3.259 \(\int (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^4(c+d x) \, dx\)

Optimal. Leaf size=116 \[ -\frac{2 b^2 (A-C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \cos (c+d x)}}{d \sqrt{\cos (c+d x)}}+\frac{2 A b^3 \sin (c+d x)}{d \sqrt{b \cos (c+d x)}}+\frac{2 b^3 B \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d \sqrt{b \cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.190206, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {16, 3021, 2748, 2642, 2641, 2640, 2639} \[ -\frac{2 b^2 (A-C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \cos (c+d x)}}{d \sqrt{\cos (c+d x)}}+\frac{2 A b^3 \sin (c+d x)}{d \sqrt{b \cos (c+d x)}}+\frac{2 b^3 B \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d \sqrt{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]

[Out]

(-2*b^2*(A - C)*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(d*Sqrt[Cos[c + d*x]]) + (2*b^3*B*Sqrt[Cos[c +
d*x]]*EllipticF[(c + d*x)/2, 2])/(d*Sqrt[b*Cos[c + d*x]]) + (2*A*b^3*Sin[c + d*x])/(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 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

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 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]

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]

Rubi steps

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

Mathematica [A]  time = 0.269255, size = 80, normalized size = 0.69 \[ \frac{2 b^3 \left (-(A-C) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+A \sin (c+d x)+B \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{d \sqrt{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]

[Out]

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

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Maple [A]  time = 3.858, size = 261, normalized size = 2.3 \begin{align*} -2\,{\frac{{b}^{3}\sqrt{-2\,b \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b} \left ( A\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 ) -2\,A\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{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 ) -C\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 ) \right ) }{\sqrt{-b \left ( 2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}- \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }\sin \left ( 1/2\,dx+c/2 \right ) \sqrt{b \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) }d}} \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)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x)

[Out]

-2*b^3*(-2*b*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2*b)^(1/2)*(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(cos(1/2*d*x+1/2*c),2^(1/2))-2*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^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))-C*(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)))/(-b*(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)/(b*(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)/d

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Maxima [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)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C b^{2} \cos \left (d x + c\right )^{4} + 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 )^{4}, 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)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="fricas")

[Out]

integral((C*b^2*cos(d*x + c)^4 + 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
)^4, 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)+C*cos(d*x+c)**2)*sec(d*x+c)**4,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} + 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 )^{4}\,{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)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^(5/2)*sec(d*x + c)^4, x)