3.244 \(\int (d \cos (a+b x))^{7/2} \csc ^3(a+b x) \, dx\)

Optimal. Leaf size=113 \[ -\frac{5 d^3 \sqrt{d \cos (a+b x)}}{2 b}+\frac{5 d^{7/2} \tan ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{4 b}+\frac{5 d^{7/2} \tanh ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{4 b}-\frac{d \csc ^2(a+b x) (d \cos (a+b x))^{5/2}}{2 b} \]

[Out]

(5*d^(7/2)*ArcTan[Sqrt[d*Cos[a + b*x]]/Sqrt[d]])/(4*b) + (5*d^(7/2)*ArcTanh[Sqrt[d*Cos[a + b*x]]/Sqrt[d]])/(4*
b) - (5*d^3*Sqrt[d*Cos[a + b*x]])/(2*b) - (d*(d*Cos[a + b*x])^(5/2)*Csc[a + b*x]^2)/(2*b)

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Rubi [A]  time = 0.0801313, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2565, 288, 321, 329, 212, 206, 203} \[ -\frac{5 d^3 \sqrt{d \cos (a+b x)}}{2 b}+\frac{5 d^{7/2} \tan ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{4 b}+\frac{5 d^{7/2} \tanh ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{4 b}-\frac{d \csc ^2(a+b x) (d \cos (a+b x))^{5/2}}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[(d*Cos[a + b*x])^(7/2)*Csc[a + b*x]^3,x]

[Out]

(5*d^(7/2)*ArcTan[Sqrt[d*Cos[a + b*x]]/Sqrt[d]])/(4*b) + (5*d^(7/2)*ArcTanh[Sqrt[d*Cos[a + b*x]]/Sqrt[d]])/(4*
b) - (5*d^3*Sqrt[d*Cos[a + b*x]])/(2*b) - (d*(d*Cos[a + b*x])^(5/2)*Csc[a + b*x]^2)/(2*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 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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

Mathematica [A]  time = 1.20431, size = 118, normalized size = 1.04 \[ \frac{(d \cos (a+b x))^{7/2} \left (-8 \sqrt{\cos (a+b x)}-\log \left (1-\sqrt{\cos (a+b x)}\right )+\log \left (\sqrt{\cos (a+b x)}+1\right )+5 \tan ^{-1}\left (\sqrt{\cos (a+b x)}\right )-2 \sqrt{\cos (a+b x)} \csc ^2(a+b x)+3 \tanh ^{-1}\left (\sqrt{\cos (a+b x)}\right )\right )}{4 b \cos ^{\frac{7}{2}}(a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Cos[a + b*x])^(7/2)*Csc[a + b*x]^3,x]

[Out]

((d*Cos[a + b*x])^(7/2)*(5*ArcTan[Sqrt[Cos[a + b*x]]] + 3*ArcTanh[Sqrt[Cos[a + b*x]]] - 8*Sqrt[Cos[a + b*x]] -
2*Sqrt[Cos[a + b*x]]*Csc[a + b*x]^2 - Log[1 - Sqrt[Cos[a + b*x]]] + Log[1 + Sqrt[Cos[a + b*x]]]))/(4*b*Cos[a
+ b*x]^(7/2))

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Maple [B]  time = 0.207, size = 327, normalized size = 2.9 \begin{align*} -2\,{\frac{{d}^{3}\sqrt{d \left ( 2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1 \right ) }}{b}}+{\frac{5}{8\,b}{d}^{{\frac{7}{2}}}\ln \left ({ \left ( 4\,d\cos \left ( 1/2\,bx+a/2 \right ) +2\,\sqrt{d}\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d+d}-2\,d \right ) \left ( \cos \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) -1 \right ) ^{-1}} \right ) }+{\frac{{d}^{3}}{16\,b}\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d+d} \left ( \cos \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) -1 \right ) ^{-1}}+{\frac{5}{8\,b}{d}^{{\frac{7}{2}}}\ln \left ({ \left ( -4\,d\cos \left ( 1/2\,bx+a/2 \right ) +2\,\sqrt{d}\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d+d}-2\,d \right ) \left ( \cos \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) +1 \right ) ^{-1}} \right ) }-{\frac{5\,{d}^{4}}{4\,b}\ln \left ({ \left ( -2\,d+2\,\sqrt{-d}\sqrt{2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d-d} \right ) \left ( \cos \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{-1}} \right ){\frac{1}{\sqrt{-d}}}}-{\frac{{d}^{3}}{16\,b}\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d+d} \left ( \cos \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) +1 \right ) ^{-1}}-{\frac{{d}^{3}}{8\,b}\sqrt{2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d-d} \left ( \cos \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{-2}} \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)^3,x)

