3.223 \(\int (d \cos (a+b x))^{7/2} \csc (a+b x) \, dx\)

Optimal. Leaf size=99 \[ \frac{2 d^3 \sqrt{d \cos (a+b x)}}{b}-\frac{d^{7/2} \tan ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{b}-\frac{d^{7/2} \tanh ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{b}+\frac{2 d (d \cos (a+b x))^{5/2}}{5 b} \]

[Out]

-((d^(7/2)*ArcTan[Sqrt[d*Cos[a + b*x]]/Sqrt[d]])/b) - (d^(7/2)*ArcTanh[Sqrt[d*Cos[a + b*x]]/Sqrt[d]])/b + (2*d
^3*Sqrt[d*Cos[a + b*x]])/b + (2*d*(d*Cos[a + b*x])^(5/2))/(5*b)

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Rubi [A]  time = 0.0710352, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {2565, 321, 329, 212, 206, 203} \[ \frac{2 d^3 \sqrt{d \cos (a+b x)}}{b}-\frac{d^{7/2} \tan ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{b}-\frac{d^{7/2} \tanh ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{b}+\frac{2 d (d \cos (a+b x))^{5/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[(d*Cos[a + b*x])^(7/2)*Csc[a + b*x],x]

[Out]

-((d^(7/2)*ArcTan[Sqrt[d*Cos[a + b*x]]/Sqrt[d]])/b) - (d^(7/2)*ArcTanh[Sqrt[d*Cos[a + b*x]]/Sqrt[d]])/b + (2*d
^3*Sqrt[d*Cos[a + b*x]])/b + (2*d*(d*Cos[a + b*x])^(5/2))/(5*b)

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int (d \cos (a+b x))^{7/2} \csc (a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x^{7/2}}{1-\frac{x^2}{d^2}} \, dx,x,d \cos (a+b x)\right )}{b d}\\ &=\frac{2 d (d \cos (a+b x))^{5/2}}{5 b}-\frac{d \operatorname{Subst}\left (\int \frac{x^{3/2}}{1-\frac{x^2}{d^2}} \, dx,x,d \cos (a+b x)\right )}{b}\\ &=\frac{2 d^3 \sqrt{d \cos (a+b x)}}{b}+\frac{2 d (d \cos (a+b x))^{5/2}}{5 b}-\frac{d^3 \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (1-\frac{x^2}{d^2}\right )} \, dx,x,d \cos (a+b x)\right )}{b}\\ &=\frac{2 d^3 \sqrt{d \cos (a+b x)}}{b}+\frac{2 d (d \cos (a+b x))^{5/2}}{5 b}-\frac{\left (2 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{x^4}{d^2}} \, dx,x,\sqrt{d \cos (a+b x)}\right )}{b}\\ &=\frac{2 d^3 \sqrt{d \cos (a+b x)}}{b}+\frac{2 d (d \cos (a+b x))^{5/2}}{5 b}-\frac{d^4 \operatorname{Subst}\left (\int \frac{1}{d-x^2} \, dx,x,\sqrt{d \cos (a+b x)}\right )}{b}-\frac{d^4 \operatorname{Subst}\left (\int \frac{1}{d+x^2} \, dx,x,\sqrt{d \cos (a+b x)}\right )}{b}\\ &=-\frac{d^{7/2} \tan ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{b}-\frac{d^{7/2} \tanh ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{b}+\frac{2 d^3 \sqrt{d \cos (a+b x)}}{b}+\frac{2 d (d \cos (a+b x))^{5/2}}{5 b}\\ \end{align*}

Mathematica [A]  time = 0.180964, size = 80, normalized size = 0.81 \[ \frac{d^3 \sqrt{d \cos (a+b x)} \left (\sqrt{\cos (a+b x)} (\cos (2 (a+b x))+11)-5 \tan ^{-1}\left (\sqrt{\cos (a+b x)}\right )-5 \tanh ^{-1}\left (\sqrt{\cos (a+b x)}\right )\right )}{5 b \sqrt{\cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Cos[a + b*x])^(7/2)*Csc[a + b*x],x]

[Out]

(d^3*Sqrt[d*Cos[a + b*x]]*(-5*ArcTan[Sqrt[Cos[a + b*x]]] - 5*ArcTanh[Sqrt[Cos[a + b*x]]] + Sqrt[Cos[a + b*x]]*
(11 + Cos[2*(a + b*x)])))/(5*b*Sqrt[Cos[a + b*x]])

