3.1185 \(\int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\)
Optimal. Leaf size=123 \[ \frac{2 B \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 d}-\frac{2 (5 A+3 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 (5 A+3 C) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 B \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 C \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)} \]
[Out]
(-2*(5*A + 3*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*B*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*C*Sin[c + d*x])/
(5*d*Cos[c + d*x]^(5/2)) + (2*B*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + (2*(5*A + 3*C)*Sin[c + d*x])/(5*d*Sqr
t[Cos[c + d*x]])
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Rubi [A] time = 0.134258, antiderivative size = 123, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used =
{4064, 3021, 2748, 2636, 2641, 2639} \[ -\frac{2 (5 A+3 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 (5 A+3 C) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 B F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 B \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 C \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[In]
Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/Cos[c + d*x]^(3/2),x]
[Out]
(-2*(5*A + 3*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*B*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*C*Sin[c + d*x])/
(5*d*Cos[c + d*x]^(5/2)) + (2*B*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + (2*(5*A + 3*C)*Sin[c + d*x])/(5*d*Sqr
t[Cos[c + d*x]])
Rule 4064
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(m_)*((A_.) + (B_.)*sec[(e_.) + (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.)*(x_)
]^2), x_Symbol] :> Dist[b^2, Int[(b*Cos[e + f*x])^(m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; F
reeQ[{b, e, f, A, B, C, m}, x] && !IntegerQ[m]
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]
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 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 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 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]
Rubi steps
\begin{align*} \int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x)} \, dx &=\int \frac{C+B \cos (c+d x)+A \cos ^2(c+d x)}{\cos ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 C \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2}{5} \int \frac{\frac{5 B}{2}+\frac{1}{2} (5 A+3 C) \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 C \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+B \int \frac{1}{\cos ^{\frac{5}{2}}(c+d x)} \, dx+\frac{1}{5} (5 A+3 C) \int \frac{1}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 C \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 B \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 (5 A+3 C) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{1}{3} B \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx+\frac{1}{5} (-5 A-3 C) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{2 (5 A+3 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 B F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 C \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 B \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 (5 A+3 C) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}\\ \end{align*}
Mathematica [A] time = 0.516547, size = 112, normalized size = 0.91 \[ \frac{10 B \cos ^{\frac{3}{2}}(c+d x) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )-6 (5 A+3 C) \cos ^{\frac{3}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+15 A \sin (2 (c+d x))+10 B \sin (c+d x)+9 C \sin (2 (c+d x))+6 C \tan (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[In]
Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/Cos[c + d*x]^(3/2),x]
[Out]
(-6*(5*A + 3*C)*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2] + 10*B*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2,
2] + 10*B*Sin[c + d*x] + 15*A*Sin[2*(c + d*x)] + 9*C*Sin[2*(c + d*x)] + 6*C*Tan[c + d*x])/(15*d*Cos[c + d*x]^(
3/2))
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Maple [B] time = 6.925, size = 799, normalized size = 6.5 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2),x)
[Out]
2/15*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(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)/sin(1/2*d*x+1/2*c)^3*(60*A*(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)^(1/2)*sin(1/2*d*x+1/2*c)^4-120*A*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)
+20*B*(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)^(1/2)*sin(
1/2*d*x+1/2*c)^4+36*C*(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)^(1/2)*sin(1/2*d*x+1/2*c)^4-72*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-60*A*(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)^(1/2)*sin(1/2*d*x+1/2*c)^2+120*A*cos(1/2*d*
x+1/2*c)*sin(1/2*d*x+1/2*c)^4-20*B*(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)^(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)^4-36*C*(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)^(1/2)*sin(1/2*d*x+1/2*c)^2+72*
C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+15*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)*E
llipticE(cos(1/2*d*x+1/2*c),2^(1/2))-30*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*A+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*sin(1/2*d*x+1/2*c)^2*cos(1/2*
d*x+1/2*c)+9*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))-24*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*C)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos
(1/2*d*x+1/2*c)^2-1)^(1/2)/d
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{\cos \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2),x, algorithm="maxima")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/cos(d*x + c)^(3/2), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{\cos \left (d x + c\right )^{\frac{3}{2}}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2),x, algorithm="fricas")
[Out]
integral((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/cos(d*x + c)^(3/2), x)
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)**2)/cos(d*x+c)**(3/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{\cos \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2),x, algorithm="giac")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/cos(d*x + c)^(3/2), x)