\(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^3(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx\) [278]

Optimal result

Integrand size = 41, antiderivative size = 212 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx=-\frac {6 B \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt {\cos (c+d x)}}+\frac {2 (5 A+7 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 b d \sqrt {b \cos (c+d x)}}+\frac {2 A b^2 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {2 b B \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {2 (5 A+7 C) \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}+\frac {6 B \sin (c+d x)}{5 b d \sqrt {b \cos (c+d x)}} \]

[Out]

2/7*A*b^2*sin(d*x+c)/d/(b*cos(d*x+c))^(7/2)+2/5*b*B*sin(d*x+c)/d/(b*cos(d*x+c))^(5/2)+2/21*(5*A+7*C)*sin(d*x+c
)/d/(b*cos(d*x+c))^(3/2)+6/5*B*sin(d*x+c)/b/d/(b*cos(d*x+c))^(1/2)+2/21*(5*A+7*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)/b/d/(b*cos(d*x+c))^(1/2)-6/5*B*(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*cos(d*x+c))^(1/2)/b^2/d/c
os(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {16, 3100, 2827, 2716, 2721, 2719, 2720} \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx=\frac {2 A b^2 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {2 (5 A+7 C) \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}+\frac {2 (5 A+7 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 b d \sqrt {b \cos (c+d x)}}-\frac {6 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{5 b^2 d \sqrt {\cos (c+d x)}}+\frac {2 b B \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {6 B \sin (c+d x)}{5 b d \sqrt {b \cos (c+d x)}} \]

[In]

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

[Out]

(-6*B*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*b^2*d*Sqrt[Cos[c + d*x]]) + (2*(5*A + 7*C)*Sqrt[Cos[c
+ d*x]]*EllipticF[(c + d*x)/2, 2])/(21*b*d*Sqrt[b*Cos[c + d*x]]) + (2*A*b^2*Sin[c + d*x])/(7*d*(b*Cos[c + d*x
])^(7/2)) + (2*b*B*Sin[c + d*x])/(5*d*(b*Cos[c + d*x])^(5/2)) + (2*(5*A + 7*C)*Sin[c + d*x])/(21*d*(b*Cos[c +
d*x])^(3/2)) + (6*B*Sin[c + d*x])/(5*b*d*Sqrt[b*Cos[c + d*x]])

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 2716

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
))), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[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 2721

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

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]

Rubi steps \begin{align*} \text {integral}& = b^3 \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(b \cos (c+d x))^{9/2}} \, dx \\ & = \frac {2 A b^2 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {2}{7} \int \frac {\frac {7 b^2 B}{2}+\frac {1}{2} b^2 (5 A+7 C) \cos (c+d x)}{(b \cos (c+d x))^{7/2}} \, dx \\ & = \frac {2 A b^2 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\left (b^2 B\right ) \int \frac {1}{(b \cos (c+d x))^{7/2}} \, dx+\frac {1}{7} (b (5 A+7 C)) \int \frac {1}{(b \cos (c+d x))^{5/2}} \, dx \\ & = \frac {2 A b^2 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {2 b B \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {2 (5 A+7 C) \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}+\frac {1}{5} (3 B) \int \frac {1}{(b \cos (c+d x))^{3/2}} \, dx+\frac {(5 A+7 C) \int \frac {1}{\sqrt {b \cos (c+d x)}} \, dx}{21 b} \\ & = \frac {2 A b^2 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {2 b B \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {2 (5 A+7 C) \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}+\frac {6 B \sin (c+d x)}{5 b d \sqrt {b \cos (c+d x)}}-\frac {(3 B) \int \sqrt {b \cos (c+d x)} \, dx}{5 b^2}+\frac {\left ((5 A+7 C) \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{21 b \sqrt {b \cos (c+d x)}} \\ & = \frac {2 (5 A+7 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 b d \sqrt {b \cos (c+d x)}}+\frac {2 A b^2 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {2 b B \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {2 (5 A+7 C) \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}+\frac {6 B \sin (c+d x)}{5 b d \sqrt {b \cos (c+d x)}}-\frac {\left (3 B \sqrt {b \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 b^2 \sqrt {\cos (c+d x)}} \\ & = -\frac {6 B \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt {\cos (c+d x)}}+\frac {2 (5 A+7 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 b d \sqrt {b \cos (c+d x)}}+\frac {2 A b^2 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {2 b B \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {2 (5 A+7 C) \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}+\frac {6 B \sin (c+d x)}{5 b d \sqrt {b \cos (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.80 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.64 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx=\frac {2 \left (-63 B \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 (5 A+7 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+63 B \sin (c+d x)+25 A \tan (c+d x)+35 C \tan (c+d x)+21 B \sec (c+d x) \tan (c+d x)+15 A \sec ^2(c+d x) \tan (c+d x)\right )}{105 b d \sqrt {b \cos (c+d x)}} \]

