3.1038 \(\int (a+b \cos (c+d x))^{3/2} (a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)) \, dx\)

Optimal. Leaf size=285 \[ \frac {2 b \left (-41 a^2 C+56 a b B+25 b^2 C\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{105 d}-\frac {2 \left (a^2-b^2\right ) \left (-41 a^2 C+56 a b B+25 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (-146 a^3 C+161 a^2 b B+82 a b^2 C+63 b^3 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 b (7 b B-2 a C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{35 d}+\frac {2 b C \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 d} \]

[Out]

2/35*b*(7*B*b-2*C*a)*(a+b*cos(d*x+c))^(3/2)*sin(d*x+c)/d+2/7*b*C*(a+b*cos(d*x+c))^(5/2)*sin(d*x+c)/d+2/105*b*(
56*B*a*b-41*C*a^2+25*C*b^2)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/d+2/105*(161*B*a^2*b+63*B*b^3-146*C*a^3+82*C*a*b
^2)*(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)-2/105*(a^2-b^2)*(56*B*a*b-41*C*a^2+25*C*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)/d/(a+b*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.66, antiderivative size = 285, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {3015, 2753, 2752, 2663, 2661, 2655, 2653} \[ \frac {2 b \left (-41 a^2 C+56 a b B+25 b^2 C\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{105 d}-\frac {2 \left (a^2-b^2\right ) \left (-41 a^2 C+56 a b B+25 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (161 a^2 b B-146 a^3 C+82 a b^2 C+63 b^3 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 b (7 b B-2 a C) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{35 d}+\frac {2 b C \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(161*a^2*b*B + 63*b^3*B - 146*a^3*C + 82*a*b^2*C)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a
+ b)])/(105*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - (2*(a^2 - b^2)*(56*a*b*B - 41*a^2*C + 25*b^2*C)*Sqrt[(a +
b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(105*d*Sqrt[a + b*Cos[c + d*x]]) + (2*b*(56*a*
b*B - 41*a^2*C + 25*b^2*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(105*d) + (2*b*(7*b*B - 2*a*C)*(a + b*Cos[c
+ d*x])^(3/2)*Sin[c + d*x])/(35*d) + (2*b*C*(a + b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*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 2752

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 2753

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)*Simp[
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]

Rule 3015

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> Dist[1/b^2, Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*B - a*C + b*C*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rubi steps

\begin {align*} \int (a+b \cos (c+d x))^{3/2} \left (a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)\right ) \, dx &=\frac {\int (a+b \cos (c+d x))^{5/2} \left (b^2 (b B-a C)+b^3 C \cos (c+d x)\right ) \, dx}{b^2}\\ &=\frac {2 b C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac {2 \int (a+b \cos (c+d x))^{3/2} \left (\frac {1}{2} b^2 \left (7 a b B-7 a^2 C+5 b^2 C\right )+\frac {1}{2} b^3 (7 b B-2 a C) \cos (c+d x)\right ) \, dx}{7 b^2}\\ &=\frac {2 b (7 b B-2 a C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 b C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac {4 \int \sqrt {a+b \cos (c+d x)} \left (\frac {1}{4} b^2 \left (35 a^2 b B+21 b^3 B-35 a^3 C+19 a b^2 C\right )+\frac {1}{4} b^3 \left (56 a b B-41 a^2 C+25 b^2 C\right ) \cos (c+d x)\right ) \, dx}{35 b^2}\\ &=\frac {2 b \left (56 a b B-41 a^2 C+25 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 d}+\frac {2 b (7 b B-2 a C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 b C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac {8 \int \frac {\frac {1}{8} b^2 \left (105 a^3 b B+119 a b^3 B-105 a^4 C+16 a^2 b^2 C+25 b^4 C\right )+\frac {1}{8} b^3 \left (161 a^2 b B+63 b^3 B-146 a^3 C+82 a b^2 C\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{105 b^2}\\ &=\frac {2 b \left (56 a b B-41 a^2 C+25 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 d}+\frac {2 b (7 b B-2 a C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 b C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}-\frac {1}{105} \left (\left (a^2-b^2\right ) \left (56 a b B-41 a^2 C+25 b^2 C\right )\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx+\frac {1}{105} \left (161 a^2 b B+63 b^3 B-146 a^3 C+82 a b^2 C\right ) \int \sqrt {a+b \cos (c+d x)} \, dx\\ &=\frac {2 b \left (56 a b B-41 a^2 C+25 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 d}+\frac {2 b (7 b B-2 a C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 b C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac {\left (\left (161 a^2 b B+63 b^3 B-146 a^3 C+82 a b^2 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}{105 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (\left (a^2-b^2\right ) \left (56 a b B-41 a^2 C+25 b^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}{105 \sqrt {a+b \cos (c+d x)}}\\ &=\frac {2 \left (161 a^2 b B+63 b^3 B-146 a^3 C+82 a b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \left (a^2-b^2\right ) \left (56 a b B-41 a^2 C+25 b^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 )}{105 d \sqrt {a+b \cos (c+d x)}}+\frac {2 b \left (56 a b B-41 a^2 C+25 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 d}+\frac {2 b (7 b B-2 a C) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 b C (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}\\ \end {align*}

