3.708 \(\int (a+b \cos (c+d x))^3 \sec ^{\frac{3}{2}}(c+d x) \, dx\)

Optimal. Leaf size=166 \[ \frac{2 a \left (3 a^2-b^2\right ) \sin (c+d x) \sqrt{\sec (c+d x)}}{3 d}+\frac{2 b \left (9 a^2+b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{2 a \left (a^2-3 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 b^2 \sin (c+d x) (a \sec (c+d x)+b)}{3 d \sqrt{\sec (c+d x)}} \]

[Out]

(-2*a*(a^2 - 3*b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*b*(9*a^2 + b^2)*Sq
rt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*a*(3*a^2 - b^2)*Sqrt[Sec[c + d*x]]*S
in[c + d*x])/(3*d) + (2*b^2*(b + a*Sec[c + d*x])*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]])

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Rubi [A]  time = 0.228518, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {3238, 3841, 4047, 3771, 2641, 4046, 2639} \[ \frac{2 a \left (3 a^2-b^2\right ) \sin (c+d x) \sqrt{\sec (c+d x)}}{3 d}+\frac{2 b \left (9 a^2+b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{2 a \left (a^2-3 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 b^2 \sin (c+d x) (a \sec (c+d x)+b)}{3 d \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^3*Sec[c + d*x]^(3/2),x]

[Out]

(-2*a*(a^2 - 3*b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*b*(9*a^2 + b^2)*Sq
rt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*a*(3*a^2 - b^2)*Sqrt[Sec[c + d*x]]*S
in[c + d*x])/(3*d) + (2*b^2*(b + a*Sec[c + d*x])*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]])

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 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 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 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))^3 \sec ^{\frac{3}{2}}(c+d x) \, dx &=\int \frac{(b+a \sec (c+d x))^3}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 b^2 (b+a \sec (c+d x)) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{2}{3} \int \frac{4 a b^2+\frac{1}{2} b \left (9 a^2+b^2\right ) \sec (c+d x)+\frac{1}{2} a \left (3 a^2-b^2\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{2 b^2 (b+a \sec (c+d x)) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{2}{3} \int \frac{4 a b^2+\frac{1}{2} a \left (3 a^2-b^2\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{3} \left (b \left (9 a^2+b^2\right )\right ) \int \sqrt{\sec (c+d x)} \, dx\\ &=\frac{2 a \left (3 a^2-b^2\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d}+\frac{2 b^2 (b+a \sec (c+d x)) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}-\left (a \left (a^2-3 b^2\right )\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{3} \left (b \left (9 a^2+b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 b \left (9 a^2+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)}}{3 d}+\frac{2 a \left (3 a^2-b^2\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d}+\frac{2 b^2 (b+a \sec (c+d x)) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}-\left (a \left (a^2-3 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 (a^2-3 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)}}{d}+\frac{2 b \left (9 a^2+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)}}{3 d}+\frac{2 a \left (3 a^2-b^2\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d}+\frac{2 b^2 (b+a \sec (c+d x)) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.551762, size = 108, normalized size = 0.65 \[ \frac{\sqrt{\sec (c+d x)} \left (2 \sin (c+d x) \left (3 a^3+b^3 \cos (c+d x)\right )+2 b \left (9 a^2+b^2\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-6 a \left (a^2-3 b^2\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^3*Sec[c + d*x]^(3/2),x]

[Out]

(Sqrt[Sec[c + d*x]]*(-6*a*(a^2 - 3*b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 2*b*(9*a^2 + b^2)*Sqrt[
Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 2*(3*a^3 + b^3*Cos[c + d*x])*Sin[c + d*x]))/(3*d)

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Maple [A]  time = 3.428, size = 303, normalized size = 1.8 \begin{align*} -{\frac{2}{3\,d} \left ( 4\,{b}^{3}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+9\,{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 ) +{b}^{3}\sqrt{ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) ,\sqrt{2} \right ) +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 EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ){a}^{3}-9\,\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}-6\,{a}^{3}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-2\,{b}^{3}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) \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)^(3/2),x)

[Out]

-2/3*(4*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+9*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))+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))+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)*El
lipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^3-9*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipt
icE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^2-6*a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-2*b^3*cos(1/2*d*x+1/2*c)*s
in(1/2*d*x+1/2*c)^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{\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{\frac{3}{2}}\,{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)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (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}\right )} \sec \left (d x + c\right )^{\frac{3}{2}}, 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)^(3/2),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)*sec(d*x + c)^(3/2), 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))**3*sec(d*x+c)**(3/2),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 )}^{3} \sec \left (d x + c\right )^{\frac{3}{2}}\,{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)^(3/2),x, algorithm="giac")

[Out]

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