3.232 \(\int (d \cos (a+b x))^{11/2} \csc ^2(a+b x) \, dx\)

Optimal. Leaf size=124 \[ -\frac{15 d^5 \sin (a+b x) \sqrt{d \cos (a+b x)}}{7 b}-\frac{9 d^3 \sin (a+b x) (d \cos (a+b x))^{5/2}}{7 b}-\frac{15 d^6 \sqrt{\cos (a+b x)} F\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{7 b \sqrt{d \cos (a+b x)}}-\frac{d \csc (a+b x) (d \cos (a+b x))^{9/2}}{b} \]

[Out]

-((d*(d*Cos[a + b*x])^(9/2)*Csc[a + b*x])/b) - (15*d^6*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2])/(7*b*Sqrt
[d*Cos[a + b*x]]) - (15*d^5*Sqrt[d*Cos[a + b*x]]*Sin[a + b*x])/(7*b) - (9*d^3*(d*Cos[a + b*x])^(5/2)*Sin[a + b
*x])/(7*b)

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Rubi [A]  time = 0.101253, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2567, 2635, 2642, 2641} \[ -\frac{15 d^5 \sin (a+b x) \sqrt{d \cos (a+b x)}}{7 b}-\frac{9 d^3 \sin (a+b x) (d \cos (a+b x))^{5/2}}{7 b}-\frac{15 d^6 \sqrt{\cos (a+b x)} F\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{7 b \sqrt{d \cos (a+b x)}}-\frac{d \csc (a+b x) (d \cos (a+b x))^{9/2}}{b} \]

Antiderivative was successfully verified.

[In]

Int[(d*Cos[a + b*x])^(11/2)*Csc[a + b*x]^2,x]

[Out]

-((d*(d*Cos[a + b*x])^(9/2)*Csc[a + b*x])/b) - (15*d^6*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2])/(7*b*Sqrt
[d*Cos[a + b*x]]) - (15*d^5*Sqrt[d*Cos[a + b*x]]*Sin[a + b*x])/(7*b) - (9*d^3*(d*Cos[a + b*x])^(5/2)*Sin[a + b
*x])/(7*b)

Rule 2567

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(a*Cos[e +
f*x])^(m - 1)*(b*Sin[e + f*x])^(n + 1))/(b*f*(n + 1)), x] + Dist[(a^2*(m - 1))/(b^2*(n + 1)), Int[(a*Cos[e +
f*x])^(m - 2)*(b*Sin[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[m, 1] && LtQ[n, -1] && (Intege
rsQ[2*m, 2*n] || EqQ[m + n, 0])

Rule 2635

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

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 (d \cos (a+b x))^{11/2} \csc ^2(a+b x) \, dx &=-\frac{d (d \cos (a+b x))^{9/2} \csc (a+b x)}{b}-\frac{1}{2} \left (9 d^2\right ) \int (d \cos (a+b x))^{7/2} \, dx\\ &=-\frac{d (d \cos (a+b x))^{9/2} \csc (a+b x)}{b}-\frac{9 d^3 (d \cos (a+b x))^{5/2} \sin (a+b x)}{7 b}-\frac{1}{14} \left (45 d^4\right ) \int (d \cos (a+b x))^{3/2} \, dx\\ &=-\frac{d (d \cos (a+b x))^{9/2} \csc (a+b x)}{b}-\frac{15 d^5 \sqrt{d \cos (a+b x)} \sin (a+b x)}{7 b}-\frac{9 d^3 (d \cos (a+b x))^{5/2} \sin (a+b x)}{7 b}-\frac{1}{14} \left (15 d^6\right ) \int \frac{1}{\sqrt{d \cos (a+b x)}} \, dx\\ &=-\frac{d (d \cos (a+b x))^{9/2} \csc (a+b x)}{b}-\frac{15 d^5 \sqrt{d \cos (a+b x)} \sin (a+b x)}{7 b}-\frac{9 d^3 (d \cos (a+b x))^{5/2} \sin (a+b x)}{7 b}-\frac{\left (15 d^6 \sqrt{\cos (a+b x)}\right ) \int \frac{1}{\sqrt{\cos (a+b x)}} \, dx}{14 \sqrt{d \cos (a+b x)}}\\ &=-\frac{d (d \cos (a+b x))^{9/2} \csc (a+b x)}{b}-\frac{15 d^6 \sqrt{\cos (a+b x)} F\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{7 b \sqrt{d \cos (a+b x)}}-\frac{15 d^5 \sqrt{d \cos (a+b x)} \sin (a+b x)}{7 b}-\frac{9 d^3 (d \cos (a+b x))^{5/2} \sin (a+b x)}{7 b}\\ \end{align*}

