3.1297 \(\int \frac{(a+b \sec (c+d x)) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt{\cos (c+d x)}} \, dx\)
Optimal. Leaf size=152 \[ \frac{2 \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) (a (3 A+C)+b B)}{3 d}-\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (5 a B+5 A b+3 b C)}{5 d}+\frac{2 \sin (c+d x) (5 a B+5 A b+3 b C)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 (a C+b B) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 b C \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)} \]
[Out]
(-2*(5*A*b + 5*a*B + 3*b*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*(b*B + a*(3*A + C))*EllipticF[(c + d*x)/2, 2
])/(3*d) + (2*b*C*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (2*(b*B + a*C)*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2
)) + (2*(5*A*b + 5*a*B + 3*b*C)*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]])
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Rubi [A] time = 0.304602, antiderivative size = 152, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 7, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used =
{4112, 3031, 3021, 2748, 2636, 2639, 2641} \[ \frac{2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (a (3 A+C)+b B)}{3 d}-\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (5 a B+5 A b+3 b C)}{5 d}+\frac{2 \sin (c+d x) (5 a B+5 A b+3 b C)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 (a C+b B) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 b C \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[In]
Int[((a + b*Sec[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqrt[Cos[c + d*x]],x]
[Out]
(-2*(5*A*b + 5*a*B + 3*b*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*(b*B + a*(3*A + C))*EllipticF[(c + d*x)/2, 2
])/(3*d) + (2*b*C*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (2*(b*B + a*C)*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2
)) + (2*(5*A*b + 5*a*B + 3*b*C)*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]])
Rule 4112
Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sec[(e_.)
+ (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[d^(m + 2), Int[(b + a*Cos[e + f*x])^m*(d*
Cos[e + f*x])^(n - m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}
, x] && !IntegerQ[n] && IntegerQ[m]
Rule 3031
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(A*b^2 - a*b*B + 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)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b
^2*d*(m + 1) - a*b*c*(m + 2)) + (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, B, C}, x] && NeQ[b*c - a*d, 0] && Ne
Q[a^2 - b^2, 0] && LtQ[m, -1]
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 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 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]
Rubi steps
\begin{align*} \int \frac{(a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx &=\int \frac{(b+a \cos (c+d x)) \left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right )}{\cos ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 b C \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}-\frac{2}{5} \int \frac{-\frac{5}{2} (b B+a C)-\frac{1}{2} (5 A b+5 a B+3 b C) \cos (c+d x)-\frac{5}{2} a A \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 b C \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 (b B+a C) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{4}{15} \int \frac{-\frac{3}{4} (5 A b+5 a B+3 b C)-\frac{5}{4} (b B+a (3 A+C)) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 b C \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 (b B+a C) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{1}{5} (-5 A b-5 a B-3 b C) \int \frac{1}{\cos ^{\frac{3}{2}}(c+d x)} \, dx-\frac{1}{3} (-b B-a (3 A+C)) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 (b B+a (3 A+C)) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 b C \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 (b B+a C) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 (5 A b+5 a B+3 b C) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}-\frac{1}{5} (5 A b+5 a B+3 b C) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{2 (5 A b+5 a B+3 b C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 (b B+a (3 A+C)) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 b C \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 (b B+a C) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 (5 A b+5 a B+3 b C) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}\\ \end{align*}
Mathematica [A] time = 1.55189, size = 136, normalized size = 0.89 \[ \frac{10 \cos ^{\frac{3}{2}}(c+d x) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) (a (3 A+C)+b B)+3 \sin (2 (c+d x)) (5 a B+5 A b+3 b C)-6 \cos ^{\frac{3}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (5 a B+5 A b+3 b C)+10 (a C+b B) \sin (c+d x)+6 b C \tan (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[In]
Integrate[((a + b*Sec[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqrt[Cos[c + d*x]],x]
[Out]
(-6*(5*A*b + 5*a*B + 3*b*C)*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2] + 10*(b*B + a*(3*A + C))*Cos[c + d*x]
^(3/2)*EllipticF[(c + d*x)/2, 2] + 10*(b*B + a*C)*Sin[c + d*x] + 3*(5*A*b + 5*a*B + 3*b*C)*Sin[2*(c + d*x)] +
6*b*C*Tan[c + d*x])/(15*d*Cos[c + d*x]^(3/2))
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Maple [B] time = 7.888, size = 742, normalized size = 4.9 \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))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(1/2),x)
[Out]
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*A*a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x
+1/2*c)^2+1)^(1/2)/(-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*(B*b+C*a)*(-1/6*cos(1/2*d*x+1/2*c)*(-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)/(-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*C*b/(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)^2*(12*(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)^4-24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-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)^(1/2)*sin(1/2*
d*x+1/2*c)^2+24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+3*(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)-8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c))*(-2*sin(1/2*d*x+1/2
*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+2*(A*b+B*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)*
(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*(-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)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d
*x+1/2*c)^2-1))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
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Maxima [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))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(1/2),x, algorithm="maxima")
[Out]
Timed out
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{C b \sec \left (d x + c\right )^{3} +{\left (C a + B b\right )} \sec \left (d x + c\right )^{2} + A a +{\left (B a + A b\right )} \sec \left (d x + c\right )}{\sqrt{\cos \left (d x + c\right )}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(1/2),x, algorithm="fricas")
[Out]
integral((C*b*sec(d*x + c)^3 + (C*a + B*b)*sec(d*x + c)^2 + A*a + (B*a + A*b)*sec(d*x + c))/sqrt(cos(d*x + c))
, x)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \sec{\left (c + d x \right )}\right ) \left (A + B \sec{\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right )}{\sqrt{\cos{\left (c + d x \right )}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/cos(d*x+c)**(1/2),x)
[Out]
Integral((a + b*sec(c + d*x))*(A + B*sec(c + d*x) + C*sec(c + d*x)**2)/sqrt(cos(c + d*x)), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )}{\left (b \sec \left (d x + c\right ) + a\right )}}{\sqrt{\cos \left (d x + c\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(1/2),x, algorithm="giac")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)/sqrt(cos(d*x + c)), x)