3.27 \(\int (A+C \cos ^2(c+d x)) (b \sec (c+d x))^{7/2} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.13362, 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, 2639} \[ \frac{2 b^3 (3 A+5 C) \sin (c+d x) \sqrt{b \sec (c+d x)}}{5 d}-\frac{2 b^4 (3 A+5 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 A b^2 \tan (c+d x) (b \sec (c+d x))^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(A + C*Cos[c + d*x]^2)*(b*Sec[c + d*x])^(7/2),x]

[Out]

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

Rubi steps

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

Mathematica [A]  time = 0.467094, size = 79, normalized size = 0.69 \[ -\frac{b^2 (b \sec (c+d x))^{3/2} \left (-(3 A+5 C) \sin (2 (c+d x))+2 (3 A+5 C) \cos ^{\frac{3}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )-2 A \tan (c+d x)\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + C*Cos[c + d*x]^2)*(b*Sec[c + d*x])^(7/2),x]

[Out]

-(b^2*(b*Sec[c + d*x])^(3/2)*(2*(3*A + 5*C)*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2] - (3*A + 5*C)*Sin[2*(
c + d*x)] - 2*A*Tan[c + d*x]))/(5*d)

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Maple [C]  time = 0.648, size = 668, normalized size = 5.8 \begin{align*} \text{result too large to display} \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))^(7/2),x)

[Out]

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

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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{7}{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))^(7/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*sec(d*x + c))^(7/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C b^{3} \cos \left (d x + c\right )^{2} + A b^{3}\right )} \sqrt{b \sec \left (d x + c\right )} \sec \left (d x + c\right )^{3}, 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))^(7/2),x, algorithm="fricas")

[Out]

integral((C*b^3*cos(d*x + c)^2 + A*b^3)*sqrt(b*sec(d*x + c))*sec(d*x + c)^3, 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))**(7/2),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{7}{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))^(7/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*sec(d*x + c))^(7/2), x)