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

Optimal. Leaf size=183 \[ -\frac {2 (3 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{5 b^2 d \sqrt {\cos (c+d x)}}+\frac {2 (3 A+5 C) \sin (c+d x)}{5 b d \sqrt {b \cos (c+d x)}}+\frac {2 A b \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {2 B \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}}+\frac {2 B \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 b d \sqrt {b \cos (c+d x)}} \]

[Out]

2/5*A*b*sin(d*x+c)/d/(b*cos(d*x+c))^(5/2)+2/3*B*sin(d*x+c)/d/(b*cos(d*x+c))^(3/2)+2/5*(3*A+5*C)*sin(d*x+c)/b/d
/(b*cos(d*x+c))^(1/2)+2/3*B*(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)-2/5*(3*A+5*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*El
lipticE(sin(1/2*d*x+1/2*c),2^(1/2))*(b*cos(d*x+c))^(1/2)/b^2/d/cos(d*x+c)^(1/2)

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Rubi [A]  time = 0.24, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {16, 3021, 2748, 2636, 2642, 2641, 2640, 2639} \[ -\frac {2 (3 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{5 b^2 d \sqrt {\cos (c+d x)}}+\frac {2 (3 A+5 C) \sin (c+d x)}{5 b d \sqrt {b \cos (c+d x)}}+\frac {2 A b \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {2 B \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}}+\frac {2 B \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 b d \sqrt {b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(3*A + 5*C)*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*b^2*d*Sqrt[Cos[c + d*x]]) + (2*B*Sqrt[Cos[c
+ d*x]]*EllipticF[(c + d*x)/2, 2])/(3*b*d*Sqrt[b*Cos[c + d*x]]) + (2*A*b*Sin[c + d*x])/(5*d*(b*Cos[c + d*x])^
(5/2)) + (2*B*Sin[c + d*x])/(3*d*(b*Cos[c + d*x])^(3/2)) + (2*(3*A + 5*C)*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 2636

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 2639

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

Rule 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2641

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

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2748

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 3021

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*} \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx &=b^2 \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(b \cos (c+d x))^{7/2}} \, dx\\ &=\frac {2 A b \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {2 \int \frac {\frac {5 b^2 B}{2}+\frac {1}{2} b^2 (3 A+5 C) \cos (c+d x)}{(b \cos (c+d x))^{5/2}} \, dx}{5 b}\\ &=\frac {2 A b \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+(b B) \int \frac {1}{(b \cos (c+d x))^{5/2}} \, dx+\frac {1}{5} (3 A+5 C) \int \frac {1}{(b \cos (c+d x))^{3/2}} \, dx\\ &=\frac {2 A b \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {2 B \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}}+\frac {2 (3 A+5 C) \sin (c+d x)}{5 b d \sqrt {b \cos (c+d x)}}+\frac {B \int \frac {1}{\sqrt {b \cos (c+d x)}} \, dx}{3 b}-\frac {(3 A+5 C) \int \sqrt {b \cos (c+d x)} \, dx}{5 b^2}\\ &=\frac {2 A b \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {2 B \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}}+\frac {2 (3 A+5 C) \sin (c+d x)}{5 b d \sqrt {b \cos (c+d x)}}+\frac {\left (B \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{3 b \sqrt {b \cos (c+d x)}}-\frac {\left ((3 A+5 C) \sqrt {b \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 b^2 \sqrt {\cos (c+d x)}}\\ &=-\frac {2 (3 A+5 C) \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 B \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 b d \sqrt {b \cos (c+d x)}}+\frac {2 A b \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {2 B \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}}+\frac {2 (3 A+5 C) \sin (c+d x)}{5 b d \sqrt {b \cos (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.36, size = 119, normalized size = 0.65 \[ \frac {2 \left (-3 (3 A+5 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+9 A \sin (c+d x)+3 A \tan (c+d x) \sec (c+d x)+5 B \tan (c+d x)+5 B \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+15 C \sin (c+d x)\right )}{15 b d \sqrt {b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-3*(3*A + 5*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 5*B*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2
, 2] + 9*A*Sin[c + d*x] + 15*C*Sin[c + d*x] + 5*B*Tan[c + d*x] + 3*A*Sec[c + d*x]*Tan[c + d*x]))/(15*b*d*Sqrt[
b*Cos[c + d*x]])

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt {b \cos \left (d x + c\right )} \sec \left (d x + c\right )^{2}}{b^{2} \cos \left (d x + c\right )^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{2}}{\left (b \cos \left (d x + c\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [B]  time = 4.62, size = 807, normalized size = 4.41 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(b*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/b^2/sin(1/2*d*x+1/2*c)^3/(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)*(36*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x
+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-72*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c
)^6+20*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*s
in(1/2*d*x+1/2*c)^4+60*C*(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)^4-120*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-36*A*EllipticE(cos(1/2*d*x+1
/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+72*A*cos(1/2
*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-20*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*si
n(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2+20*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-60*C*(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+
120*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+9*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))-24*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+5*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))-10*B*cos(1/2*d*x+1/2*c)*sin(1/2
*d*x+1/2*c)^2+15*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))-30*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)*(-2*sin(1/2*d*x+1/2*c)^4*b+sin(1/2*d*x+1/2*c)^2*b)^(1/2
)/(b*(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{2}}{\left (b \cos \left (d x + c\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\cos \left (c+d\,x\right )}^2\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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