3.1318 \(\int \frac{\cos ^{\frac{5}{2}}(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{a+b \sec (c+d x)} \, dx\)
Optimal. Leaf size=209 \[ -\frac{2 \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (a^2 b (A+3 C)+a^3 (-B)-3 a b^2 B+3 A b^3\right )}{3 a^4 d}+\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{5 a^3 d}+\frac{2 b^2 \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^4 d (a+b)}-\frac{2 (A b-a B) \sin (c+d x) \sqrt{\cos (c+d x)}}{3 a^2 d}+\frac{2 A \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 a d} \]
[Out]
(2*(5*A*b^2 - 5*a*b*B + a^2*(3*A + 5*C))*EllipticE[(c + d*x)/2, 2])/(5*a^3*d) - (2*(3*A*b^3 - a^3*B - 3*a*b^2*
B + a^2*b*(A + 3*C))*EllipticF[(c + d*x)/2, 2])/(3*a^4*d) + (2*b^2*(A*b^2 - a*(b*B - a*C))*EllipticPi[(2*a)/(a
+ b), (c + d*x)/2, 2])/(a^4*(a + b)*d) - (2*(A*b - a*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*a^2*d) + (2*A*Cos
[c + d*x]^(3/2)*Sin[c + d*x])/(5*a*d)
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Rubi [A] time = 0.965389, antiderivative size = 209, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 7, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.163, Rules used =
{4112, 3049, 3059, 2639, 3002, 2641, 2805} \[ -\frac{2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 b (A+3 C)+a^3 (-B)-3 a b^2 B+3 A b^3\right )}{3 a^4 d}+\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{5 a^3 d}+\frac{2 b^2 \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^4 d (a+b)}-\frac{2 (A b-a B) \sin (c+d x) \sqrt{\cos (c+d x)}}{3 a^2 d}+\frac{2 A \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 a d} \]
Antiderivative was successfully verified.
[In]
Int[(Cos[c + d*x]^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x]),x]
[Out]
(2*(5*A*b^2 - 5*a*b*B + a^2*(3*A + 5*C))*EllipticE[(c + d*x)/2, 2])/(5*a^3*d) - (2*(3*A*b^3 - a^3*B - 3*a*b^2*
B + a^2*b*(A + 3*C))*EllipticF[(c + d*x)/2, 2])/(3*a^4*d) + (2*b^2*(A*b^2 - a*(b*B - a*C))*EllipticPi[(2*a)/(a
+ b), (c + d*x)/2, 2])/(a^4*(a + b)*d) - (2*(A*b - a*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*a^2*d) + (2*A*Cos
[c + d*x]^(3/2)*Sin[c + d*x])/(5*a*d)
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 3049
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e +
f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 0] && !(IGtQ[n, 0] && ( !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{5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx &=\int \frac{\cos ^{\frac{3}{2}}(c+d x) \left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right )}{b+a \cos (c+d x)} \, dx\\ &=\frac{2 A \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 a d}+\frac{2 \int \frac{\sqrt{\cos (c+d x)} \left (\frac{3 A b}{2}+\frac{1}{2} a (3 A+5 C) \cos (c+d x)-\frac{5}{2} (A b-a B) \cos ^2(c+d x)\right )}{b+a \cos (c+d x)} \, dx}{5 a}\\ &=-\frac{2 (A b-a B) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 a^2 d}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 a d}+\frac{4 \int \frac{-\frac{5}{4} b (A b-a B)+\frac{1}{4} a (4 A b+5 a B) \cos (c+d x)+\frac{3}{4} \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{15 a^2}\\ &=-\frac{2 (A b-a B) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 a^2 d}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 a d}-\frac{4 \int \frac{\frac{5}{4} a b (A b-a B)+\frac{5}{4} \left (3 A b^3-a^3 B-3 a b^2 B+a^2 b (A+3 C)\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{15 a^3}+\frac{\left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 a^3}\\ &=\frac{2 \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^3 d}-\frac{2 (A b-a B) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 a^2 d}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 a d}-\frac{\left (3 A b^3-a^3 B-3 a b^2 B+a^2 b (A+3 C)\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{3 a^4}+\frac{\left (b^2 \left (A b^2-a (b B-a C)\right )\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{a^4}\\ &=\frac{2 \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^3 d}-\frac{2 \left (3 A b^3-a^3 B-3 a b^2 B+a^2 b (A+3 C)\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^4 d}+\frac{2 b^2 \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^4 (a+b) d}-\frac{2 (A b-a B) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 a^2 d}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 a d}\\ \end{align*}
Mathematica [A] time = 2.39304, size = 274, normalized size = 1.31 \[ \frac{\frac{6 \sin (c+d x) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right ) \left (2 b (a+b) \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right ),-1\right )-\left (a^2-2 b^2\right ) \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{b \sqrt{\sin ^2(c+d x)}}+2 a^2 (5 a B+4 A b) \left (2 \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )-\frac{2 b \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}\right )+\frac{2 a^2 \left (3 a^2 (3 A+5 C)-5 a b B+5 A b^2\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}+4 a^2 \sin (c+d x) \sqrt{\cos (c+d x)} (3 a A \cos (c+d x)+5 a B-5 A b)}{30 a^4 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(Cos[c + d*x]^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x]),x]
[Out]
((2*a^2*(5*A*b^2 - 5*a*b*B + 3*a^2*(3*A + 5*C))*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b) + 2*a^2*(4*
A*b + 5*a*B)*(2*EllipticF[(c + d*x)/2, 2] - (2*b*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b)) + 4*a^2*S
qrt[Cos[c + d*x]]*(-5*A*b + 5*a*B + 3*a*A*Cos[c + d*x])*Sin[c + d*x] + (6*(5*A*b^2 - 5*a*b*B + a^2*(3*A + 5*C)
)*(-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])/(b*Sqrt[Sin[c + d*x]^2]))/(30
*a^4*d)
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Maple [B] time = 6.517, size = 801, normalized size = 3.8 \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)^(5/2)*(A+B*sec(d*x+c)+C*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)*(4/5*A/a*(-4*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c
)+14*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+5*(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
llipticF(cos(1/2*d*x+1/2*c),2^(1/2))-9*(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)-6*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)-4/3/a^2*(3*A*a+A*b-B*a)*(2*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+2*(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))-3*(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))*(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
*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+2/a^3*(3*A*a^2+2*A*a*b+A*b^2-2*B*a^2-B*a*b+C*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)*(Ell
ipticF(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*a^3+A*a^2*b+A*a*b^2+A*b^3-B*a^3
-B*a^2*b-B*a*b^2+C*a^3+C*a^2*b)/a^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)/(-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^2*(A*b^2-B*a*b+C*a^2)/a^3/
(a^2-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)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),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(-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)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, algorithm="maxima")
[Out]
Timed out
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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)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(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)**(5/2)*(A+B*sec(d*x+c)+C*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 \frac{{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac{5}{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)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, algorithm="giac")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*cos(d*x + c)^(5/2)/(b*sec(d*x + c) + a), x)