3.571 \(\int \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx\)
Optimal. Leaf size=126 \[ \frac{2 \left (3 a^2 B+6 a A b+b^2 B\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 d}+\frac{2 \left (a^2 A-2 a b B-A b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 b (2 a B+A b) \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{2 b^2 B \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)} \]
[Out]
(2*(a^2*A - A*b^2 - 2*a*b*B)*EllipticE[(c + d*x)/2, 2])/d + (2*(6*a*A*b + 3*a^2*B + b^2*B)*EllipticF[(c + d*x)
/2, 2])/(3*d) + (2*b^2*B*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + (2*b*(A*b + 2*a*B)*Sin[c + d*x])/(d*Sqrt[Cos
[c + d*x]])
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Rubi [A] time = 0.335161, antiderivative size = 126, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used =
{2954, 2988, 3021, 2748, 2641, 2639} \[ \frac{2 \left (3 a^2 B+6 a A b+b^2 B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 \left (a^2 A-2 a b B-A b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 b (2 a B+A b) \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{2 b^2 B \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[In]
Int[Sqrt[Cos[c + d*x]]*(a + b*Sec[c + d*x])^2*(A + B*Sec[c + d*x]),x]
[Out]
(2*(a^2*A - A*b^2 - 2*a*b*B)*EllipticE[(c + d*x)/2, 2])/d + (2*(6*a*A*b + 3*a^2*B + b^2*B)*EllipticF[(c + d*x)
/2, 2])/(3*d) + (2*b^2*B*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + (2*b*(A*b + 2*a*B)*Sin[c + d*x])/(d*Sqrt[Cos
[c + d*x]])
Rule 2954
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(
d + c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]
Rule 2988
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((B*c - A*d)*(b*c - a*d)^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/
(f*d^2*(n + 1)*(c^2 - d^2)), x] - Dist[1/(d^2*(n + 1)*(c^2 - d^2)), Int[(c + d*Sin[e + f*x])^(n + 1)*Simp[d*(n
+ 1)*(B*(b*c - a*d)^2 - A*d*(a^2*c + b^2*c - 2*a*b*d)) - ((B*c - A*d)*(a^2*d^2*(n + 2) + b^2*(c^2 + d^2*(n +
1))) + 2*a*b*d*(A*c*d*(n + 2) - B*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b^2*B*d*(n + 1)*(c^2 - d^2)*Sin[e + f*x
]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d
^2, 0] && LtQ[n, -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 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 \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx &=\int \frac{(b+a \cos (c+d x))^2 (B+A \cos (c+d x))}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 b^2 B \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{2}{3} \int \frac{-\frac{3}{2} b (A b+2 a B)-\frac{1}{2} \left (6 a A b+3 a^2 B+b^2 B\right ) \cos (c+d x)-\frac{3}{2} a^2 A \cos ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 b^2 B \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 b (A b+2 a B) \sin (c+d x)}{d \sqrt{\cos (c+d x)}}-\frac{4}{3} \int \frac{\frac{1}{4} \left (-6 a A b-3 a^2 B-b^2 B\right )-\frac{3}{4} \left (a^2 A-A b^2-2 a b B\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 b^2 B \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 b (A b+2 a B) \sin (c+d x)}{d \sqrt{\cos (c+d x)}}-\left (-a^2 A+A b^2+2 a b B\right ) \int \sqrt{\cos (c+d x)} \, dx-\frac{1}{3} \left (-6 a A b-3 a^2 B-b^2 B\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 \left (a^2 A-A b^2-2 a b B\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 \left (6 a A b+3 a^2 B+b^2 B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 b^2 B \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 b (A b+2 a B) \sin (c+d x)}{d \sqrt{\cos (c+d x)}}\\ \end{align*}
Mathematica [A] time = 1.19733, size = 105, normalized size = 0.83 \[ \frac{2 \left (\left (3 a^2 B+6 a A b+b^2 B\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+3 \left (a^2 A-2 a b B-A b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+\frac{b \sin (c+d x) (3 (2 a B+A b) \cos (c+d x)+b B)}{\cos ^{\frac{3}{2}}(c+d x)}\right )}{3 d} \]
Antiderivative was successfully verified.
[In]
Integrate[Sqrt[Cos[c + d*x]]*(a + b*Sec[c + d*x])^2*(A + B*Sec[c + d*x]),x]
[Out]
(2*(3*(a^2*A - A*b^2 - 2*a*b*B)*EllipticE[(c + d*x)/2, 2] + (6*a*A*b + 3*a^2*B + b^2*B)*EllipticF[(c + d*x)/2,
2] + (b*(b*B + 3*(A*b + 2*a*B)*Cos[c + d*x])*Sin[c + d*x])/Cos[c + d*x]^(3/2)))/(3*d)
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Maple [B] time = 4.912, size = 677, normalized size = 5.4 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)),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^2*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
))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-2*a^2*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))+4*A*a*b*(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)*El
lipticF(cos(1/2*d*x+1/2*c),2^(1/2))+2*B*a^2*(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^2*(-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)*Ellipti
cF(cos(1/2*d*x+1/2*c),2^(1/2)))+2*b*(A*b+2*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
)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(-2*sin(1/2*d*x+1/2*c)^4+sin(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] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sec \left (d x + c\right ) + A\right )}{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \sqrt{\cos \left (d x + c\right )}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)),x, algorithm="maxima")
[Out]
integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^2*sqrt(cos(d*x + c)), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (B b^{2} \sec \left (d x + c\right )^{3} + A a^{2} +{\left (2 \, B a b + A b^{2}\right )} \sec \left (d x + c\right )^{2} +{\left (B a^{2} + 2 \, A a b\right )} \sec \left (d x + c\right )\right )} \sqrt{\cos \left (d x + c\right )}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)),x, algorithm="fricas")
[Out]
integral((B*b^2*sec(d*x + c)^3 + A*a^2 + (2*B*a*b + A*b^2)*sec(d*x + c)^2 + (B*a^2 + 2*A*a*b)*sec(d*x + c))*sq
rt(cos(d*x + c)), 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(cos(d*x+c)**(1/2)*(a+b*sec(d*x+c))**2*(A+B*sec(d*x+c)),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sec \left (d x + c\right ) + A\right )}{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \sqrt{\cos \left (d x + c\right )}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)),x, algorithm="giac")
[Out]
integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^2*sqrt(cos(d*x + c)), x)