3.575 \(\int \frac{\cos ^{\frac{3}{2}}(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx\)
Optimal. Leaf size=136 \[ \frac{2 \left (a^2 A-3 a b B+3 A b^2\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 a^3 d}-\frac{2 b^2 (A b-a B) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 d (a+b)}-\frac{2 (A b-a B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac{2 A \sin (c+d x) \sqrt{\cos (c+d x)}}{3 a d} \]
[Out]
(-2*(A*b - a*B)*EllipticE[(c + d*x)/2, 2])/(a^2*d) + (2*(a^2*A + 3*A*b^2 - 3*a*b*B)*EllipticF[(c + d*x)/2, 2])
/(3*a^3*d) - (2*b^2*(A*b - a*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a^3*(a + b)*d) + (2*A*Sqrt[Cos[c +
d*x]]*Sin[c + d*x])/(3*a*d)
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Rubi [A] time = 0.586128, antiderivative size = 136, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used =
{2954, 2990, 3059, 2639, 3002, 2641, 2805} \[ \frac{2 \left (a^2 A-3 a b B+3 A b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^3 d}-\frac{2 b^2 (A b-a B) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 d (a+b)}-\frac{2 (A b-a B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac{2 A \sin (c+d x) \sqrt{\cos (c+d x)}}{3 a d} \]
Antiderivative was successfully verified.
[In]
Int[(Cos[c + d*x]^(3/2)*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x]),x]
[Out]
(-2*(A*b - a*B)*EllipticE[(c + d*x)/2, 2])/(a^2*d) + (2*(a^2*A + 3*A*b^2 - 3*a*b*B)*EllipticF[(c + d*x)/2, 2])
/(3*a^3*d) - (2*b^2*(A*b - a*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a^3*(a + b)*d) + (2*A*Sqrt[Cos[c +
d*x]]*Sin[c + d*x])/(3*a*d)
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 2990
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x
])^n*Simp[a^2*A*d*(m + n + 1) + b*B*(b*c*(m - 1) + a*d*(n + 1)) + (a*d*(2*A*b + a*B)*(m + n + 1) - b*B*(a*c -
b*d*(m + n)))*Sin[e + f*x] + b*(A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x]^2, x], x], x] /; F
reeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m,
1] && !(IGtQ[n, 1] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
Rule 3059
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) +
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[C/(b*d), Int[Sqrt[a + b*Sin[e + f*x]]
, x], x] - Dist[1/(b*d), Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[e + f*x], x]/(Sqrt[a + b*Sin[e +
f*x]]*(c + d*Sin[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0]
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 3002
Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
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 2805
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
Rubi steps
\begin{align*} \int \frac{\cos ^{\frac{3}{2}}(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx &=\int \frac{\cos ^{\frac{3}{2}}(c+d x) (B+A \cos (c+d x))}{b+a \cos (c+d x)} \, dx\\ &=\frac{2 A \sqrt{\cos (c+d x)} \sin (c+d x)}{3 a d}+\frac{2 \int \frac{\frac{A b}{2}+\frac{1}{2} a A \cos (c+d x)-\frac{3}{2} (A b-a B) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{3 a}\\ &=\frac{2 A \sqrt{\cos (c+d x)} \sin (c+d x)}{3 a d}-\frac{2 \int \frac{-\frac{1}{2} a A b-\frac{1}{2} \left (a^2 A+3 A b^2-3 a b B\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{3 a^2}-\frac{(A b-a B) \int \sqrt{\cos (c+d x)} \, dx}{a^2}\\ &=-\frac{2 (A b-a B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac{2 A \sqrt{\cos (c+d x)} \sin (c+d x)}{3 a d}-\frac{\left (b^2 (A b-a B)\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{a^3}+\frac{\left (a^2 A+3 A b^2-3 a b B\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{3 a^3}\\ &=-\frac{2 (A b-a B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac{2 \left (a^2 A+3 A b^2-3 a b B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^3 d}-\frac{2 b^2 (A b-a B) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 (a+b) d}+\frac{2 A \sqrt{\cos (c+d x)} \sin (c+d x)}{3 a d}\\ \end{align*}
Mathematica [A] time = 1.20836, size = 210, normalized size = 1.54 \[ \frac{-\frac{6 (A b-a B) \sin (c+d x) \left (2 b (a+b) \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right ),-1\right )-\left (a^2-2 b^2\right ) \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a^2 b \sqrt{\sin ^2(c+d x)}}+4 A \left (\text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )-\frac{b \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}\right )+\frac{2 (3 a B-A b) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}+4 A \sin (c+d x) \sqrt{\cos (c+d x)}}{6 a d} \]
Antiderivative was successfully verified.
[In]
Integrate[(Cos[c + d*x]^(3/2)*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x]),x]
[Out]
((2*(-(A*b) + 3*a*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b) + 4*A*(EllipticF[(c + d*x)/2, 2] - (b*
EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b)) + 4*A*Sqrt[Cos[c + d*x]]*Sin[c + d*x] - (6*(A*b - a*B)*(-2
*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*b*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] - (a^2
- 2*b^2)*EllipticPi[-(a/b), -ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a^2*b*Sqrt[Sin[c + d*x]^2]))/(6*a
*d)
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Maple [B] time = 2.597, size = 786, normalized size = 5.8 \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)^(3/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x)
[Out]
-2/3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*((4*A*a^3-4*A*a^2*b)*cos(1/2*d*x+1/2*c)*sin(1/2*d
*x+1/2*c)^4+(-2*A*a^3+2*A*a^2*b)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^3-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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b+3*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^2-3*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^3+3*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))*a^2*b-3*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))*a*b^2+3*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)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))*b^3-3*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))*a^2*b+3*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))*a*b^2-3*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^3+3*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*si
n(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b-3*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*si
n(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))*a*b^2)/a^3/(a-b)/(-2*sin(1/2*d*x+
1/2*c)^4+sin(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)^(1/2)/d
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac{3}{2}}}{b \sec \left (d x + c\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x, algorithm="maxima")
[Out]
integrate((B*sec(d*x + c) + A)*cos(d*x + c)^(3/2)/(b*sec(d*x + c) + a), x)
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Fricas [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)^(3/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x, algorithm="fricas")
[Out]
Timed out
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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)**(3/2)*(A+B*sec(d*x+c))/(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 \frac{{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac{3}{2}}}{b \sec \left (d x + c\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x, algorithm="giac")
[Out]
integrate((B*sec(d*x + c) + A)*cos(d*x + c)^(3/2)/(b*sec(d*x + c) + a), x)