3.310 \(\int (a+b \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx\)
Optimal. Leaf size=375 \[ \frac{\left (16 a^2 A+30 a b B+3 A b^2\right ) \tan (c+d x) \sqrt{a+b \cos (c+d x)}}{24 a d}+\frac{\left (16 a^2 A+42 a b B+17 A b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{24 d \sqrt{a+b \cos (c+d x)}}-\frac{\left (16 a^2 A+30 a b B+3 A b^2\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{24 a d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{\left (12 a^2 A b+8 a^3 B+6 a b^2 B-A b^3\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{8 a d \sqrt{a+b \cos (c+d x)}}+\frac{(6 a B+7 A b) \tan (c+d x) \sec (c+d x) \sqrt{a+b \cos (c+d x)}}{12 d}+\frac{a A \tan (c+d x) \sec ^2(c+d x) \sqrt{a+b \cos (c+d x)}}{3 d} \]
[Out]
-((16*a^2*A + 3*A*b^2 + 30*a*b*B)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(24*a*d*Sqrt
[(a + b*Cos[c + d*x])/(a + b)]) + ((16*a^2*A + 17*A*b^2 + 42*a*b*B)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*Ellipti
cF[(c + d*x)/2, (2*b)/(a + b)])/(24*d*Sqrt[a + b*Cos[c + d*x]]) + ((12*a^2*A*b - A*b^3 + 8*a^3*B + 6*a*b^2*B)*
Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/(8*a*d*Sqrt[a + b*Cos[c + d*x]])
+ ((16*a^2*A + 3*A*b^2 + 30*a*b*B)*Sqrt[a + b*Cos[c + d*x]]*Tan[c + d*x])/(24*a*d) + ((7*A*b + 6*a*B)*Sqrt[a
+ b*Cos[c + d*x]]*Sec[c + d*x]*Tan[c + d*x])/(12*d) + (a*A*Sqrt[a + b*Cos[c + d*x]]*Sec[c + d*x]^2*Tan[c + d*x
])/(3*d)
________________________________________________________________________________________
Rubi [A] time = 1.44994, antiderivative size = 375, normalized size of antiderivative
= 1., number of steps used = 11, number of rules used = 10, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\)
= 0.303, Rules used = {2989, 3055, 3059, 2655, 2653, 3002, 2663, 2661, 2807, 2805}
\[ \frac{\left (16 a^2 A+30 a b B+3 A b^2\right ) \tan (c+d x) \sqrt{a+b \cos (c+d x)}}{24 a d}+\frac{\left (16 a^2 A+42 a b B+17 A b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{24 d \sqrt{a+b \cos (c+d x)}}-\frac{\left (16 a^2 A+30 a b B+3 A b^2\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{24 a d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{\left (12 a^2 A b+8 a^3 B+6 a b^2 B-A b^3\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{8 a d \sqrt{a+b \cos (c+d x)}}+\frac{(6 a B+7 A b) \tan (c+d x) \sec (c+d x) \sqrt{a+b \cos (c+d x)}}{12 d}+\frac{a A \tan (c+d x) \sec ^2(c+d x) \sqrt{a+b \cos (c+d x)}}{3 d} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^4,x]
[Out]
-((16*a^2*A + 3*A*b^2 + 30*a*b*B)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(24*a*d*Sqrt
[(a + b*Cos[c + d*x])/(a + b)]) + ((16*a^2*A + 17*A*b^2 + 42*a*b*B)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*Ellipti
cF[(c + d*x)/2, (2*b)/(a + b)])/(24*d*Sqrt[a + b*Cos[c + d*x]]) + ((12*a^2*A*b - A*b^3 + 8*a^3*B + 6*a*b^2*B)*
Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/(8*a*d*Sqrt[a + b*Cos[c + d*x]])
+ ((16*a^2*A + 3*A*b^2 + 30*a*b*B)*Sqrt[a + b*Cos[c + d*x]]*Tan[c + d*x])/(24*a*d) + ((7*A*b + 6*a*B)*Sqrt[a
+ b*Cos[c + d*x]]*Sec[c + d*x]*Tan[c + d*x])/(12*d) + (a*A*Sqrt[a + b*Cos[c + d*x]]*Sec[c + d*x]^2*Tan[c + d*x
])/(3*d)
Rule 2989
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[((b*c - a*d)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)
*(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 - 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(b*c - a*d)*(B*c - A*d)*(m - 1) + a*d*(a*A*c + b*B*c - (
A*b + a*B)*d)*(n + 1) + (b*(b*d*(B*c - A*d) + a*(A*c*d + B*(c^2 - 2*d^2)))*(n + 1) - a*(b*c - a*d)*(B*c - A*d)
*(n + 2))*Sin[e + f*x] + b*(d*(A*b*c + a*B*c - a*A*d)*(m + n + 1) - b*B*(c^2*m + d^2*(n + 1)))*Sin[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2,
0] && GtQ[m, 1] && LtQ[n, -1]
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 (a+b \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx &=\frac{a A \sqrt{a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{3} \int \frac{\left (\frac{1}{2} a (7 A b+6 a B)+\left (2 a^2 A+3 A b^2+6 a b B\right ) \cos (c+d x)+\frac{3}{2} b (a A+2 b B) \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx\\ &=\frac{(7 A b+6 a B) \sqrt{a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{12 d}+\frac{a A \sqrt{a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{\int \frac{\left (\frac{1}{4} a \left (16 a^2 A+3 A b^2+30 a b B\right )+\frac{1}{2} a \left (13 a A b+6 a^2 B+12 b^2 B\right ) \cos (c+d x)+\frac{1}{4} a b (7 A b+6 a B) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{6 a}\\ &=\frac{\left (16 a^2 