3.690 \(\int (A+B \cos (c+d x)) \sec ^{\frac{7}{2}}(c+d x) \, dx\)

Optimal. Leaf size=151 \[ \frac{2 A \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{5 d}+\frac{6 A \sin (c+d x) \sqrt{\sec (c+d x)}}{5 d}-\frac{6 A \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 B \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{2 B \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d} \]

[Out]

(-6*A*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*B*Sqrt[Cos[c + d*x]]*Ellipti
cF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (6*A*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (2*B*Sec[c + d*x]
^(3/2)*Sin[c + d*x])/(3*d) + (2*A*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(5*d)

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Rubi [A]  time = 0.112296, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3238, 3787, 3768, 3771, 2641, 2639} \[ \frac{2 A \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{5 d}+\frac{6 A \sin (c+d x) \sqrt{\sec (c+d x)}}{5 d}-\frac{6 A \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 B \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{2 B \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*A*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*B*Sqrt[Cos[c + d*x]]*Ellipti
cF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (6*A*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (2*B*Sec[c + d*x]
^(3/2)*Sin[c + d*x])/(3*d) + (2*A*Sec[c + d*x]^(5/2)*Sin[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 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

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]

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 (A+B \cos (c+d x)) \sec ^{\frac{7}{2}}(c+d x) \, dx &=\int \sec ^{\frac{5}{2}}(c+d x) (B+A \sec (c+d x)) \, dx\\ &=A \int \sec ^{\frac{7}{2}}(c+d x) \, dx+B \int \sec ^{\frac{5}{2}}(c+d x) \, dx\\ &=\frac{2 B \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac{2 A \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{1}{5} (3 A) \int \sec ^{\frac{3}{2}}(c+d x) \, dx+\frac{1}{3} B \int \sqrt{\sec (c+d x)} \, dx\\ &=\frac{6 A \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d}+\frac{2 B \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac{2 A \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d}-\frac{1}{5} (3 A) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{3} \left (B \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 B \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 d}+\frac{6 A \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d}+\frac{2 B \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac{2 A \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d}-\frac{1}{5} \left (3 A \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{6 A \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{5 d}+\frac{2 B \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 d}+\frac{6 A \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d}+\frac{2 B \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac{2 A \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.32215, size = 97, normalized size = 0.64 \[ \frac{\sec ^{\frac{5}{2}}(c+d x) \left (21 A \sin (c+d x)+9 A \sin (3 (c+d x))-36 A \cos ^{\frac{5}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+10 B \sin (2 (c+d x))+20 B \cos ^{\frac{5}{2}}(c+d x) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{30 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^(5/2)*(-36*A*Cos[c + d*x]^(5/2)*EllipticE[(c + d*x)/2, 2] + 20*B*Cos[c + d*x]^(5/2)*EllipticF[(c
+ d*x)/2, 2] + 21*A*Sin[c + d*x] + 10*B*Sin[2*(c + d*x)] + 9*A*Sin[3*(c + d*x)]))/(30*d)

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Maple [B]  time = 6.905, size = 502, normalized size = 3.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c))*sec(d*x+c)^(7/2),x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2/5*A/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*
c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)^2*(12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1
/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-
12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2
*d*x+1/2*c)^2+24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^
2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-8*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)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+2*B*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)
^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(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))))/sin(1/2*d*x+1/2*c)/(2*co
s(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac{7}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)*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 (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac{7}{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

integral((B*cos(d*x + c) + A)*sec(d*x + c)^(7/2), 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+B*cos(d*x+c))*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 (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac{7}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(7/2),x, algorithm="giac")

[Out]

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