3.3.23 \(\int (d \cos (a+b x))^{7/2} \csc (a+b x) \, dx\) [223]

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

[Out]

-d^(7/2)*arctan((d*cos(b*x+a))^(1/2)/d^(1/2))/b-d^(7/2)*arctanh((d*cos(b*x+a))^(1/2)/d^(1/2))/b+2/5*d*(d*cos(b
*x+a))^(5/2)/b+2*d^3*(d*cos(b*x+a))^(1/2)/b

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Rubi [A]
time = 0.05, antiderivative size = 99, normalized size of antiderivative = 1.00, 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 = {2645, 327, 335, 218, 212, 209} \begin {gather*} -\frac {d^{7/2} \text {ArcTan}\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 {gather*}

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 209

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

Rule 212

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

Rule 218

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] &&  !Gt
Q[a/b, 0]

Rule 327

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 335

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 2645

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])

Rubi steps

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

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Mathematica [A]
time = 0.15, size = 80, normalized size = 0.81 \begin {gather*} \frac {d^3 \sqrt {d \cos (a+b x)} \left (-5 \tan ^{-1}\left (\sqrt {\cos (a+b x)}\right )-5 \tanh ^{-1}\left (\sqrt {\cos (a+b x)}\right )+\sqrt {\cos (a+b x)} (11+\cos (2 (a+b x)))\right )}{5 b \sqrt {\cos (a+b x)}} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(292\) vs. \(2(81)=162\).
time = 0.36, size = 293, normalized size = 2.96

method	result	size
default	\(-\frac {5 d^{\frac {7}{2}} \ln \left (\frac {2 \sqrt {d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}+4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right ) \sqrt {-d}-16 \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}\, \sqrt {-d}\, d^{3} \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+5 d^{\frac {7}{2}} \ln \left (\frac {2 \sqrt {d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}-4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right ) \sqrt {-d}+16 \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}\, \sqrt {-d}\, d^{3} \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-24 d^{3} \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}\, \sqrt {-d}-10 d^{4} \ln \left (\frac {2 \sqrt {-d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right )}{10 \sqrt {-d}\, b}\)	\(293\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10/(-d)^(1/2)*(5*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))*(-d)^(1/2)-16*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)*(-d)^(1/2)*d^3*sin(1/2*b*x+1/2*a)^4+5*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))*(-d)^(
1/2)+16*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)*(-d)^(1/2)*d^3*sin(1/2*b*x+1/2*a)^2-24*d^3*(-2*sin(1/2*b*x+1/2*a)^
2*d+d)^(1/2)*(-d)^(1/2)-10*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)))/b

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Maxima [A]
time = 0.52, size = 98, normalized size = 0.99 \begin {gather*} -\frac {10 \, d^{\frac {9}{2}} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right ) - 5 \, d^{\frac {9}{2}} \log \left (\frac {\sqrt {d \cos \left (b x + a\right )} - \sqrt {d}}{\sqrt {d \cos \left (b x + a\right )} + \sqrt {d}}\right ) - 4 \, \left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}} d^{2} - 20 \, \sqrt {d \cos \left (b x + a\right )} d^{4}}{10 \, b d} \end {gather*}

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]

-1/10*(10*d^(9/2)*arctan(sqrt(d*cos(b*x + a))/sqrt(d)) - 5*d^(9/2)*log((sqrt(d*cos(b*x + a)) - sqrt(d))/(sqrt(
d*cos(b*x + a)) + sqrt(d))) - 4*(d*cos(b*x + a))^(5/2)*d^2 - 20*sqrt(d*cos(b*x + a))*d^4)/(b*d)

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Fricas [A]
time = 0.47, size = 299, normalized size = 3.02 \begin {gather*} \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 {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d\,\cos \left (a+b\,x\right )\right )}^{7/2}}{\sin \left (a+b\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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