3.243 \(\int (d \cos (a+b x))^{9/2} \csc ^3(a+b x) \, dx\)

Optimal. Leaf size=113 \[ -\frac {7 d^{9/2} \tan ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b}+\frac {7 d^{9/2} \tanh ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b}-\frac {7 d^3 (d \cos (a+b x))^{3/2}}{6 b}-\frac {d \csc ^2(a+b x) (d \cos (a+b x))^{7/2}}{2 b} \]

[Out]

-7/4*d^(9/2)*arctan((d*cos(b*x+a))^(1/2)/d^(1/2))/b+7/4*d^(9/2)*arctanh((d*cos(b*x+a))^(1/2)/d^(1/2))/b-7/6*d^
3*(d*cos(b*x+a))^(3/2)/b-1/2*d*(d*cos(b*x+a))^(7/2)*csc(b*x+a)^2/b

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Rubi [A]  time = 0.08, antiderivative size = 113, normalized size of antiderivative = 1.00, 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, 298, 203, 206} \[ -\frac {7 d^3 (d \cos (a+b x))^{3/2}}{6 b}-\frac {7 d^{9/2} \tan ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b}+\frac {7 d^{9/2} \tanh ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b}-\frac {d \csc ^2(a+b x) (d \cos (a+b x))^{7/2}}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[(d*Cos[a + b*x])^(9/2)*Csc[a + b*x]^3,x]

[Out]

(-7*d^(9/2)*ArcTan[Sqrt[d*Cos[a + b*x]]/Sqrt[d]])/(4*b) + (7*d^(9/2)*ArcTanh[Sqrt[d*Cos[a + b*x]]/Sqrt[d]])/(4
*b) - (7*d^3*(d*Cos[a + b*x])^(3/2))/(6*b) - (d*(d*Cos[a + b*x])^(7/2)*Csc[a + b*x]^2)/(2*b)

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])

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 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 298

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

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 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])

Rubi steps

\begin {align*} \int (d \cos (a+b x))^{9/2} \csc ^3(a+b x) \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {x^{9/2}}{\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))^{7/2} \csc ^2(a+b x)}{2 b}+\frac {(7 d) \operatorname {Subst}\left (\int \frac {x^{5/2}}{1-\frac {x^2}{d^2}} \, dx,x,d \cos (a+b x)\right )}{4 b}\\ &=-\frac {7 d^3 (d \cos (a+b x))^{3/2}}{6 b}-\frac {d (d \cos (a+b x))^{7/2} \csc ^2(a+b x)}{2 b}+\frac {\left (7 d^3\right ) \operatorname {Subst}\left (\int \frac {\sqrt {x}}{1-\frac {x^2}{d^2}} \, dx,x,d \cos (a+b x)\right )}{4 b}\\ &=-\frac {7 d^3 (d \cos (a+b x))^{3/2}}{6 b}-\frac {d (d \cos (a+b x))^{7/2} \csc ^2(a+b x)}{2 b}+\frac {\left (7 d^3\right ) \operatorname {Subst}\left (\int \frac {x^2}{1-\frac {x^4}{d^2}} \, dx,x,\sqrt {d \cos (a+b x)}\right )}{2 b}\\ &=-\frac {7 d^3 (d \cos (a+b x))^{3/2}}{6 b}-\frac {d (d \cos (a+b x))^{7/2} \csc ^2(a+b x)}{2 b}+\frac {\left (7 d^5\right ) \operatorname {Subst}\left (\int \frac {1}{d-x^2} \, dx,x,\sqrt {d \cos (a+b x)}\right )}{4 b}-\frac {\left (7 d^5\right ) \operatorname {Subst}\left (\int \frac {1}{d+x^2} \, dx,x,\sqrt {d \cos (a+b x)}\right )}{4 b}\\ &=-\frac {7 d^{9/2} \tan ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b}+\frac {7 d^{9/2} \tanh ^{-1}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b}-\frac {7 d^3 (d \cos (a+b x))^{3/2}}{6 b}-\frac {d (d \cos (a+b x))^{7/2} \csc ^2(a+b x)}{2 b}\\ \end {align*}

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Mathematica [C]  time = 0.64, size = 78, normalized size = 0.69 \[ \frac {d^5 \left (21 \sqrt [4]{-\cot ^2(a+b x)} \, _2F_1\left (\frac {1}{4},\frac {1}{4};\frac {5}{4};\csc ^2(a+b x)\right )+(2 \cos (2 (a+b x))-5) \cot ^2(a+b x)\right )}{6 b \sqrt {d \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Cos[a + b*x])^(9/2)*Csc[a + b*x]^3,x]

[Out]

(d^5*((-5 + 2*Cos[2*(a + b*x)])*Cot[a + b*x]^2 + 21*(-Cot[a + b*x]^2)^(1/4)*Hypergeometric2F1[1/4, 1/4, 5/4, C
sc[a + b*x]^2]))/(6*b*Sqrt[d*Cos[a + b*x]])

