\(\int (a+b \cos (c+d x)) (A+C \cos ^2(c+d x)) \sec ^{\frac {5}{2}}(c+d x) \, dx\) [1362]

Optimal result

Integrand size = 33, antiderivative size = 135 \[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=-\frac {2 b (A-C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {2 a (A+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 A b \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {2 a A \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4306, 3111, 3100, 2827, 2720, 2719} \[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\frac {2 a (A+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {2 b (A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 A b \sin (c+d x) \sqrt {\sec (c+d x)}}{d} \]

[In]

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

[Out]

(-2*b*(A - C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*a*(A + 3*C)*Sqrt[Cos[c +
d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*A*b*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d + (2*a*
A*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*d)

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3100

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

Rule 3111

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

Rule 4306

Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u,
x]

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 a A \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {1}{3} \left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {3 A b}{2}+\frac {1}{2} a (A+3 C) \cos (c+d x)+\frac {3}{2} b C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 A b \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {2 a A \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {1}{3} \left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{4} a (A+3 C)-\frac {3}{4} b (A-C) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {2 A b \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {2 a A \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}-\left (b (A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{3} \left (a (A+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = -\frac {2 b (A-C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {2 a (A+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 A b \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {2 a A \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.50 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.71 \[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\frac {\sec ^{\frac {3}{2}}(c+d x) \left (-6 b (A-C) \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+2 a (A+3 C) \cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+2 A (a+3 b \cos (c+d x)) \sin (c+d x)\right )}{3 d} \]

[In]

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

[Out]

(Sec[c + d*x]^(3/2)*(-6*b*(A - C)*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2] + 2*a*(A + 3*C)*Cos[c + d*x]^(3
/2)*EllipticF[(c + d*x)/2, 2] + 2*A*(a + 3*b*Cos[c + d*x])*Sin[c + d*x]))/(3*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(612\) vs. \(2(171)=342\).

Time = 326.65 (sec) , antiderivative size = 613, normalized size of antiderivative = 4.54

method	result	size
default	\(-\frac {2 \sqrt {-\left (-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (12 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -2 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -6 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -6 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +6 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -2 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -6 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b +3 C a \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b \right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}}{3 \left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {-1+2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(613\)
parts	\(\text {Expression too large to display}\)	\(667\)

[In]

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

[Out]

-2/3*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(4*sin(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+
1)/sin(1/2*d*x+1/2*c)^3*(12*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*b-2*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*si
n(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2*a-6*A*(sin(1/2*d*x+1/2*
c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2*b-6*C*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*
x+1/2*c)^2*a+6*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^
(1/2))*sin(1/2*d*x+1/2*c)^2*b-2*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a-6*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x
+1/2*c)^2*b+a*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(
1/2))+3*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*
b+3*C*a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*
C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b)*(-2*s
in(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(-1+2*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

Fricas [C] (verification not implemented)

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

Time = 0.11 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.41 \[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\frac {-i \, \sqrt {2} {\left (A + 3 \, C\right )} a \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} {\left (A + 3 \, C\right )} a \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 i \, \sqrt {2} {\left (A - C\right )} b \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 i \, \sqrt {2} {\left (A - C\right )} b \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (3 \, A b \cos \left (d x + c\right ) + A a\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{3 \, d \cos \left (d x + c\right )} \]

[In]

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

[Out]

1/3*(-I*sqrt(2)*(A + 3*C)*a*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + I*sqrt(2)
*(A + 3*C)*a*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*I*sqrt(2)*(A - C)*b*co
s(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*I*sqrt(2)*(A
- C)*b*cos(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(3*A
*b*cos(d*x + c) + A*a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c))

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {5}{2}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {5}{2}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (a+b\,\cos \left (c+d\,x\right )\right ) \,d x \]

[In]

int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(5/2)*(a + b*cos(c + d*x)),x)

[Out]

int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(5/2)*(a + b*cos(c + d*x)), x)