3.338 \(\int \frac{(A+B \cos (c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx\)
Optimal. Leaf size=349 \[ \frac{2 b \left (7 a^2 A b-4 a^3 B-3 A b^3\right ) \sin (c+d x)}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt{a+b \cos (c+d x)}}+\frac{2 b (A b-a B) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}+\frac{2 (A b-a B) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 a d \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}-\frac{2 \left (7 a^2 A b-4 a^3 B-3 A b^3\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 A \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{a^2 d \sqrt{a+b \cos (c+d x)}} \]
[Out]
(-2*(7*a^2*A*b - 3*A*b^3 - 4*a^3*B)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(3*a^2*(a^
2 - b^2)^2*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + (2*(A*b - a*B)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF
[(c + d*x)/2, (2*b)/(a + b)])/(3*a*(a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]]) + (2*A*Sqrt[(a + b*Cos[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*b*(A*b - a*B)*Sin[c +
d*x])/(3*a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^(3/2)) + (2*b*(7*a^2*A*b - 3*A*b^3 - 4*a^3*B)*Sin[c + d*x])/(3*a
^2*(a^2 - b^2)^2*d*Sqrt[a + b*Cos[c + d*x]])
________________________________________________________________________________________
Rubi [A] time = 1.09905, antiderivative size = 349, normalized size of antiderivative
= 1., number of steps used = 10, number of rules used = 10, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\)
= 0.323, Rules used = {3000, 3055, 3059, 2655, 2653, 3002, 2663, 2661, 2807, 2805}
\[ \frac{2 b \left (7 a^2 A b-4 a^3 B-3 A b^3\right ) \sin (c+d x)}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt{a+b \cos (c+d x)}}+\frac{2 b (A b-a B) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}+\frac{2 (A b-a B) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 a d \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}-\frac{2 \left (7 a^2 A b-4 a^3 B-3 A b^3\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 A \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{a^2 d \sqrt{a+b \cos (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Int[((A + B*Cos[c + d*x])*Sec[c + d*x])/(a + b*Cos[c + d*x])^(5/2),x]
[Out]
(-2*(7*a^2*A*b - 3*A*b^3 - 4*a^3*B)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(3*a^2*(a^
2 - b^2)^2*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + (2*(A*b - a*B)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF
[(c + d*x)/2, (2*b)/(a + b)])/(3*a*(a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]]) + (2*A*Sqrt[(a + b*Cos[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*b*(A*b - a*B)*Sin[c +
d*x])/(3*a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^(3/2)) + (2*b*(7*a^2*A*b - 3*A*b^3 - 4*a^3*B)*Sin[c + d*x])/(3*a
^2*(a^2 - b^2)^2*d*Sqrt[a + b*Cos[c + d*x]])
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 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{(A+B \cos (c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx &=\frac{2 b (A b-a B) \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 (A b-a B) \cos (c+d x)+\frac{1}{2} b (A b-a B) \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 b (A b-a B) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac{2 b \left (7 a^2 A b-3 A b^3-4 a^3 B\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}{4} a \left (6 a^2 A b-2 A b^3-3 a^3 B-a b^2 B\right ) \cos (c+d x)-\frac{1}{4} b \left (7 a^2 A b-3 A b^3-4 a^3 B\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 b (A b-a B) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac{2 b \left (7 a^2 A b-3 A b^3-4 a^3 B\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 b \left (a^2-b^2\right ) (A b-a B) \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 (7 a^2 A b-3 A b^3-4 a^3 B\right ) \int \sqrt{a+b \cos (c+d x)} \, dx}{3 a^2 \left (a^2-b^2\right )^2}\\ &=\frac{2 b (A b-a B) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac{2 b \left (7 a^2 A b-3 A b^3-4 a^3 B\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{(A b-a B) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx}{3 a \left (a^2-b^2\right )}-\frac{\left (\left (7 a^2 A b-3 A b^3-4 a^3 B\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 \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}\\ &=-\frac{2 \left (7 a^2 A b-3 A b^3-4 a^3 B\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 \left (a^2-b^2\right )^2 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 b (A b-a B) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac{2 b \left (7 a^2 A b-3 A b^3-4 a^3 B\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 ((A b-a B) \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 \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}\\ &=-\frac{2 \left (7 a^2 A b-3 A b^3-4 a^3 B\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 \left (a^2-b^2\right )^2 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 (A b-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 )}{3 a \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 b (A b-a B) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac{2 b \left (7 a^2 A b-3 A b^3-4 a^3 B\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 [C] time = 6.72774, size = 743, normalized size = 2.13 \[ \frac{\cos (c+d x) \sqrt{a+b \cos (c+d x)} (A \sec (c+d x)+B) \left (-\frac{2 \left (a b B \sin (c+d x)-A b^2 \sin (c+d x)\right )}{3 a \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac{2 \left (-7 a^2 A b^2 \sin (c+d x)+4 a^3 b B \sin (c+d x)+3 A b^4 \sin (c+d x)\right )}{3 a^2 \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}\right )}{d (A+B \cos (c+d x))}+\frac{\cos (c+d x) (A \sec (c+d x)+B) \left (\frac{2 \left (-12 a^3 A b+2 a^2 b^2 B+6 a^4 B+4 a A b^3\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{\sqrt{a+b \cos (c+d x)}}+\frac{2 \left (-19 a^2 A b^2+6 a^4 A+4 a^3 b B+9 A b^4\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{\sqrt{a+b \cos (c+d x)}}-\frac{2 i \left (-7 a^2 A b^2+4 a^3 b B+3 A b^4\right ) \sin (c+d x) \cos (2 (c+d x)) \sqrt{\frac{b-b \cos (c+d x)}{a+b}} \sqrt{-\frac{b \cos (c+d x)+b}{a-b}} \left (2 a (a-b) E\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \cos (c+d x)}\right )|\frac{a+b}{a-b}\right )+b \left (2 a F\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \cos (c+d x)}\right )|\frac{a+b}{a-b}\right )-b \Pi \left (\frac{a+b}{a};i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \cos (c+d x)}\right )|\frac{a+b}{a-b}\right )\right )\right )}{a \sqrt{-\frac{1}{a+b}} \sqrt{1-\cos ^2(c+d x)} \sqrt{-\frac{a^2-2 a (a+b \cos (c+d x))+(a+b \cos (c+d x))^2-b^2}{b^2}} \left (2 a^2-4 a (a+b \cos (c+d x))+2 (a+b \cos (c+d x))^2-b^2\right )}\right )}{6 a^2 d (a-b)^2 (a+b)^2 (A+B \cos (c+d x))} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[((A + B*Cos[c + d*x])*Sec[c + d*x])/(a + b*Cos[c + d*x])^(5/2),x]
[Out]
(Cos[c + d*x]*(B + A*Sec[c + d*x])*((2*(-12*a^3*A*b + 4*a*A*b^3 + 6*a^4*B + 2*a^2*b^2*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*(6*a^4*A - 19*a^2*A*b^2 + 9
*A*b^4 + 4*a^3*b*B)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/Sqrt[a + b*C
os[c + d*x]] - ((2*I)*(-7*a^2*A*b^2 + 3*A*b^4 + 4*a^3*b*B)*Sqrt[(b - b*Cos[c + d*x])/(a + b)]*Sqrt[-((b + b*Co
s[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))))/(6*a^2*(a
- b)^2*(a + b)^2*d*(A + B*Cos[c + d*x])) + (Cos[c + d*x]*Sqrt[a + b*Cos[c + d*x]]*(B + A*Sec[c + d*x])*((-2*(
-(A*b^2*Sin[c + d*x]) + a*b*B*Sin[c + d*x]))/(3*a*(a^2 - b^2)*(a + b*Cos[c + d*x])^2) - (2*(-7*a^2*A*b^2*Sin[c
+ d*x] + 3*A*b^4*Sin[c + d*x] + 4*a^3*b*B*Sin[c + d*x]))/(3*a^2*(a^2 - b^2)^2*(a + b*Cos[c + d*x]))))/(d*(A +
B*Cos[c + d*x]))
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Maple [B] time = 14.143, size = 854, normalized size = 2.5 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+B*cos(d*x+c))*sec(d*x+c)/(a+b*cos(d*x+c))^(5/2),x)
[Out]
-(-(-2*b*cos(1/2*d*x+1/2*c)^2-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*A*b/a^2/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*
b*cos(1/2*d*x+1/2*c)^2+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)*Elliptic
Pi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))+2*(-A*b+B*a)/a*(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*sin(1/2*d*x+1/2*c)^
2/(a-b)^2/(a+b)^2*cos(1/2*d*x+1/2*c)*a/(-(-2*b*cos(1/2*d*x+1/2*c)^2-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*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*b*s
in(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)/(a+b)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+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))-EllipticE(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 (B \cos \left (d x + c\right ) + 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+B*cos(d*x+c))*sec(d*x+c)/(a+b*cos(d*x+c))^(5/2),x, algorithm="maxima")
[Out]
integrate((B*cos(d*x + c) + 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+B*cos(d*x+c))*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+B*cos(d*x+c))*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 (B \cos \left (d x + c\right ) + 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+B*cos(d*x+c))*sec(d*x+c)/(a+b*cos(d*x+c))^(5/2),x, algorithm="giac")
[Out]
integrate((B*cos(d*x + c) + A)*sec(d*x + c)/(b*cos(d*x + c) + a)^(5/2), x)