\(\int \frac {(a+b \cos (c+d x)) (A+C \cos ^2(c+d x))}{\sqrt {\sec (c+d x)}} \, dx\) [1365]

Optimal result

Integrand size = 33, antiderivative size = 174 \[ \int \frac {(a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\frac {2 a (5 A+3 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 b (7 A+5 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 b C \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 a C \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 b (7 A+5 C) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}} \]

[Out]

2/7*b*C*sin(d*x+c)/d/sec(d*x+c)^(5/2)+2/5*a*C*sin(d*x+c)/d/sec(d*x+c)^(3/2)+2/21*b*(7*A+5*C)*sin(d*x+c)/d/sec(
d*x+c)^(1/2)+2/5*a*(5*A+3*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+2/21*b*(7*A+5*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elli
pticF(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 = 0.29 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {4306, 3113, 3102, 2827, 2719, 2715, 2720} \[ \int \frac {(a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\frac {2 a (5 A+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a C \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 b (7 A+5 C) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {2 b (7 A+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {2 b C \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)} \]

[In]

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

[Out]

(2*a*(5*A + 3*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*b*(7*A + 5*C)*Sqr
t[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*b*C*Sin[c + d*x])/(7*d*Sec[c + d*x]^
(5/2)) + (2*a*C*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (2*b*(7*A + 5*C)*Sin[c + d*x])/(21*d*Sqrt[Sec[c + d*x
]])

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

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 3102

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

Rule 3113

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (C_.)*sin[(e
_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m
+ 3))), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*(m + 3) + b*d*(C*(m + 2) + A*(
m + 3))*Sin[e + f*x] - (2*a*C*d - b*c*C*(m + 3))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C,
m}, 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 \sqrt {\cos (c+d x)} (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \, dx \\ & = \frac {2 b C \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {1}{7} \left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \left (\frac {7 a A}{2}+\frac {1}{2} b (7 A+5 C) \cos (c+d x)+\frac {7}{2} a C \cos ^2(c+d x)\right ) \, dx \\ & = \frac {2 b C \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 a C \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {1}{35} \left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \left (\frac {7}{4} a (5 A+3 C)+\frac {5}{4} b (7 A+5 C) \cos (c+d x)\right ) \, dx \\ & = \frac {2 b C \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 a C \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {1}{5} \left (a (5 A+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{7} \left (b (7 A+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \cos ^{\frac {3}{2}}(c+d x) \, dx \\ & = \frac {2 a (5 A+3 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 b C \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 a C \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 b (7 A+5 C) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {1}{21} \left (b (7 A+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {2 a (5 A+3 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 b (7 A+5 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 b C \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 a C \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 b (7 A+5 C) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.61 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.69 \[ \int \frac {(a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\frac {\sqrt {\sec (c+d x)} \left (84 a (5 A+3 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+20 b (7 A+5 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+(70 A b+65 b C+42 a C \cos (c+d x)+15 b C \cos (2 (c+d x))) \sin (2 (c+d x))\right )}{210 d} \]

[In]

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

[Out]

(Sqrt[Sec[c + d*x]]*(84*a*(5*A + 3*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 20*b*(7*A + 5*C)*Sqrt[Cos
[c + d*x]]*EllipticF[(c + d*x)/2, 2] + (70*A*b + 65*b*C + 42*a*C*Cos[c + d*x] + 15*b*C*Cos[2*(c + d*x)])*Sin[2
*(c + d*x)]))/(210*d)

Maple [A] (verified)

Time = 7.84 (sec) , antiderivative size = 401, normalized size of antiderivative = 2.30

method	result	size
default	\(-\frac {2 \sqrt {\left (-1+2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (240 C b \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-168 C a -360 C b \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (140 A b +168 C a +280 C b \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-70 A b -42 C a -80 C b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+35 A b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-105 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a +25 C b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-63 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a \right )}{105 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-1+2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(401\)
parts	\(\text {Expression too large to display}\)	\(719\)

[In]

int((a+b*cos(d*x+c))*(A+C*cos(d*x+c)^2)/sec(d*x+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/105*((-1+2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(240*C*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)
^8+(-168*C*a-360*C*b)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(140*A*b+168*C*a+280*C*b)*sin(1/2*d*x+1/2*c)^4*c
os(1/2*d*x+1/2*c)+(-70*A*b-42*C*a-80*C*b)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+35*A*b*(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))-105*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))*a+25*C*b*(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))-63*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))*a)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1
/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(-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 = 0.10 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.11 \[ \int \frac {(a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\frac {-5 i \, \sqrt {2} {\left (7 \, A + 5 \, C\right )} b {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} {\left (7 \, A + 5 \, C\right )} b {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 i \, \sqrt {2} {\left (5 \, A + 3 \, C\right )} a {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 21 i \, \sqrt {2} {\left (5 \, A + 3 \, C\right )} a {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (15 \, C b \cos \left (d x + c\right )^{3} + 21 \, C a \cos \left (d x + c\right )^{2} + 5 \, {\left (7 \, A + 5 \, C\right )} b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{105 \, d} \]

[In]

integrate((a+b*cos(d*x+c))*(A+C*cos(d*x+c)^2)/sec(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

1/105*(-5*I*sqrt(2)*(7*A + 5*C)*b*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*I*sqrt(2)*(7*A
+ 5*C)*b*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 21*I*sqrt(2)*(5*A + 3*C)*a*weierstrassZe
ta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 21*I*sqrt(2)*(5*A + 3*C)*a*weierstrassZ
eta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(15*C*b*cos(d*x + c)^3 + 21*C*a*cos(
d*x + c)^2 + 5*(7*A + 5*C)*b*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/d

Sympy [F]

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

[In]

integrate((a+b*cos(d*x+c))*(A+C*cos(d*x+c)**2)/sec(d*x+c)**(1/2),x)

[Out]

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

Maxima [F]

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

[In]

integrate((a+b*cos(d*x+c))*(A+C*cos(d*x+c)^2)/sec(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate((a+b*cos(d*x+c))*(A+C*cos(d*x+c)^2)/sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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