[In]

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

[Out]

(2*(-63*B*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 5*(5*A + 7*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/
2, 2] + 63*B*Sin[c + d*x] + 25*A*Tan[c + d*x] + 35*C*Tan[c + d*x] + 21*B*Sec[c + d*x]*Tan[c + d*x] + 15*A*Sec[
c + d*x]^2*Tan[c + d*x]))/(105*b*d*Sqrt[b*Cos[c + d*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(729\) vs. \(2(236)=472\).

Time = 19.62 (sec) , antiderivative size = 730, normalized size of antiderivative = 3.44

method	result	size
default	\(\text {Expression too large to display}\)	\(730\)
parts	\(\text {Expression too large to display}\)	\(1008\)

[In]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*b*sin(1/2*d*x+1/2*c)^2)^(1/2)/b*(2*A*(-1/56*cos(1/2*d*x+1/2*c)/b*(-b*(2*sin(1/2
*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^4-5/42*cos(1/2*d*x+1/2*c)/b*(-b*(2*sin(1
/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+5/21*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-
2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-b*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)*EllipticF(cos(1/2*d*x
+1/2*c),2^(1/2)))+2/5*B/b/sin(1/2*d*x+1/2*c)^2/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1
/2*c)^2-1)*(24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-12*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)
^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c
)+12*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1
/2*d*x+1/2*c)^2+8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-3*(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)))*(-2*sin(1/2*d*x+1/2*c)^4*b+b*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*
C*(-1/6*cos(1/2*d*x+1/2*c)/b*(-b*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/
2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-b*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*
x+1/2*c)^2))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))))/sin(1/2*d*x+1/2*c)/((2*cos(1/2*d*x+1/2*c)^2-1)*b)^(
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 = 234, normalized size of antiderivative = 1.10 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx=-\frac {5 \, \sqrt {2} {\left (5 i \, A + 7 i \, C\right )} \sqrt {b} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-5 i \, A - 7 i \, C\right )} \sqrt {b} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 63 i \, \sqrt {2} B \sqrt {b} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 63 i \, \sqrt {2} B \sqrt {b} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (63 \, B \cos \left (d x + c\right )^{3} + 5 \, {\left (5 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{2} + 21 \, B \cos \left (d x + c\right ) + 15 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{105 \, b^{2} d \cos \left (d x + c\right )^{4}} \]

[In]

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

[Out]

-1/105*(5*sqrt(2)*(5*I*A + 7*I*C)*sqrt(b)*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x +
c)) + 5*sqrt(2)*(-5*I*A - 7*I*C)*sqrt(b)*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x +
c)) + 63*I*sqrt(2)*B*sqrt(b)*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) +
I*sin(d*x + c))) - 63*I*sqrt(2)*B*sqrt(b)*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos
(d*x + c) - I*sin(d*x + c))) - 2*(63*B*cos(d*x + c)^3 + 5*(5*A + 7*C)*cos(d*x + c)^2 + 21*B*cos(d*x + c) + 15*
A)*sqrt(b*cos(d*x + c))*sin(d*x + c))/(b^2*d*cos(d*x + c)^4)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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