3.4.13 \(\int (a+b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx\) [313]

Optimal. Leaf size=288 \[ \frac {2 \left (161 a^2 A b+63 A b^3+15 a^3 B+145 a b^2 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \left (a^2-b^2\right ) \left (56 a A b+15 a^2 B+25 b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (56 a A b+15 a^2 B+25 b^2 B\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 d}+\frac {2 (7 A b+5 a B) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 B (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d} \]

[Out]

2/35*(7*A*b+5*B*a)*(a+b*cos(d*x+c))^(3/2)*sin(d*x+c)/d+2/7*B*(a+b*cos(d*x+c))^(5/2)*sin(d*x+c)/d+2/105*(56*A*a
*b+15*B*a^2+25*B*b^2)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/d+2/105*(161*A*a^2*b+63*A*b^3+15*B*a^3+145*B*a*b^2)*(c
os(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)/b/d/((a+b*cos(d*x+c))/(a+b))^(1/2)-2/105*(a^2-b^2)*(56*A*a*b+15*B*a^2+25*B*b^2)*(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)/b/d/(a+b*cos(d*x+c))^(1/2)

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Rubi [A]
time = 0.31, antiderivative size = 288, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2832, 2831, 2742, 2740, 2734, 2732} \begin {gather*} \frac {2 \left (15 a^2 B+56 a A b+25 b^2 B\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{105 d}-\frac {2 \left (a^2-b^2\right ) \left (15 a^2 B+56 a A b+25 b^2 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 )}{105 b d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (15 a^3 B+161 a^2 A b+145 a b^2 B+63 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 )}{105 b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 (5 a B+7 A b) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{35 d}+\frac {2 B \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x]),x]

[Out]

(2*(161*a^2*A*b + 63*A*b^3 + 15*a^3*B + 145*a*b^2*B)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a
+ b)])/(105*b*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - (2*(a^2 - b^2)*(56*a*A*b + 15*a^2*B + 25*b^2*B)*Sqrt[(a
+ b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(105*b*d*Sqrt[a + b*Cos[c + d*x]]) + (2*(56*
a*A*b + 15*a^2*B + 25*b^2*B)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(105*d) + (2*(7*A*b + 5*a*B)*(a + b*Cos[c
+ d*x])^(3/2)*Sin[c + d*x])/(35*d) + (2*B*(a + b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*d)

Rule 2732

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

Rule 2734

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

Rule 2740

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

Rule 2742

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

Rule 2831

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

Rule 2832

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d
)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(f*(m + 1))), x] + Dist[1/(m + 1), Int[(a + b*Sin[e + f*x])^(m - 1)*Sim
p[b*d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && IntegerQ[2*m]

Rubi steps

\begin {align*} \int (a+b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx &=\frac {2 B (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac {2}{7} \int (a+b \cos (c+d x))^{3/2} \left (\frac {1}{2} (7 a A+5 b B)+\frac {1}{2} (7 A b+5 a B) \cos (c+d x)\right ) \, dx\\ &=\frac {2 (7 A b+5 a B) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 B (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac {4}{35} \int \sqrt {a+b \cos (c+d x)} \left (\frac {1}{4} \left (35 a^2 A+21 A b^2+40 a b B\right )+\frac {1}{4} \left (56 a A b+15 a^2 B+25 b^2 B\right ) \cos (c+d x)\right ) \, dx\\ &=\frac {2 \left (56 a A b+15 a^2 B+25 b^2 B\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 d}+\frac {2 (7 A b+5 a B) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 B (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac {8}{105} \int \frac {\frac {1}{8} \left (105 a^3 A+119 a A b^2+135 a^2 b B+25 b^3 B\right )+\frac {1}{8} \left (161 a^2 A b+63 A b^3+15 a^3 B+145 a b^2 B\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx\\ &=\frac {2 \left (56 a A b+15 a^2 B+25 b^2 B\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 d}+\frac {2 (7 A b+5 a B) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 B (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}-\frac {\left (\left (a^2-b^2\right ) \left (56 a A b+15 a^2 B+25 b^2 B\right )\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{105 b}+\frac {\left (161 a^2 A b+63 A b^3+15 a^3 B+145 a b^2 B\right ) \int \sqrt {a+b \cos (c+d x)} \, dx}{105 b}\\ &=\frac {2 \left (56 a A b+15 a^2 B+25 b^2 B\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 d}+\frac {2 (7 A b+5 a B) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 B (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac {\left (\left (161 a^2 A b+63 A b^3+15 a^3 B+145 a b^2 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}{105 b \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (\left (a^2-b^2\right ) \left (56 a A b+15 a^2 B+25 b^2 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}{105 b \sqrt {a+b \cos (c+d x)}}\\ &=\frac {2 \left (161 a^2 A b+63 A b^3+15 a^3 B+145 a b^2 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \left (a^2-b^2\right ) \left (56 a A b+15 a^2 B+25 b^2 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 )}{105 b d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (56 a A b+15 a^2 B+25 b^2 B\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 d}+\frac {2 (7 A b+5 a B) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 B (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}\\ \end {align*}

