3.710 \(\int \frac{(a+b \cos (c+d x))^3}{\sqrt{\sec (c+d x)}} \, dx\)

Optimal. Leaf size=199 \[ \frac{2 b \left (21 a^2+5 b^2\right ) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{2 b \left (21 a^2+5 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{2 a \left (5 a^2+9 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 b^2 \sin (c+d x) (a \sec (c+d x)+b)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{32 a b^2 \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)} \]

[Out]

(2*a*(5*a^2 + 9*b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*b*(21*a^2 + 5
*b^2)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (32*a*b^2*Sin[c + d*x])/(35*d*
Sec[c + d*x]^(3/2)) + (2*b*(21*a^2 + 5*b^2)*Sin[c + d*x])/(21*d*Sqrt[Sec[c + d*x]]) + (2*b^2*(b + a*Sec[c + d*
x])*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2))

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Rubi [A]  time = 0.247532, antiderivative size = 199, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {3238, 3841, 4047, 3769, 3771, 2641, 4045, 2639} \[ \frac{2 b \left (21 a^2+5 b^2\right ) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{2 b \left (21 a^2+5 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{2 a \left (5 a^2+9 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 b^2 \sin (c+d x) (a \sec (c+d x)+b)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{32 a b^2 \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^3/Sqrt[Sec[c + d*x]],x]

[Out]

(2*a*(5*a^2 + 9*b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*b*(21*a^2 + 5
*b^2)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (32*a*b^2*Sin[c + d*x])/(35*d*
Sec[c + d*x]^(3/2)) + (2*b*(21*a^2 + 5*b^2)*Sin[c + d*x])/(21*d*Sqrt[Sec[c + d*x]]) + (2*b^2*(b + a*Sec[c + d*
x])*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2))

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 3841

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(a^2*C
ot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[1/(d*n), Int[(a + b*Csc[e + f*x]
)^(m - 3)*(d*Csc[e + f*x])^(n + 1)*Simp[a^2*b*(m - 2*n - 2) - a*(3*b^2*n + a^2*(n + 1))*Csc[e + f*x] - b*(b^2*
n + a^2*(m + n - 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 2]
&& ((IntegerQ[m] && LtQ[n, -1]) || (IntegersQ[m + 1/2, 2*n] && LeQ[n, -1]))

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rule 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[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 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
+ f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

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 \frac{(a+b \cos (c+d x))^3}{\sqrt{\sec (c+d x)}} \, dx &=\int \frac{(b+a \sec (c+d x))^3}{\sec ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 b^2 (b+a \sec (c+d x)) \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2}{7} \int \frac{8 a b^2+\frac{1}{2} b \left (21 a^2+5 b^2\right ) \sec (c+d x)+\frac{1}{2} a \left (7 a^2+3 b^2\right ) \sec ^2(c+d x)}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 b^2 (b+a \sec (c+d x)) \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2}{7} \int \frac{8 a b^2+\frac{1}{2} a \left (7 a^2+3 b^2\right ) \sec ^2(c+d x)}{\sec ^{\frac{5}{2}}(c+d x)} \, dx+\frac{1}{7} \left (b \left (21 a^2+5 b^2\right )\right ) \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{32 a b^2 \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 b \left (21 a^2+5 b^2\right ) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{2 b^2 (b+a \sec (c+d x)) \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{1}{21} \left (b \left (21 a^2+5 b^2\right )\right ) \int \sqrt{\sec (c+d x)} \, dx+\frac{1}{5} \left (a \left (5 a^2+9 b^2\right )\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{32 a b^2 \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 b \left (21 a^2+5 b^2\right ) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{2 b^2 (b+a \sec (c+d x)) \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{1}{21} \left (b \left (21 a^2+5 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx+\frac{1}{5} \left (a \left (5 a^2+9 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{2 a \left (5 a^2+9 b^2\right ) \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 \left (21 a^2+5 b^2\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{21 d}+\frac{32 a b^2 \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 b \left (21 a^2+5 b^2\right ) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{2 b^2 (b+a \sec (c+d x)) \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 0.875641, size = 132, normalized size = 0.66 \[ \frac{\sqrt{\sec (c+d x)} \left (b \sin (2 (c+d x)) \left (210 a^2+126 a b \cos (c+d x)+15 b^2 \cos (2 (c+d x))+65 b^2\right )+20 b \left (21 a^2+5 b^2\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+84 a \left (5 a^2+9 b^2\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{210 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^3/Sqrt[Sec[c + d*x]],x]

[Out]

(Sqrt[Sec[c + d*x]]*(84*a*(5*a^2 + 9*b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 20*b*(21*a^2 + 5*b^2)
*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + b*(210*a^2 + 65*b^2 + 126*a*b*Cos[c + d*x] + 15*b^2*Cos[2*(c +
d*x)])*Sin[2*(c + d*x)]))/(210*d)

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Maple [A]  time = 3.052, size = 421, normalized size = 2.1 \begin{align*} -{\frac{2}{105\,d}\sqrt{ \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 240\,{b}^{3}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}+ \left ( -504\,a{b}^{2}-360\,{b}^{3} \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{6}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) + \left ( 420\,{a}^{2}b+504\,a{b}^{2}+280\,{b}^{3} \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) + \left ( -210\,{a}^{2}b-126\,a{b}^{2}-80\,{b}^{3} \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +105\,{a}^{2}b\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +25\,{b}^{3}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -105\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ){a}^{3}-189\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) a{b}^{2} \right ){\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^3/sec(d*x+c)^(1/2),x)

[Out]

-2/105*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(240*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^
8+(-504*a*b^2-360*b^3)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(420*a^2*b+504*a*b^2+280*b^3)*sin(1/2*d*x+1/2*c
)^4*cos(1/2*d*x+1/2*c)+(-210*a^2*b-126*a*b^2-80*b^3)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+105*a^2*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))+25*b^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-105*(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))*a^3-189*(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))*a*b^2)/(-2*sin(1/2*d*x+1/2*c
)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(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 \frac{{\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\sqrt{\sec \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3/sec(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((b*cos(d*x + c) + a)^3/sqrt(sec(d*x + c)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3/sec(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

integral((b^3*cos(d*x + c)^3 + 3*a*b^2*cos(d*x + c)^2 + 3*a^2*b*cos(d*x + c) + a^3)/sqrt(sec(d*x + c)), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \cos{\left (c + d x \right )}\right )^{3}}{\sqrt{\sec{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))**3/sec(d*x+c)**(1/2),x)

[Out]

Integral((a + b*cos(c + d*x))**3/sqrt(sec(c + d*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\sqrt{\sec \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3/sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c) + a)^3/sqrt(sec(d*x + c)), x)