A+3 A b^2+30 a b B\right ) \sqrt{a+b \cos (c+d x)} \tan (c+d x)}{24 a d}+\frac{(7 A b+6 a B) \sqrt{a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{12 d}+\frac{a A \sqrt{a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{\int \frac{\left (\frac{3}{8} a \left (12 a^2 A b-A b^3+8 a^3 B+6 a b^2 B\right )+\frac{1}{4} a^2 b (7 A b+6 a B) \cos (c+d x)-\frac{1}{8} a b \left (16 a^2 A+3 A b^2+30 a b B\right ) \cos ^2(c+d x)\right ) \sec (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{6 a^2}\\ &=\frac{\left (16 a^2 A+3 A b^2+30 a b B\right ) \sqrt{a+b \cos (c+d x)} \tan (c+d x)}{24 a d}+\frac{(7 A b+6 a B) \sqrt{a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{12 d}+\frac{a A \sqrt{a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac{\int \frac{\left (-\frac{3}{8} a b \left (12 a^2 A b-A b^3+8 a^3 B+6 a b^2 B\right )-\frac{1}{8} a^2 b \left (16 a^2 A+17 A b^2+42 a b B\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{6 a^2 b}-\frac{\left (16 a^2 A+3 A b^2+30 a b B\right ) \int \sqrt{a+b \cos (c+d x)} \, dx}{48 a}\\ &=\frac{\left (16 a^2 A+3 A b^2+30 a b B\right ) \sqrt{a+b \cos (c+d x)} \tan (c+d x)}{24 a d}+\frac{(7 A b+6 a B) \sqrt{a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{12 d}+\frac{a A \sqrt{a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac{1}{48} \left (-16 a^2 A-17 A b^2-42 a b B\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx+\frac{\left (12 a^2 A b-A b^3+8 a^3 B+6 a b^2 B\right ) \int \frac{\sec (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{16 a}-\frac{\left (\left (16 a^2 A+3 A b^2+30 a b 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}{48 a \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}\\ &=-\frac{\left (16 a^2 A+3 A b^2+30 a b B\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{24 a d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{\left (16 a^2 A+3 A b^2+30 a b B\right ) \sqrt{a+b \cos (c+d x)} \tan (c+d x)}{24 a d}+\frac{(7 A b+6 a B) \sqrt{a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{12 d}+\frac{a A \sqrt{a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac{\left (\left (-16 a^2 A-17 A b^2-42 a b B\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 (\left (12 a^2 A b-A b^3+8 a^3 B+6 a b^2 B\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 \sqrt{a+b \cos (c+d x)}}\\ &=-\frac{\left (16 a^2 A+3 A b^2+30 a b B\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{24 a d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{\left (16 a^2 A+17 A b^2+42 a b B\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{\left (12 a^2 A b-A b^3+8 a^3 B+6 a b^2 B\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 d \sqrt{a+b \cos (c+d x)}}+\frac{\left (16 a^2 A+3 A b^2+30 a b B\right ) \sqrt{a+b \cos (c+d x)} \tan (c+d x)}{24 a d}+\frac{(7 A b+6 a B) \sqrt{a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{12 d}+\frac{a A \sqrt{a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end{align*}
Mathematica [C] time = 6.59094, size = 634, normalized size = 1.69 \[ \frac{\sqrt{a+b \cos (c+d x)} \left (\frac{\sec (c+d x) \left (16 a^2 A \sin (c+d x)+30 a b B \sin (c+d x)+3 A b^2 \sin (c+d x)\right )}{24 a}+\frac{1}{12} \sec ^2(c+d x) (6 a B \sin (c+d x)+7 A b \sin (c+d x))+\frac{1}{3} a A \tan (c+d x) \sec ^2(c+d x)\right )}{d}+\frac{\frac{2 \left (24 a^2 b B+28 a A b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{\sqrt{a+b \cos (c+d x)}}+\frac{2 \left (56 a^2 A b+48 a^3 B+6 a b^2 B-9 A b^3\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{\sqrt{a+b \cos (c+d x)}}-\frac{2 i \left (-16 a^2 A b-30 a b^2 B-3 A b^3\right ) \sin (c+d x) \cos (2 (c+d x)) \sqrt{\frac{b-b \cos (c+d x)}{a+b}} \sqrt{-\frac{b \cos (c+d x)+b}{a-b}} \left (2 a (a-b) E\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \cos (c+d x)}\right )|\frac{a+b}{a-b}\right )+b \left (2 a F\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \cos (c+d x)}\right )|\frac{a+b}{a-b}\right )-b \Pi \left (\frac{a+b}{a};i \sinh ^{-1}\left (\sqrt{-\frac{1}{a+b}} \sqrt{a+b \cos (c+d x)}\right )|\frac{a+b}{a-b}\right )\right )\right )}{a \sqrt{-\frac{1}{a+b}} \sqrt{1-\cos ^2(c+d x)} \sqrt{-\frac{a^2-2 a (a+b \cos (c+d x))+(a+b \cos (c+d x))^2-b^2}{b^2}} \left (2 a^2-4 a (a+b \cos (c+d x))+2 (a+b \cos (c+d x))^2-b^2\right )}}{96 a d} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^4,x]
[Out]
((2*(28*a*A*b^2 + 24*a^2*b*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*(56*a^2*A*b - 9*A*b^3 + 48*a^3*B + 6*a*b^2*B)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*Ellip
