3.819 \(\int (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx\)

Optimal. Leaf size=176 \[ \frac{2 A b^5 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac{6 A b^3 \sin (c+d x)}{5 d \sqrt{b \cos (c+d x)}}-\frac{6 A b^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \cos (c+d x)}}{5 d \sqrt{\cos (c+d x)}}+\frac{2 b^4 B \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}}+\frac{2 b^3 B \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d \sqrt{b \cos (c+d x)}} \]

[Out]

(-6*A*b^2*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*d*Sqrt[Cos[c + d*x]]) + (2*b^3*B*Sqrt[Cos[c + d*x
]]*EllipticF[(c + d*x)/2, 2])/(3*d*Sqrt[b*Cos[c + d*x]]) + (2*A*b^5*Sin[c + d*x])/(5*d*(b*Cos[c + d*x])^(5/2))
+ (2*b^4*B*Sin[c + d*x])/(3*d*(b*Cos[c + d*x])^(3/2)) + (6*A*b^3*Sin[c + d*x])/(5*d*Sqrt[b*Cos[c + d*x]])

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Rubi [A]  time = 0.162002, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {16, 2748, 2636, 2640, 2639, 2642, 2641} \[ \frac{2 A b^5 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac{6 A b^3 \sin (c+d x)}{5 d \sqrt{b \cos (c+d x)}}-\frac{6 A b^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \cos (c+d x)}}{5 d \sqrt{\cos (c+d x)}}+\frac{2 b^4 B \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}}+\frac{2 b^3 B \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d \sqrt{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^6,x]

[Out]

(-6*A*b^2*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*d*Sqrt[Cos[c + d*x]]) + (2*b^3*B*Sqrt[Cos[c + d*x
]]*EllipticF[(c + d*x)/2, 2])/(3*d*Sqrt[b*Cos[c + d*x]]) + (2*A*b^5*Sin[c + d*x])/(5*d*(b*Cos[c + d*x])^(5/2))
+ (2*b^4*B*Sin[c + d*x])/(3*d*(b*Cos[c + d*x])^(3/2)) + (6*A*b^3*Sin[c + d*x])/(5*d*Sqrt[b*Cos[c + d*x]])

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2636

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

Rule 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, 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]

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

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 (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx &=b^6 \int \frac{A+B \cos (c+d x)}{(b \cos (c+d x))^{7/2}} \, dx\\ &=\left (A b^6\right ) \int \frac{1}{(b \cos (c+d x))^{7/2}} \, dx+\left (b^5 B\right ) \int \frac{1}{(b \cos (c+d x))^{5/2}} \, dx\\ &=\frac{2 A b^5 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac{2 b^4 B \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}}+\frac{1}{5} \left (3 A b^4\right ) \int \frac{1}{(b \cos (c+d x))^{3/2}} \, dx+\frac{1}{3} \left (b^3 B\right ) \int \frac{1}{\sqrt{b \cos (c+d x)}} \, dx\\ &=\frac{2 A b^5 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac{2 b^4 B \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}}+\frac{6 A b^3 \sin (c+d x)}{5 d \sqrt{b \cos (c+d x)}}-\frac{1}{5} \left (3 A b^2\right ) \int \sqrt{b \cos (c+d x)} \, dx+\frac{\left (b^3 B \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{3 \sqrt{b \cos (c+d x)}}\\ &=\frac{2 b^3 B \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d \sqrt{b \cos (c+d x)}}+\frac{2 A b^5 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac{2 b^4 B \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}}+\frac{6 A b^3 \sin (c+d x)}{5 d \sqrt{b \cos (c+d x)}}-\frac{\left (3 A b^2 \sqrt{b \cos (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 \sqrt{\cos (c+d x)}}\\ &=-\frac{6 A b^2 \sqrt{b \cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d \sqrt{\cos (c+d x)}}+\frac{2 b^3 B \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d \sqrt{b \cos (c+d x)}}+\frac{2 A b^5 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac{2 b^4 B \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}}+\frac{6 A b^3 \sin (c+d x)}{5 d \sqrt{b \cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.235969, size = 102, normalized size = 0.58 \[ -\frac{2 b^4 \left (-\frac{9}{2} A \sin (2 (c+d x))-3 A \tan (c+d x)+9 A \cos ^{\frac{3}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )-5 B \sin (c+d x)-5 B \cos ^{\frac{3}{2}}(c+d x) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{15 d (b \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^6,x]

[Out]

(-2*b^4*(9*A*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2] - 5*B*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2] -
5*B*Sin[c + d*x] - (9*A*Sin[2*(c + d*x)])/2 - 3*A*Tan[c + d*x]))/(15*d*(b*Cos[c + d*x])^(3/2))

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Maple [B]  time = 7.057, size = 578, 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((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^6,x)

[Out]

2/15*(b*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*b^2/sin(1/2*d*x+1/2*c)^3/(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)*(36*A*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)^4-72*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c
)^6+20*B*EllipticF(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)*s
in(1/2*d*x+1/2*c)^4-36*A*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+72*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-20*B*EllipticF(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+20*B*cos(1/2*
d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+9*A*(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(co
s(1/2*d*x+1/2*c),2^(1/2))-24*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+5*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin
(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-10*B*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c))
*(-2*b*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2*b)^(1/2)/(b*(2*cos(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 )} \left (b \cos \left (d x + c\right )\right )^{\frac{5}{2}} \sec \left (d x + c\right )^{6}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^6,x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c))^(5/2)*sec(d*x + c)^6, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (B b^{2} \cos \left (d x + c\right )^{3} + A b^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt{b \cos \left (d x + c\right )} \sec \left (d x + c\right )^{6}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^6,x, algorithm="fricas")

[Out]

integral((B*b^2*cos(d*x + c)^3 + A*b^2*cos(d*x + c)^2)*sqrt(b*cos(d*x + c))*sec(d*x + c)^6, 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((b*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)**6,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 )} \left (b \cos \left (d x + c\right )\right )^{\frac{5}{2}} \sec \left (d x + c\right )^{6}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^6,x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c))^(5/2)*sec(d*x + c)^6, x)