3.14.18 \(\int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{a+b \sec (c+d x)} \, dx\) [1318]

Optimal. Leaf size=209 \[ \frac {2 \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^3 d}-\frac {2 \left (3 A b^3-a^3 B-3 a b^2 B+a^2 b (A+3 C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^4 d}+\frac {2 b^2 \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^4 (a+b) d}-\frac {2 (A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a^2 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 a d} \]

[Out]

2/5*(5*A*b^2-5*a*b*B+a^2*(3*A+5*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*
c),2^(1/2))/a^3/d-2/3*(3*A*b^3-a^3*B-3*a*b^2*B+a^2*b*(A+3*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*
EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/a^4/d+2*b^2*(A*b^2-a*(B*b-C*a))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x
+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*a/(a+b),2^(1/2))/a^4/(a+b)/d+2/5*A*cos(d*x+c)^(3/2)*sin(d*x+c)/a/d-2/3
*(A*b-B*a)*sin(d*x+c)*cos(d*x+c)^(1/2)/a^2/d

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Rubi [A]
time = 0.65, antiderivative size = 209, normalized size of antiderivative = 1.00, 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 = {4197, 3128, 3138, 2719, 3081, 2720, 2884} \begin {gather*} \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 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 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^3 (-B)+a^2 b (A+3 C)-3 a b^2 B+3 A b^3\right )}{3 a^4 d}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 a d} \end {gather*}

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 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2884

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(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]

Rule 3081

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 3128

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 3138

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 4197

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]

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*}

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Mathematica [A]
time = 12.46, size = 271, normalized size = 1.30 \begin {gather*} \frac {\frac {2 a^2 \left (5 A b^2-5 a b B+3 a^2 (3 A+5 C)\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}+2 a^2 (4 A b+5 a B) \left (2 F\left (\left .\frac {1}{2} (c+d x)\right |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 )+4 a^2 \sqrt {\cos (c+d x)} (-5 A b+5 a B+3 a A \cos (c+d x)) \sin (c+d x)+\frac {6 \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) \left (-2 a b E\left (\left .\text {ArcSin}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) F\left (\left .\text {ArcSin}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+\left (a^2-2 b^2\right ) \Pi \left (-\frac {a}{b};\left .\text {ArcSin}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )\right ) \sin (c+d x)}{b \sqrt {\sin ^2(c+d x)}}}{30 a^4 d} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(800\) vs. \(2(275)=550\).
time = 0.23, size = 801, normalized size = 3.83

method	result	size
default	\(\text {Expression too large to display}\)	\(801\)

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,method=_RETURNVERBOSE)

[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/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*
c)^2)^(1/2)*(4*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-14*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*
x+1/2*c)^2*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)*EllipticF(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)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/
2*c),2^(1/2)))-4/3/a^2*(3*A*a+A*b-B*a)*(2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-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)^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)*(sin(1/2*d*x+1/2*c)^2)^(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/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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

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)

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\cos \left (c+d\,x\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{a+\frac {b}{\cos \left (c+d\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + b/cos(c + d*x)),x)

[Out]

int((cos(c + d*x)^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + b/cos(c + d*x)), x)

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