x]
[Out]
(2*(3*A*b^4 - a^4*C - a^2*b^2*(7*A + 3*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(3*
a^2*b*(a^2 - b^2)^2*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + (2*(A*b^2 + a^2*C)*Sqrt[(a + b*Cos[c + d*x])/(a +
b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(3*a*b*(a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]]) + (2*A*Sqrt[(a + b*C
os[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/(a^2*d*Sqrt[a + b*Cos[c + d*x]]) + (2*(A*b^2
+ a^2*C)*Sin[c + d*x])/(3*a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^(3/2)) - (2*(3*A*b^4 - a^4*C - a^2*b^2*(7*A + 3
*C))*Sin[c + d*x])/(3*a^2*(a^2 - b^2)^2*d*Sqrt[a + b*Cos[c + d*x]])
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 (c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx &=\frac{2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac{2 \int \frac{\left (\frac{3}{2} A \left (a^2-b^2\right )-\frac{3}{2} a b (A+C) \cos (c+d x)+\frac{1}{2} \left (A b^2+a^2 C\right ) \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx}{3 a \left (a^2-b^2\right )}\\ &=\frac{2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac{2 \left (3 A b^4-a^4 C-a^2 b^2 (7 A+3 C)\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \cos (c+d x)}}+\frac{4 \int \frac{\left (\frac{3}{4} A \left (a^2-b^2\right )^2+\frac{1}{2} a b \left (A b^2-a^2 (3 A+2 C)\right ) \cos (c+d x)+\frac{1}{4} \left (3 A b^4-a^4 C-a^2 b^2 (7 A+3 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2}\\ &=\frac{2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac{2 \left (3 A b^4-a^4 C-a^2 b^2 (7 A+3 C)\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \cos (c+d x)}}-\frac{4 \int \frac{\left (-\frac{3}{4} A b \left (a^2-b^2\right )^2-\frac{1}{4} a \left (a^2-b^2\right ) \left (A b^2+a^2 C\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{3 a^2 b \left (a^2-b^2\right )^2}+\frac{\left (3 A b^4-a^4 C-a^2 b^2 (7 A+3 C)\right ) \int \sqrt{a+b \cos (c+d x)} \, dx}{3 a^2 b \left (a^2-b^2\right )^2}\\ &=\frac{2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac{2 \left (3 A b^4-a^4 C-a^2 b^2 (7 A+3 C)\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \cos (c+d x)}}+\frac{A \int \frac{\sec (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{a^2}+\frac{\left (A b^2+a^2 C\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx}{3 a b \left (a^2-b^2\right )}+\frac{\left (\left (3 A b^4-a^4 C-a^2 b^2 (7 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}{3 a^2 b \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}\\ &=\frac{2 \left (3 A b^4-a^4 C-a^2 b^2 (7 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 )}{3 a^2 b \left (a^2-b^2\right )^2 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac{2 \left (3 A b^4-a^4 C-a^2 b^2 (7 A+3 C)\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \cos (c+d x)}}+\frac{\left (A \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}{a^2 \sqrt{a+b \cos (c+d x)}}+\frac{\left (\left (A b^2+a^2 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}{3 a b \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}\\ &=\frac{2 \left (3 A b^4-a^4 C-a^2 b^2 (7 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 )}{3 a^2 b \left (a^2-b^2\right )^2 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 \left (A b^2+a^2 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 a b \left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}+\frac{2 A \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 )}{a^2 d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac{2 \left (3 A b^4-a^4 C-a^2 b^2 (7 A+3 C)\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \cos (c+d x)}}\\ \end{align*}
Mathematica [F] time = 36.6464, size = 0, normalized size = 0. \[ \int \frac{\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx \]
Verification is Not applicable to the result.
[In]
Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x])/(a + b*Cos[c + d*x])^(5/2),x]
[Out]
Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x])/(a + b*Cos[c + d*x])^(5/2), x]
________________________________________________________________________________________
Maple [A] time = 1.57, size = 875, normalized size = 2.3 \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)/(a+b*cos(d*x+c))^(5/2),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*(-A*b^2+C*a^2)/a^2/b/sin(1/2*d*x+1/2*c)^2/(-
2*sin(1/2*d*x+1/2*c)^2*b+a+b)/(a^2-b^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)*((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-(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))*b+2*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)-2*A/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)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))+2*(-A*b^2-C*a^2)/a/b*(1/6/b/(a-b)/(a+b)*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)/(cos(1/2*d*x+1/2*c)^2+1/2*(a-b)/b)^2+8/3*b*s
in(1/2*d*x+1/2*c)^2/(a-b)^2/(a+b)^2*cos(1/2*d*x+1/2*c)*a/(-(-2*cos(1/2*d*x+1/2*c)^2*b-a+b)*sin(1/2*d*x+1/2*c)^
2)^(1/2)+(3*a-b)/(3*a^3+3*a^2*b-3*a*b^2-3*b^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))-4/3*a/(a+b)^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)*(EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-Elli
pticE(cos(1/2*d*x+1/2*c),(-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] 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 )}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{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)/(a+b*cos(d*x+c))^(5/2),x, algorithm="maxima")
[Out]
integrate((C*cos(d*x + c)^2 + A)*sec(d*x + c)/(b*cos(d*x + c) + a)^(5/2), x)
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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)/(a+b*cos(d*x+c))^(5/2),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)/(a+b*cos(d*x+c))**(5/2),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 )}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{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)/(a+b*cos(d*x+c))^(5/2),x, algorithm="giac")
[Out]
integrate((C*cos(d*x + c)^2 + A)*sec(d*x + c)/(b*cos(d*x + c) + a)^(5/2), x)