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Mathematica [A]  time = 1.36, size = 259, normalized size = 0.91 \[ \frac {b \sin (c+d x) (a+b \cos (c+d x)) \left (-64 a^2 C+6 b (8 a C+7 b B) \cos (c+d x)+154 a b B+15 b^2 C \cos (2 (c+d x))+65 b^2 C\right )+2 \left (-146 a^3 C+161 a^2 b B+82 a b^2 C+63 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \left ((a+b) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-a F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )\right )+2 \left (-105 a^4 C+105 a^3 b B+16 a^2 b^2 C+119 a b^3 B+25 b^4 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 d \sqrt {a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(105*a^3*b*B + 119*a*b^3*B - 105*a^4*C + 16*a^2*b^2*C + 25*b^4*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*Ellipt
icF[(c + d*x)/2, (2*b)/(a + b)] + 2*(161*a^2*b*B + 63*b^3*B - 146*a^3*C + 82*a*b^2*C)*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*b*B - 64*a^2*C + 65*b^2*C + 6*b*(7*b*B + 8*a*C)*Cos[c + d*x] + 15*b^2*C*Cos[2*(c + d*x)
])*Sin[c + d*x])/(105*d*Sqrt[a + b*Cos[c + d*x]])

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fricas [F]  time = 1.14, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b^{3} \cos \left (d x + c\right )^{3} - C a^{3} + B a^{2} b + {\left (C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{2} - {\left (C a^{2} b - 2 \, B a b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {b \cos \left (d x + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*b^3*cos(d*x + c)^3 - C*a^3 + B*a^2*b + (C*a*b^2 + B*b^3)*cos(d*x + c)^2 - (C*a^2*b - 2*B*a*b^2)*co
s(d*x + c))*sqrt(b*cos(d*x + c) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 3.32, size = 1302, normalized size = 4.57 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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*C*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1
/2*c)^8+(-168*B*b^4-312*C*a*b^3-360*C*b^4)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(392*B*a*b^3+168*B*b^4-32*C
*a^2*b^2+312*C*a*b^3+280*C*b^4)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-154*B*a^2*b^2-196*B*a*b^3-42*B*b^4+6
4*C*a^3*b+16*C*a^2*b^2-128*C*a*b^3-80*C*b^4)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+161*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-161*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+63*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*b^3-63*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))*b^4-56*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^3*b+56*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)*
EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^3-146*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^4+146*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
^3*b+82*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*b^2-82*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^3+41*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))*a^4-66*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))*a^2*b^2+25*C*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)))/(-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 \[ \int {\left (C b^{2} \cos \left (d x + c\right )^{2} + B b^{2} \cos \left (d x + c\right ) - C a^{2} + B a b\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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