3.1.28 \(\int (A+C \cos ^2(c+d x)) (b \sec (c+d x))^{5/2} \, dx\) [28]

Optimal. Leaf size=78 \[ \frac {2 b^2 (A+3 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{3 d}+\frac {2 A b^2 \sqrt {b \sec (c+d x)} \tan (c+d x)}{3 d} \]

[Out]

2/3*b^2*(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))*cos(d*x+
c)^(1/2)*(b*sec(d*x+c))^(1/2)/d+2/3*A*b^2*(b*sec(d*x+c))^(1/2)*tan(d*x+c)/d

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Rubi [A]
time = 0.07, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3317, 4131, 3856, 2720} \begin {gather*} \frac {2 b^2 (A+3 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{3 d}+\frac {2 A b^2 \tan (c+d x) \sqrt {b \sec (c+d x)}}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + C*Cos[c + d*x]^2)*(b*Sec[c + d*x])^(5/2),x]

[Out]

(2*b^2*(A + 3*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(3*d) + (2*A*b^2*Sqrt[b*Se
c[c + d*x]]*Tan[c + d*x])/(3*d)

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 3317

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Csc[e + f*x])^(m - n*p)*(b + a*Csc[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] &&  !IntegerQ[m] && IntegersQ[n, p]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 4131

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot
[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1))), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x
] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

Rubi steps

\begin {align*} \int \left (A+C \cos ^2(c+d x)\right ) (b \sec (c+d x))^{5/2} \, dx &=b^2 \int \sqrt {b \sec (c+d x)} \left (C+A \sec ^2(c+d x)\right ) \, dx\\ &=\frac {2 A b^2 \sqrt {b \sec (c+d x)} \tan (c+d x)}{3 d}+\frac {1}{3} \left (b^2 (A+3 C)\right ) \int \sqrt {b \sec (c+d x)} \, dx\\ &=\frac {2 A b^2 \sqrt {b \sec (c+d x)} \tan (c+d x)}{3 d}+\frac {1}{3} \left (b^2 (A+3 C) \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 b^2 (A+3 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{3 d}+\frac {2 A b^2 \sqrt {b \sec (c+d x)} \tan (c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]
time = 0.25, size = 58, normalized size = 0.74 \begin {gather*} \frac {2 b^2 \sqrt {b \sec (c+d x)} \left ((A+3 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+A \tan (c+d x)\right )}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + C*Cos[c + d*x]^2)*(b*Sec[c + d*x])^(5/2),x]

[Out]

(2*b^2*Sqrt[b*Sec[c + d*x]]*((A + 3*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + A*Tan[c + d*x]))/(3*d)

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Maple [C] Result contains complex when optimal does not.
time = 0.40, size = 199, normalized size = 2.55

method	result	size
default	\(-\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (i A \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+3 i C \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-A \cos \left (d x +c \right )+A \right ) \cos \left (d x +c \right ) \left (1+\cos \left (d x +c \right )\right )^{2} \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}}}{3 d \sin \left (d x +c \right )^{3}}\)	\(199\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*cos(d*x+c)^2)*(b*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/3/d*(-1+cos(d*x+c))*(I*A*cos(d*x+c)*sin(d*x+c)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*E
llipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)+3*I*C*cos(d*x+c)*sin(d*x+c)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+c
os(d*x+c)))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)-A*cos(d*x+c)+A)*cos(d*x+c)*(1+cos(d*x+c))^2*(b/cos
(d*x+c))^(5/2)/sin(d*x+c)^3

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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((A+C*cos(d*x+c)^2)*(b*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*sec(d*x + c))^(5/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 112, normalized size = 1.44 \begin {gather*} \frac {-i \, \sqrt {2} {\left (A + 3 \, C\right )} b^{\frac {5}{2}} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} {\left (A + 3 \, C\right )} b^{\frac {5}{2}} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, A b^{2} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*(b*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/3*(-I*sqrt(2)*(A + 3*C)*b^(5/2)*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + I*s
qrt(2)*(A + 3*C)*b^(5/2)*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*A*b^2*sqrt
(b/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c))

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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((A+C*cos(d*x+c)**2)*(b*sec(d*x+c))**(5/2),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((A+C*cos(d*x+c)^2)*(b*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*sec(d*x + c))^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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