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Mathematica [A]
time = 1.14, size = 254, normalized size = 0.88 \begin {gather*} \frac {2 b \left (105 a^3 A+119 a A b^2+135 a^2 b B+25 b^3 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 )+2 \left (161 a^2 A b+63 A b^3+15 a^3 B+145 a b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \left ((a+b) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-a F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )\right )+b (a+b \cos (c+d x)) \left (154 a A b+90 a^2 B+65 b^2 B+6 b (7 A b+15 a B) \cos (c+d x)+15 b^2 B \cos (2 (c+d x))\right ) \sin (c+d x)}{105 b d \sqrt {a+b \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x]),x]

[Out]

(2*b*(105*a^3*A + 119*a*A*b^2 + 135*a^2*b*B + 25*b^3*B)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)
/2, (2*b)/(a + b)] + 2*(161*a^2*A*b + 63*A*b^3 + 15*a^3*B + 145*a*b^2*B)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*((
a + b)*EllipticE[(c + d*x)/2, (2*b)/(a + b)] - a*EllipticF[(c + d*x)/2, (2*b)/(a + b)]) + b*(a + b*Cos[c + d*x
])*(154*a*A*b + 90*a^2*B + 65*b^2*B + 6*b*(7*A*b + 15*a*B)*Cos[c + d*x] + 15*b^2*B*Cos[2*(c + d*x)])*Sin[c + d
*x])/(105*b*d*Sqrt[a + b*Cos[c + d*x]])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1304\) vs. \(2(322)=644\).
time = 0.38, size = 1305, normalized size = 4.53

method	result	size
default	\(\text {Expression too large to display}\)	\(1305\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-2/105*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(240*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c
)^8*b^4+(-168*A*b^4-480*B*a*b^3-360*B*b^4)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(392*A*a*b^3+168*A*b^4+360*
B*a^2*b^2+480*B*a*b^3+280*B*b^4)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-154*A*a^2*b^2-196*A*a*b^3-42*A*b^4-
90*B*a^3*b-180*B*a^2*b^2-170*B*a*b^3-80*B*b^4)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-56*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^3*b+56*a*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))*b^3+161*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^3*b-161*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*b^2+63*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^3-63*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^4-15*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)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^4-10*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)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a
^2*b^2+25*B*b^4*(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))+15*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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^4-15*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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3*b+145*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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a
-b))^(1/2))*a^2*b^2-145*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)*Ell
ipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^3)/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)/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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^(5/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.15, size = 562, normalized size = 1.95 \begin {gather*} \frac {\sqrt {2} {\left (30 i \, B a^{4} + 7 i \, A a^{3} b - 115 i \, B a^{2} b^{2} - 231 i \, A a b^{3} - 75 i \, B b^{4}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right ) + \sqrt {2} {\left (-30 i \, B a^{4} - 7 i \, A a^{3} b + 115 i \, B a^{2} b^{2} + 231 i \, A a b^{3} + 75 i \, B b^{4}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right ) - 3 \, \sqrt {2} {\left (-15 i \, B a^{3} b - 161 i \, A a^{2} b^{2} - 145 i \, B a b^{3} - 63 i \, A b^{4}\right )} \sqrt {b} {\rm weierstrassZeta}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right )\right ) - 3 \, \sqrt {2} {\left (15 i \, B a^{3} b + 161 i \, A a^{2} b^{2} + 145 i \, B a b^{3} + 63 i \, A b^{4}\right )} \sqrt {b} {\rm weierstrassZeta}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right )\right ) + 6 \, {\left (15 \, B b^{4} \cos \left (d x + c\right )^{2} + 45 \, B a^{2} b^{2} + 77 \, A a b^{3} + 25 \, B b^{4} + 3 \, {\left (15 \, B a b^{3} + 7 \, A b^{4}\right )} \cos \left (d x + c\right )\right )} \sqrt {b \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{315 \, b^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)),x, algorithm="fricas")

[Out]

1/315*(sqrt(2)*(30*I*B*a^4 + 7*I*A*a^3*b - 115*I*B*a^2*b^2 - 231*I*A*a*b^3 - 75*I*B*b^4)*sqrt(b)*weierstrassPI
nverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*a)
/b) + sqrt(2)*(-30*I*B*a^4 - 7*I*A*a^3*b + 115*I*B*a^2*b^2 + 231*I*A*a*b^3 + 75*I*B*b^4)*sqrt(b)*weierstrassPI
nverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x + c) + 2*a)
/b) - 3*sqrt(2)*(-15*I*B*a^3*b - 161*I*A*a^2*b^2 - 145*I*B*a*b^3 - 63*I*A*b^4)*sqrt(b)*weierstrassZeta(4/3*(4*
a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a
*b^2)/b^3, 1/3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*a)/b)) - 3*sqrt(2)*(15*I*B*a^3*b + 161*I*A*a^2*b^2 +
145*I*B*a*b^3 + 63*I*A*b^4)*sqrt(b)*weierstrassZeta(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, wei
erstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x +
c) + 2*a)/b)) + 6*(15*B*b^4*cos(d*x + c)^2 + 45*B*a^2*b^2 + 77*A*a*b^3 + 25*B*b^4 + 3*(15*B*a*b^3 + 7*A*b^4)*
cos(d*x + c))*sqrt(b*cos(d*x + c) + a)*sin(d*x + c))/(b^2*d)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c)),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3062 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*cos(c + d*x))*(a + b*cos(c + d*x))^(5/2),x)

[Out]

int((A + B*cos(c + d*x))*(a + b*cos(c + d*x))^(5/2), x)

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