\(\int (a+b \cos (c+d x))^{3/2} (B \cos (c+d x)+C \cos ^2(c+d x)) \sec (c+d x) \, dx\) [822]

Optimal result

Integrand size = 40, antiderivative size = 225 \[ \int (a+b \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {2 \left (20 a b B+3 a^2 C+9 b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \left (a^2-b^2\right ) (5 b B+3 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{15 b d \sqrt {a+b \cos (c+d x)}}+\frac {2 (5 b B+3 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d} \]

[Out]

2/5*C*(a+b*cos(d*x+c))^(3/2)*sin(d*x+c)/d+2/15*(5*B*b+3*C*a)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/d+2/15*(20*B*a*
b+3*C*a^2+9*C*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)/b/d/((a+b*cos(d*x+c))/(a+b))^(1/2)-2/15*(a^2-b^2)*(5*B*b+3*C*a)*(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)

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {3108, 2832, 2831, 2742, 2740, 2734, 2732} \[ \int (a+b \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=-\frac {2 \left (a^2-b^2\right ) (3 a C+5 b B) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{15 b d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (3 a^2 C+20 a b B+9 b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 (3 a C+5 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{15 d}+\frac {2 C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d} \]

[In]

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

[Out]

(2*(20*a*b*B + 3*a^2*C + 9*b^2*C)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(15*b*d*Sqrt
[(a + b*Cos[c + d*x])/(a + b)]) - (2*(a^2 - b^2)*(5*b*B + 3*a*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[
(c + d*x)/2, (2*b)/(a + b)])/(15*b*d*Sqrt[a + b*Cos[c + d*x]]) + (2*(5*b*B + 3*a*C)*Sqrt[a + b*Cos[c + d*x]]*S
in[c + d*x])/(15*d) + (2*C*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*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]

Rule 3108

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((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)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rubi steps \begin{align*} \text {integral}& = \int (a+b \cos (c+d x))^{3/2} (B+C \cos (c+d x)) \, dx \\ & = \frac {2 C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {2}{5} \int \sqrt {a+b \cos (c+d x)} \left (\frac {1}{2} (5 a B+3 b C)+\frac {1}{2} (5 b B+3 a C) \cos (c+d x)\right ) \, dx \\ & = \frac {2 (5 b B+3 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {4}{15} \int \frac {\frac {1}{4} \left (15 a^2 B+5 b^2 B+12 a b C\right )+\frac {1}{4} \left (20 a b B+3 a^2 C+9 b^2 C\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx \\ & = \frac {2 (5 b B+3 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}-\frac {\left (\left (a^2-b^2\right ) (5 b B+3 a C)\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{15 b}+\frac {\left (20 a b B+3 a^2 C+9 b^2 C\right ) \int \sqrt {a+b \cos (c+d x)} \, dx}{15 b} \\ & = \frac {2 (5 b B+3 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {\left (\left (20 a b B+3 a^2 C+9 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}{15 b \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (\left (a^2-b^2\right ) (5 b B+3 a C) \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}{15 b \sqrt {a+b \cos (c+d x)}} \\ & = \frac {2 \left (20 a b B+3 a^2 C+9 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 )}{15 b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \left (a^2-b^2\right ) (5 b B+3 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{15 b d \sqrt {a+b \cos (c+d x)}}+\frac {2 (5 b B+3 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.47 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.90 \[ \int (a+b \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {2 \left (b \left (15 a^2 B+5 b^2 B+12 a b C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )+\left (20 a b B+3 a^2 C+9 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \left ((a+b) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-a \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )\right )+b (a+b \cos (c+d x)) (5 b B+6 a C+3 b C \cos (c+d x)) \sin (c+d x)\right )}{15 b d \sqrt {a+b \cos (c+d x)}} \]

[In]

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

[Out]

(2*(b*(15*a^2*B + 5*b^2*B + 12*a*b*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)]
+ (20*a*b*B + 3*a^2*C + 9*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])*(5*b*B + 6*a*C + 3*b*C*Cos[c + d*x])
*Sin[c + d*x]))/(15*b*d*Sqrt[a + b*Cos[c + d*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(992\) vs. \(2(263)=526\).

Time = 12.59 (sec) , antiderivative size = 993, normalized size of antiderivative = 4.41

method	result	size
default	\(\text {Expression too large to display}\)	\(993\)
parts	\(\text {Expression too large to display}\)	\(1115\)

[In]

int((a+cos(d*x+c)*b)^(3/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c),x,method=_RETURNVERBOSE)

[Out]

-2/15*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-24*C*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*
c)*b^3+(20*B*b^3+36*C*a*b^2+24*C*b^3)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-10*B*a*b^2-10*B*b^3-12*C*a^2*b
-18*C*a*b^2-6*C*b^3)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-5*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+5*B*b^3*(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))+20*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-20*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^2-3*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^3+3*a*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))*b^2+3*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)*Ellipt
icE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3-3*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+9*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^2-9*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))*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*b*sin(1/2*d*x+1/2*c)^2+a+b)^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.13 (sec) , antiderivative size = 493, normalized size of antiderivative = 2.19 \[ \int (a+b \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {\sqrt {2} {\left (6 i \, C a^{3} - 5 i \, B a^{2} b - 18 i \, C a b^{2} - 15 i \, B b^{3}\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 (-6 i \, C a^{3} + 5 i \, B a^{2} b + 18 i \, C a b^{2} + 15 i \, B b^{3}\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 (-3 i \, C a^{2} b - 20 i \, B a b^{2} - 9 i \, C b^{3}\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 (3 i \, C a^{2} b + 20 i \, B a b^{2} + 9 i \, C b^{3}\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 (3 \, C b^{3} \cos \left (d x + c\right ) + 6 \, C a b^{2} + 5 \, B b^{3}\right )} \sqrt {b \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{45 \, b^{2} d} \]

[In]

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

[Out]

1/45*(sqrt(2)*(6*I*C*a^3 - 5*I*B*a^2*b - 18*I*C*a*b^2 - 15*I*B*b^3)*sqrt(b)*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) + sqrt(2)*(-6*I*C
*a^3 + 5*I*B*a^2*b + 18*I*C*a*b^2 + 15*I*B*b^3)*sqrt(b)*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)*(-3*I*C*a^2*b - 20*I*B*a*
b^2 - 9*I*C*b^3)*sqrt(b)*weierstrassZeta(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, weierstrassPInv
erse(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)*(3*I*C*a^2*b + 20*I*B*a*b^2 + 9*I*C*b^3)*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)) + 6*(3*C*b^3*cos(d*x + c) + 6*C*a*b^2 + 5*B*b^3)*sqrt(b*cos(d*x + c)
+ a)*sin(d*x + c))/(b^2*d)

Sympy [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int (a+b \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right ) \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int (a+b \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right ) \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2}}{\cos \left (c+d\,x\right )} \,d x \]

[In]

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

[Out]

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