3.26 \(\int (A+C \cos ^2(c+d x)) (b \sec (c+d x))^{9/2} \, dx\)

Optimal. Leaf size=115 \[ \frac {2 b^4 (5 A+7 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{21 d}+\frac {2 b^3 (5 A+7 C) \sin (c+d x) (b \sec (c+d x))^{3/2}}{21 d}+\frac {2 A b^2 \tan (c+d x) (b \sec (c+d x))^{5/2}}{7 d} \]

[Out]

2/21*b^3*(5*A+7*C)*(b*sec(d*x+c))^(3/2)*sin(d*x+c)/d+2/21*b^4*(5*A+7*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+2/7*A*b^2*(b*sec(d*x+c
))^(5/2)*tan(d*x+c)/d

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Rubi [A]  time = 0.12, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3238, 4046, 3768, 3771, 2641} \[ \frac {2 b^3 (5 A+7 C) \sin (c+d x) (b \sec (c+d x))^{3/2}}{21 d}+\frac {2 b^4 (5 A+7 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{21 d}+\frac {2 A b^2 \tan (c+d x) (b \sec (c+d x))^{5/2}}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[(A + C*Cos[c + d*x]^2)*(b*Sec[c + d*x])^(9/2),x]

[Out]

(2*b^4*(5*A + 7*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(21*d) + (2*b^3*(5*A + 7
*C)*(b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(21*d) + (2*A*b^2*(b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*d)

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 3238

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 3768

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

Rule 3771

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 4046

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))^{9/2} \, dx &=b^2 \int (b \sec (c+d x))^{5/2} \left (C+A \sec ^2(c+d x)\right ) \, dx\\ &=\frac {2 A b^2 (b \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac {1}{7} \left (b^2 (5 A+7 C)\right ) \int (b \sec (c+d x))^{5/2} \, dx\\ &=\frac {2 b^3 (5 A+7 C) (b \sec (c+d x))^{3/2} \sin (c+d x)}{21 d}+\frac {2 A b^2 (b \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac {1}{21} \left (b^4 (5 A+7 C)\right ) \int \sqrt {b \sec (c+d x)} \, dx\\ &=\frac {2 b^3 (5 A+7 C) (b \sec (c+d x))^{3/2} \sin (c+d x)}{21 d}+\frac {2 A b^2 (b \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac {1}{21} \left (b^4 (5 A+7 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^4 (5 A+7 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{21 d}+\frac {2 b^3 (5 A+7 C) (b \sec (c+d x))^{3/2} \sin (c+d x)}{21 d}+\frac {2 A b^2 (b \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}\\ \end {align*}

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Mathematica [A]  time = 0.87, size = 78, normalized size = 0.68 \[ \frac {b^2 (b \sec (c+d x))^{5/2} \left ((5 A+7 C) \sin (2 (c+d x))+2 (5 A+7 C) \cos ^{\frac {5}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+6 A \tan (c+d x)\right )}{21 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + C*Cos[c + d*x]^2)*(b*Sec[c + d*x])^(9/2),x]

[Out]

(b^2*(b*Sec[c + d*x])^(5/2)*(2*(5*A + 7*C)*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + (5*A + 7*C)*Sin[2*(c
+ d*x)] + 6*A*Tan[c + d*x]))/(21*d)

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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b^{4} \cos \left (d x + c\right )^{2} + A b^{4}\right )} \sqrt {b \sec \left (d x + c\right )} \sec \left (d x + c\right )^{4}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*(b*sec(d*x+c))^(9/2),x, algorithm="fricas")

[Out]

integral((C*b^4*cos(d*x + c)^2 + A*b^4)*sqrt(b*sec(d*x + c))*sec(d*x + c)^4, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {9}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*(b*sec(d*x+c))^(9/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*sec(d*x + c))^(9/2), x)

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maple [C]  time = 0.37, size = 249, normalized size = 2.17 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (5 i A \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+7 i C \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )-5 A \left (\cos ^{3}\left (d x +c \right )\right )-7 C \left (\cos ^{3}\left (d x +c \right )\right )+5 A \left (\cos ^{2}\left (d x +c \right )\right )+7 C \left (\cos ^{2}\left (d x +c \right )\right )-3 A \cos \left (d x +c \right )+3 A \right ) \cos \left (d x +c \right ) \left (1+\cos \left (d x +c \right )\right )^{2} \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {9}{2}}}{21 d \sin \left (d x +c \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*cos(d*x+c)^2)*(b*sec(d*x+c))^(9/2),x)

[Out]

-2/21/d*(-1+cos(d*x+c))*(5*I*A*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticF(I*(-1+cos(
d*x+c))/sin(d*x+c),I)*sin(d*x+c)*cos(d*x+c)^3+7*I*C*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)
*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)*cos(d*x+c)^3-5*A*cos(d*x+c)^3-7*C*cos(d*x+c)^3+5*A*cos(d
*x+c)^2+7*C*cos(d*x+c)^2-3*A*cos(d*x+c)+3*A)*cos(d*x+c)*(1+cos(d*x+c))^2*(b/cos(d*x+c))^(9/2)/sin(d*x+c)^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {9}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*(b*sec(d*x+c))^(9/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*sec(d*x + c))^(9/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{9/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C*cos(c + d*x)^2)*(b/cos(c + d*x))^(9/2),x)

[Out]

int((A + C*cos(c + d*x)^2)*(b/cos(c + d*x))^(9/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)**2)*(b*sec(d*x+c))**(9/2),x)

[Out]

Timed out

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