ticPi[2, (c + d*x)/2, (2*b)/(a + b)])/Sqrt[a + b*Cos[c + d*x]] - ((2*I)*(-16*a^2*A*b - 3*A*b^3 - 30*a*b^2*B)*S
qrt[(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)*Elliptic
E[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*d) + (Sqrt[a + b*Cos[c + d*x]]*((Sec[c + d*x]^2*(7*A*b*Sin[c + d
*x] + 6*a*B*Sin[c + d*x]))/12 + (Sec[c + d*x]*(16*a^2*A*Sin[c + d*x] + 3*A*b^2*Sin[c + d*x] + 30*a*b*B*Sin[c +
d*x]))/(24*a) + (a*A*Sec[c + d*x]^2*Tan[c + d*x])/3))/d
________________________________________________________________________________________
Maple [B] time = 12.028, size = 2327, normalized size = 6.2 \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))^(3/2)*(A+B*cos(d*x+c))*sec(d*x+c)^4,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*b^2*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*co
s(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)*EllipticPi(c
os(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))+2*b*(A*b+2*B*a)*(-cos(1/2*d*x+1/2*c)/a*(-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*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))-1/2*(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*s
in(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*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)*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*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)*El
lipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2)))+2*a^2*A*(-1/3*cos(1/2*d*x+1/2*c)/a*(-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/a^2*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)/(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*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)*si
n(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*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)*Ellipt
icF(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*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)*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*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)*b*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-5/16/a^2*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)*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*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)*Ellipt
icE(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*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)*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*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)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2)
))+2*a*(2*A*b+B*a)*(-1/2*cos(1/2*d*x+1/2*c)/a*(-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+3/4/a^2*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)/(2*cos(1/2*d*x+1/2*c)^2-1)-1/8*b/a*(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))+3/8/a*(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)*b*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-3/8/a^2*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)*EllipticE(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*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)*EllipticPi(cos(1/2
*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))-3/8/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)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b)
)^(1/2))*b^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
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \cos \left (d x + c\right ) + A\right )}{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \sec \left (d x + c\right )^{4}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c))*sec(d*x+c)^4,x, algorithm="maxima")
[Out]
integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^(3/2)*sec(d*x + c)^4, x)
________________________________________________________________________________________
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))^(3/2)*(A+B*cos(d*x+c))*sec(d*x+c)^4,x, algorithm="fricas")
[Out]
Timed out
________________________________________________________________________________________
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))**(3/2)*(A+B*cos(d*x+c))*sec(d*x+c)**4,x)
[Out]
Timed out
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \cos \left (d x + c\right ) + A\right )}{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \sec \left (d x + c\right )^{4}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c))*sec(d*x+c)^4,x, algorithm="giac")
[Out]
integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^(3/2)*sec(d*x + c)^4, x)