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Maple [B]  time = 0.109, size = 280, normalized size = 2.8 \begin{align*}{\frac{8\,{d}^{3}}{5\,b}\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d+d} \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{4}}-{\frac{1}{2\,b}{d}^{{\frac{7}{2}}}\ln \left ( 2\,{\frac{\sqrt{d}\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d+d}+2\,d\cos \left ( 1/2\,bx+a/2 \right ) -d}{\cos \left ( 1/2\,bx+a/2 \right ) -1}} \right ) }-{\frac{1}{2\,b}{d}^{{\frac{7}{2}}}\ln \left ( 2\,{\frac{\sqrt{d}\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d+d}-2\,d\cos \left ( 1/2\,bx+a/2 \right ) -d}{\cos \left ( 1/2\,bx+a/2 \right ) +1}} \right ) }-{\frac{8\,{d}^{3}}{5\,b}\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d+d} \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}}+{\frac{{d}^{4}}{b}\ln \left ( 2\,{\frac{\sqrt{-d}\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d+d}-d}{\cos \left ( 1/2\,bx+a/2 \right ) }} \right ){\frac{1}{\sqrt{-d}}}}+{\frac{12\,{d}^{3}}{5\,b}\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d+d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*cos(b*x+a))^(7/2)*csc(b*x+a),x)

[Out]

8/5/b*d^3*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)*sin(1/2*b*x+1/2*a)^4-1/2/b*d^(7/2)*ln(2/(cos(1/2*b*x+1/2*a)-1)*(
d^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)+2*d*cos(1/2*b*x+1/2*a)-d))-1/2/b*d^(7/2)*ln(2/(cos(1/2*b*x+1/2*a)+
1)*(d^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-2*d*cos(1/2*b*x+1/2*a)-d))-8/5/b*d^3*(-2*sin(1/2*b*x+1/2*a)^2*
d+d)^(1/2)*sin(1/2*b*x+1/2*a)^2+1/(-d)^(1/2)/b*d^4*ln(2/cos(1/2*b*x+1/2*a)*((-d)^(1/2)*(-2*sin(1/2*b*x+1/2*a)^
2*d+d)^(1/2)-d))+12/5/b*d^3*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(b*x+a))^(7/2)*csc(b*x+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.82196, size = 791, normalized size = 7.99 \begin{align*} \left [\frac{10 \, \sqrt{-d} d^{3} \arctan \left (\frac{2 \, \sqrt{d \cos \left (b x + a\right )} \sqrt{-d}}{d \cos \left (b x + a\right ) + d}\right ) + 5 \, \sqrt{-d} d^{3} \log \left (-\frac{d \cos \left (b x + a\right )^{2} - 4 \, \sqrt{d \cos \left (b x + a\right )} \sqrt{-d}{\left (\cos \left (b x + a\right ) - 1\right )} - 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \,{\left (d^{3} \cos \left (b x + a\right )^{2} + 5 \, d^{3}\right )} \sqrt{d \cos \left (b x + a\right )}}{20 \, b}, \frac{10 \, d^{\frac{7}{2}} \arctan \left (\frac{2 \, \sqrt{d \cos \left (b x + a\right )} \sqrt{d}}{d \cos \left (b x + a\right ) - d}\right ) + 5 \, d^{\frac{7}{2}} \log \left (-\frac{d \cos \left (b x + a\right )^{2} - 4 \, \sqrt{d \cos \left (b x + a\right )} \sqrt{d}{\left (\cos \left (b x + a\right ) + 1\right )} + 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \,{\left (d^{3} \cos \left (b x + a\right )^{2} + 5 \, d^{3}\right )} \sqrt{d \cos \left (b x + a\right )}}{20 \, b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(b*x+a))^(7/2)*csc(b*x+a),x, algorithm="fricas")

[Out]

[1/20*(10*sqrt(-d)*d^3*arctan(2*sqrt(d*cos(b*x + a))*sqrt(-d)/(d*cos(b*x + a) + d)) + 5*sqrt(-d)*d^3*log(-(d*c
os(b*x + a)^2 - 4*sqrt(d*cos(b*x + a))*sqrt(-d)*(cos(b*x + a) - 1) - 6*d*cos(b*x + a) + d)/(cos(b*x + a)^2 + 2
*cos(b*x + a) + 1)) + 8*(d^3*cos(b*x + a)^2 + 5*d^3)*sqrt(d*cos(b*x + a)))/b, 1/20*(10*d^(7/2)*arctan(2*sqrt(d
*cos(b*x + a))*sqrt(d)/(d*cos(b*x + a) - d)) + 5*d^(7/2)*log(-(d*cos(b*x + a)^2 - 4*sqrt(d*cos(b*x + a))*sqrt(
d)*(cos(b*x + a) + 1) + 6*d*cos(b*x + a) + d)/(cos(b*x + a)^2 - 2*cos(b*x + a) + 1)) + 8*(d^3*cos(b*x + a)^2 +
5*d^3)*sqrt(d*cos(b*x + a)))/b]

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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))**(7/2)*csc(b*x+a),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{7}{2}} \csc \left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(b*x+a))^(7/2)*csc(b*x+a),x, algorithm="giac")

[Out]

integrate((d*cos(b*x + a))^(7/2)*csc(b*x + a), x)