\(\int \frac {(a+a \cos (c+d x))^{3/2} (A+C \cos ^2(c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\) [183]

Optimal result

Integrand size = 37, antiderivative size = 161 \[ \int \frac {(a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {3 a^{3/2} C \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{d}-\frac {a^2 (8 A-3 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a A \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {2 A (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)} \]

[Out]

3*a^(3/2)*C*arcsin(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/d+2/3*A*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d/cos(
d*x+c)^(3/2)-1/3*a^2*(8*A-3*C)*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a+a*cos(d*x+c))^(1/2)+2*a*A*sin(d*x+c)*(a+a*cos(
d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {3123, 3054, 3060, 2853, 222} \[ \int \frac {(a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {3 a^{3/2} C \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {a^2 (8 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d \sqrt {a \cos (c+d x)+a}}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d \sqrt {\cos (c+d x)}} \]

[In]

Int[((a + a*Cos[c + d*x])^(3/2)*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(5/2),x]

[Out]

(3*a^(3/2)*C*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d - (a^2*(8*A - 3*C)*Sqrt[Cos[c + d*x]]*
Sin[c + d*x])/(3*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*A*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[Cos[c + d
*x]]) + (2*A*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2))

Rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 2853

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

Rule 3054

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^2)*(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)*(b*c + a*d))), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x
])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n
+ 1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a
*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*
n] || EqQ[c, 0])

Rule 3060

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt
[a + b*Sin[e + f*x]])), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -1]

Rule 3123

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C + A*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Si
n[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Dist[1/(b*d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x]
)^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + c*C*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(m + n
+ 2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && NeQ
[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2,
0])

Rubi steps \begin{align*} \text {integral}& = \frac {2 A (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \int \frac {(a+a \cos (c+d x))^{3/2} \left (\frac {3 a A}{2}-\frac {1}{2} a (2 A-3 C) \cos (c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{3 a} \\ & = \frac {2 a A \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {2 A (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 \int \frac {\sqrt {a+a \cos (c+d x)} \left (\frac {1}{4} a^2 (4 A+3 C)-\frac {1}{4} a^2 (8 A-3 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx}{3 a} \\ & = -\frac {a^2 (8 A-3 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a A \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {2 A (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {1}{2} (3 a C) \int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\cos (c+d x)}} \, dx \\ & = -\frac {a^2 (8 A-3 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a A \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {2 A (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {(3 a C) \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{a}}} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{d} \\ & = \frac {3 a^{3/2} C \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{d}-\frac {a^2 (8 A-3 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a A \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {2 A (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.72 \[ \int \frac {(a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {a \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \left (9 \sqrt {2} C \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^{\frac {3}{2}}(c+d x)+(4 A+3 C+20 A \cos (c+d x)+3 C \cos (2 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{6 d \cos ^{\frac {3}{2}}(c+d x)} \]

[In]

Integrate[((a + a*Cos[c + d*x])^(3/2)*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(5/2),x]

[Out]

(a*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(9*Sqrt[2]*C*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^(3/2
) + (4*A + 3*C + 20*A*Cos[c + d*x] + 3*C*Cos[2*(c + d*x)])*Sin[(c + d*x)/2]))/(6*d*Cos[c + d*x]^(3/2))

Maple [A] (verified)

Time = 32.49 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.20

method	result	size
default	\(\frac {a \left (3 C \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+10 A \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+9 C \left (\cos ^{2}\left (d x +c \right )\right ) \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+2 A \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{3 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )^{\frac {3}{2}}}\)	\(193\)
parts	\(\frac {2 A \sin \left (d x +c \right ) \left (5 \cos \left (d x +c \right )+1\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, a}{3 d \left (1+\cos \left (d x +c \right )\right ) \cos \left (d x +c \right )^{\frac {3}{2}}}+\frac {C \left (3 \cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+\cos \left (d x +c \right ) \sin \left (d x +c \right )+3 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, a}{d \sqrt {\cos \left (d x +c \right )}\, \left (1+\cos \left (d x +c \right )\right )}\)	\(207\)

[In]

int((a+cos(d*x+c)*a)^(3/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/3*a/d*(3*C*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+10*A*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/
(1+cos(d*x+c)))^(1/2)+9*C*cos(d*x+c)^2*arctan(tan(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+2*A*(cos(d*x+c)/(1
+cos(d*x+c)))^(1/2)*sin(d*x+c))*(a*(1+cos(d*x+c)))^(1/2)/(1+cos(d*x+c))/(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(
d*x+c)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.88 \[ \int \frac {(a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {{\left (3 \, C a \cos \left (d x + c\right )^{2} + 10 \, A a \cos \left (d x + c\right ) + 2 \, A a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 9 \, {\left (C a \cos \left (d x + c\right )^{3} + C a \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{3 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}} \]

[In]

integrate((a+a*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

1/3*((3*C*a*cos(d*x + c)^2 + 10*A*a*cos(d*x + c) + 2*A*a)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x
+ c) - 9*(C*a*cos(d*x + c)^3 + C*a*cos(d*x + c)^2)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/
(sqrt(a)*sin(d*x + c))))/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2)

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+a*cos(d*x+c))**(3/2)*(A+C*cos(d*x+c)**2)/cos(d*x+c)**(5/2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 930 vs. \(2 (139) = 278\).

Time = 0.47 (sec) , antiderivative size = 930, normalized size of antiderivative = 5.78 \[ \int \frac {(a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\text {Too large to display} \]

[In]

integrate((a+a*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

1/12*(3*(2*(a*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - (a*cos(d*x + c) - a)*sin
(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x
+ 2*c) + 1)^(1/4)*sqrt(a) + 3*(a*arctan2(-(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(
1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - cos(d*x + c)*sin(1/2*arctan2(sin
(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/
4)*(cos(d*x + c)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c)*sin(1/2*arctan2(sin(2
*d*x + 2*c), cos(2*d*x + 2*c) + 1))) + 1) - a*arctan2(-(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x
+ 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - cos(d*x + c)*sin(1/
2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x +
2*c) + 1)^(1/4)*(cos(d*x + c)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c)*sin(1/2*
arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))) - 1) - a*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 +
2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 +
sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) +
1) + a*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*
d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*c
os(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 1))*sqrt(a))*C + 16*(3*sqrt(2)*a^(3/2)*sin(d*x + c)/
(cos(d*x + c) + 1) - 5*sqrt(2)*a^(3/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 2*sqrt(2)*a^(3/2)*sin(d*x + c)^5/
(cos(d*x + c) + 1)^5)*A/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(5
/2)))/d

Giac [F(-1)]

Timed out. \[ \int \frac {(a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+a*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^{5/2}} \,d x \]

[In]

int(((A + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^(3/2))/cos(c + d*x)^(5/2),x)

[Out]

int(((A + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^(3/2))/cos(c + d*x)^(5/2), x)