[Out]

-2/b*d^3*(d*(2*cos(1/2*b*x+1/2*a)^2-1))^(1/2)+5/8/b*d^(7/2)*ln((4*d*cos(1/2*b*x+1/2*a)+2*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)-1))+1/16/b*d^3/(cos(1/2*b*x+1/2*a)-1)*(-2*sin(1/2*b*x+1/2*a)^2
*d+d)^(1/2)+5/8/b*d^(7/2)*ln((-4*d*cos(1/2*b*x+1/2*a)+2*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)+1))-5/4/b*d^4/(-d)^(1/2)*ln((-2*d+2*(-d)^(1/2)*(2*cos(1/2*b*x+1/2*a)^2*d-d)^(1/2))/cos(1/2*b*x+
1/2*a))-1/16/b*d^3/(cos(1/2*b*x+1/2*a)+1)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-1/8/b*d^3/cos(1/2*b*x+1/2*a)^2*(
2*cos(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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.51047, size = 991, normalized size = 8.77 \begin{align*} \left [-\frac{10 \,{\left (d^{3} \cos \left (b x + a\right )^{2} - d^{3}\right )} \sqrt{-d} \arctan \left (\frac{2 \, \sqrt{d \cos \left (b x + a\right )} \sqrt{-d}}{d \cos \left (b x + a\right ) + d}\right ) - 5 \,{\left (d^{3} \cos \left (b x + a\right )^{2} - d^{3}\right )} \sqrt{-d} \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 (4 \, d^{3} \cos \left (b x + a\right )^{2} - 5 \, d^{3}\right )} \sqrt{d \cos \left (b x + a\right )}}{16 \,{\left (b \cos \left (b x + a\right )^{2} - b\right )}}, -\frac{10 \,{\left (d^{3} \cos \left (b x + a\right )^{2} - d^{3}\right )} \sqrt{d} \arctan \left (\frac{2 \, \sqrt{d \cos \left (b x + a\right )} \sqrt{d}}{d \cos \left (b x + a\right ) - d}\right ) - 5 \,{\left (d^{3} \cos \left (b x + a\right )^{2} - d^{3}\right )} \sqrt{d} \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 (4 \, d^{3} \cos \left (b x + a\right )^{2} - 5 \, d^{3}\right )} \sqrt{d \cos \left (b x + a\right )}}{16 \,{\left (b \cos \left (b x + a\right )^{2} - b\right )}}\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)^3,x, algorithm="fricas")

[Out]

[-1/16*(10*(d^3*cos(b*x + a)^2 - d^3)*sqrt(-d)*arctan(2*sqrt(d*cos(b*x + a))*sqrt(-d)/(d*cos(b*x + a) + d)) -
5*(d^3*cos(b*x + a)^2 - d^3)*sqrt(-d)*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*(4*d^3*cos(b*x + a)^2 - 5*d^3)*sqrt(d*c
os(b*x + a)))/(b*cos(b*x + a)^2 - b), -1/16*(10*(d^3*cos(b*x + a)^2 - d^3)*sqrt(d)*arctan(2*sqrt(d*cos(b*x + a
))*sqrt(d)/(d*cos(b*x + a) - d)) - 5*(d^3*cos(b*x + a)^2 - d^3)*sqrt(d)*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*(4*d^3
*cos(b*x + a)^2 - 5*d^3)*sqrt(d*cos(b*x + a)))/(b*cos(b*x + a)^2 - 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)**3,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 )^{3}\,{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)^3,x, algorithm="giac")

[Out]

integrate((d*cos(b*x + a))^(7/2)*csc(b*x + a)^3, x)