Mathematica [A]  time = 0.35911, size = 89, normalized size = 0.72 \[ \frac{d^5 \csc (a+b x) \sqrt{d \cos (a+b x)} \left (\sqrt{\cos (a+b x)} (16 \cos (2 (a+b x))+\cos (4 (a+b x))-45)-60 \sin (a+b x) F\left (\left .\frac{1}{2} (a+b x)\right |2\right )\right )}{28 b \sqrt{\cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Cos[a + b*x])^(11/2)*Csc[a + b*x]^2,x]

[Out]

(d^5*Sqrt[d*Cos[a + b*x]]*Csc[a + b*x]*(Sqrt[Cos[a + b*x]]*(-45 + 16*Cos[2*(a + b*x)] + Cos[4*(a + b*x)]) - 60
*EllipticF[(a + b*x)/2, 2]*Sin[a + b*x]))/(28*b*Sqrt[Cos[a + b*x]])

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Maple [A]  time = 0.204, size = 242, normalized size = 2. \begin{align*} -{\frac{{d}^{7}}{14\,b}\sqrt{d \left ( 2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}}\sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \left ( -128\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{12}+384\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{10}-576\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{8}+30\, \left ( 2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1 \right ) ^{3/2}\sqrt{ \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}}{\it EllipticF} \left ( \cos \left ( 1/2\,bx+a/2 \right ) ,\sqrt{2} \right ) \cos \left ( 1/2\,bx+a/2 \right ) +512\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{6}-204\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}+12\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}+7 \right ) \left ( -2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}d+ \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}d \right ) ^{-{\frac{3}{2}}} \left ( \cos \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{d \left ( 2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1 \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*cos(b*x+a))^(11/2)*csc(b*x+a)^2,x)

[Out]

-1/14*(d*(2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)*d^7/(-2*sin(1/2*b*x+1/2*a)^4*d+sin(1/2*b*x+1/2
*a)^2*d)^(3/2)/cos(1/2*b*x+1/2*a)*sin(1/2*b*x+1/2*a)*(-128*sin(1/2*b*x+1/2*a)^12+384*sin(1/2*b*x+1/2*a)^10-576
*sin(1/2*b*x+1/2*a)^8+30*(2*sin(1/2*b*x+1/2*a)^2-1)^(3/2)*(sin(1/2*b*x+1/2*a)^2)^(1/2)*EllipticF(cos(1/2*b*x+1
/2*a),2^(1/2))*cos(1/2*b*x+1/2*a)+512*sin(1/2*b*x+1/2*a)^6-204*sin(1/2*b*x+1/2*a)^4+12*sin(1/2*b*x+1/2*a)^2+7)
/(d*(2*cos(1/2*b*x+1/2*a)^2-1))^(1/2)/b

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \cos \left (b x + a\right )\right )^{\frac{11}{2}} \csc \left (b x + a\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(b*x+a))^(11/2)*csc(b*x+a)^2,x, algorithm="maxima")

[Out]

integrate((d*cos(b*x + a))^(11/2)*csc(b*x + a)^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{d \cos \left (b x + a\right )} d^{5} \cos \left (b x + a\right )^{5} \csc \left (b x + a\right )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(b*x+a))^(11/2)*csc(b*x+a)^2,x, algorithm="fricas")

[Out]

integral(sqrt(d*cos(b*x + a))*d^5*cos(b*x + a)^5*csc(b*x + a)^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((d*cos(b*x+a))**(11/2)*csc(b*x+a)**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \cos \left (b x + a\right )\right )^{\frac{11}{2}} \csc \left (b x + a\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(b*x+a))^(11/2)*csc(b*x+a)^2,x, algorithm="giac")

[Out]

integrate((d*cos(b*x + a))^(11/2)*csc(b*x + a)^2, x)