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fricas [B]  time = 0.65, size = 405, normalized size = 3.58 \[ \left [-\frac {42 \, {\left (d^{4} \cos \left (b x + a\right )^{2} - d^{4}\right )} \sqrt {-d} \arctan \left (\frac {2 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d}}{d \cos \left (b x + a\right ) + d}\right ) - 21 \, {\left (d^{4} \cos \left (b x + a\right )^{2} - d^{4}\right )} \sqrt {-d} \log \left (-\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) - 1\right )} - 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, {\left (4 \, d^{4} \cos \left (b x + a\right )^{3} - 7 \, d^{4} \cos \left (b x + a\right )\right )} \sqrt {d \cos \left (b x + a\right )}}{48 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )}}, \frac {42 \, {\left (d^{4} \cos \left (b x + a\right )^{2} - d^{4}\right )} \sqrt {d} \arctan \left (\frac {2 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d}}{d \cos \left (b x + a\right ) - d}\right ) + 21 \, {\left (d^{4} \cos \left (b x + a\right )^{2} - d^{4}\right )} \sqrt {d} \log \left (-\frac {d \cos \left (b x + a\right )^{2} + 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d} {\left (\cos \left (b x + a\right ) + 1\right )} + 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1}\right ) - 8 \, {\left (4 \, d^{4} \cos \left (b x + a\right )^{3} - 7 \, d^{4} \cos \left (b x + a\right )\right )} \sqrt {d \cos \left (b x + a\right )}}{48 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(b*x+a))^(9/2)*csc(b*x+a)^3,x, algorithm="fricas")

[Out]

[-1/48*(42*(d^4*cos(b*x + a)^2 - d^4)*sqrt(-d)*arctan(2*sqrt(d*cos(b*x + a))*sqrt(-d)/(d*cos(b*x + a) + d)) -
21*(d^4*cos(b*x + a)^2 - d^4)*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^4*cos(b*x + a)^3 - 7*d^4*cos(b*x
+ a))*sqrt(d*cos(b*x + a)))/(b*cos(b*x + a)^2 - b), 1/48*(42*(d^4*cos(b*x + a)^2 - d^4)*sqrt(d)*arctan(2*sqrt(
d*cos(b*x + a))*sqrt(d)/(d*cos(b*x + a) - d)) + 21*(d^4*cos(b*x + a)^2 - d^4)*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^4*cos(b*x + a)^3 - 7*d^4*cos(b*x + a))*sqrt(d*cos(b*x + a)))/(b*cos(b*x + a)^2 - b)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \cos \left (b x + a\right )\right )^{\frac {9}{2}} \csc \left (b x + a\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(b*x+a))^(9/2)*csc(b*x+a)^3,x, algorithm="giac")

[Out]

integrate((d*cos(b*x + a))^(9/2)*csc(b*x + a)^3, x)

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maple [B]  time = 0.31, size = 394, normalized size = 3.49 \[ -\frac {4 d^{4} \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d -d}}{3 b}-\frac {4 d^{4} \sqrt {2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d -d}}{3 b}+\frac {2 d^{4} \sqrt {d \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right )}}{b}+\frac {7 d^{\frac {9}{2}} \ln \left (\frac {2 \sqrt {d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}+4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right )}{8 b}-\frac {d^{4} \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}}{16 b \left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}+\frac {7 d^{\frac {9}{2}} \ln \left (\frac {2 \sqrt {d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}-4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right )}{8 b}+\frac {7 d^{5} \ln \left (\frac {-2 d +2 \sqrt {-d}\, \sqrt {2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d -d}}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right )}{4 b \sqrt {-d}}+\frac {d^{4} \sqrt {2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d -d}}{8 b \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}+\frac {d^{4} \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}}{16 b \left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*cos(b*x+a))^(9/2)*csc(b*x+a)^3,x)

[Out]

-4/3/b*d^4*cos(1/2*b*x+1/2*a)^2*(2*cos(1/2*b*x+1/2*a)^2*d-d)^(1/2)-4/3/b*d^4*(2*cos(1/2*b*x+1/2*a)^2*d-d)^(1/2
)+2/b*d^4*(d*(2*cos(1/2*b*x+1/2*a)^2-1))^(1/2)+7/8/b*d^(9/2)*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^4/(cos(1/2*b*x+1/2*a)+1)*(-2*sin(1/2*b*x+1/2*a)^
2*d+d)^(1/2)+7/8/b*d^(9/2)*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))+7/4/b*d^5/(-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))+1/8/b*d^4/cos(1/2*b*x+1/2*a)^2*(2*cos(1/2*b*x+1/2*a)^2*d-d)^(1/2)+1/16/b*d^4/(cos(1/2*b*x+1/2*a)-1)*(
-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)

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maxima [A]  time = 0.71, size = 118, normalized size = 1.04 \[ \frac {\frac {12 \, \left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}} d^{6}}{d^{2} \cos \left (b x + a\right )^{2} - d^{2}} - 42 \, d^{\frac {11}{2}} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right ) - 21 \, d^{\frac {11}{2}} \log \left (\frac {\sqrt {d \cos \left (b x + a\right )} - \sqrt {d}}{\sqrt {d \cos \left (b x + a\right )} + \sqrt {d}}\right ) - 16 \, \left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}} d^{4}}{24 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(b*x+a))^(9/2)*csc(b*x+a)^3,x, algorithm="maxima")

[Out]

1/24*(12*(d*cos(b*x + a))^(3/2)*d^6/(d^2*cos(b*x + a)^2 - d^2) - 42*d^(11/2)*arctan(sqrt(d*cos(b*x + a))/sqrt(
d)) - 21*d^(11/2)*log((sqrt(d*cos(b*x + a)) - sqrt(d))/(sqrt(d*cos(b*x + a)) + sqrt(d))) - 16*(d*cos(b*x + a))
^(3/2)*d^4)/(b*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d\,\cos \left (a+b\,x\right )\right )}^{9/2}}{{\sin \left (a+b\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*cos(a + b*x))^(9/2)/sin(a + b*x)^3,x)

[Out]

int((d*cos(a + b*x))^(9/2)/sin(a + b*x)^3, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(b*x+a))**(9/2)*csc(b*x+a)**3,x)

[Out]

Timed out

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