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

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

[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)

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Rubi [A]  time = 0.124562, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, 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 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 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]

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 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]

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*}

Mathematica [A]  time = 0.80782, 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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Maple [C]  time = 0.6, size = 249, normalized size = 2.2 \begin{align*} -{\frac{ \left ( -2+2\,\cos \left ( dx+c \right ) \right ) \cos \left ( dx+c \right ) \left ( 1+\cos \left ( dx+c \right ) \right ) ^{2}}{21\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}} \left ( 5\,iA \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) \sqrt{ \left ( 1+\cos \left ( dx+c \right ) \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}{\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }},i \right ) +7\,iC \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) \sqrt{ \left ( 1+\cos \left ( dx+c \right ) \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}{\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }},i \right ) -5\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}-7\,C \left ( \cos \left ( dx+c \right ) \right ) ^{3}+5\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+7\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}-3\,A\cos \left ( dx+c \right ) +3\,A \right ) \left ({\frac{b}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{9}{2}}}} \end{align*}

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*cos(d*x+c)^3*sin(d*x+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)+7*I*C*cos(d*x+c)^3*sin(d*x+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)-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., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)