Integrand size = 21, antiderivative size = 113 \[ \int (d \cos (a+b x))^{7/2} \csc ^3(a+b x) \, dx=\frac {5 d^{7/2} \arctan \left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b}+\frac {5 d^{7/2} \text {arctanh}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b}-\frac {5 d^3 \sqrt {d \cos (a+b x)}}{2 b}-\frac {d (d \cos (a+b x))^{5/2} \csc ^2(a+b x)}{2 b} \]
5/4*d^(7/2)*arctan((d*cos(b*x+a))^(1/2)/d^(1/2))/b+5/4*d^(7/2)*arctanh((d* cos(b*x+a))^(1/2)/d^(1/2))/b-1/2*d*(d*cos(b*x+a))^(5/2)*csc(b*x+a)^2/b-5/2 *d^3*(d*cos(b*x+a))^(1/2)/b
Time = 1.04 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.04 \[ \int (d \cos (a+b x))^{7/2} \csc ^3(a+b x) \, dx=\frac {(d \cos (a+b x))^{7/2} \left (5 \arctan \left (\sqrt {\cos (a+b x)}\right )+3 \text {arctanh}\left (\sqrt {\cos (a+b x)}\right )-8 \sqrt {\cos (a+b x)}-2 \sqrt {\cos (a+b x)} \csc ^2(a+b x)-\log \left (1-\sqrt {\cos (a+b x)}\right )+\log \left (1+\sqrt {\cos (a+b x)}\right )\right )}{4 b \cos ^{\frac {7}{2}}(a+b x)} \]
((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))
Time = 0.29 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3045, 27, 252, 262, 266, 756, 216, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \csc ^3(a+b x) (d \cos (a+b x))^{7/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(d \cos (a+b x))^{7/2}}{\sin (a+b x)^3}dx\) |
\(\Big \downarrow \) 3045 |
\(\displaystyle -\frac {\int \frac {d^4 (d \cos (a+b x))^{7/2}}{\left (d^2-d^2 \cos ^2(a+b x)\right )^2}d(d \cos (a+b x))}{b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {d^3 \int \frac {(d \cos (a+b x))^{7/2}}{\left (d^2-d^2 \cos ^2(a+b x)\right )^2}d(d \cos (a+b x))}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle -\frac {d^3 \left (\frac {(d \cos (a+b x))^{5/2}}{2 \left (d^2-d^2 \cos ^2(a+b x)\right )}-\frac {5}{4} \int \frac {(d \cos (a+b x))^{3/2}}{d^2-d^2 \cos ^2(a+b x)}d(d \cos (a+b x))\right )}{b}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle -\frac {d^3 \left (\frac {(d \cos (a+b x))^{5/2}}{2 \left (d^2-d^2 \cos ^2(a+b x)\right )}-\frac {5}{4} \left (d^2 \int \frac {1}{\sqrt {d \cos (a+b x)} \left (d^2-d^2 \cos ^2(a+b x)\right )}d(d \cos (a+b x))-2 \sqrt {d \cos (a+b x)}\right )\right )}{b}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle -\frac {d^3 \left (\frac {(d \cos (a+b x))^{5/2}}{2 \left (d^2-d^2 \cos ^2(a+b x)\right )}-\frac {5}{4} \left (2 d^2 \int \frac {1}{d^2-d^4 \cos ^4(a+b x)}d\sqrt {d \cos (a+b x)}-2 \sqrt {d \cos (a+b x)}\right )\right )}{b}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle -\frac {d^3 \left (\frac {(d \cos (a+b x))^{5/2}}{2 \left (d^2-d^2 \cos ^2(a+b x)\right )}-\frac {5}{4} \left (2 d^2 \left (\frac {\int \frac {1}{d-d^2 \cos ^2(a+b x)}d\sqrt {d \cos (a+b x)}}{2 d}+\frac {\int \frac {1}{d^2 \cos ^2(a+b x)+d}d\sqrt {d \cos (a+b x)}}{2 d}\right )-2 \sqrt {d \cos (a+b x)}\right )\right )}{b}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {d^3 \left (\frac {(d \cos (a+b x))^{5/2}}{2 \left (d^2-d^2 \cos ^2(a+b x)\right )}-\frac {5}{4} \left (2 d^2 \left (\frac {\int \frac {1}{d-d^2 \cos ^2(a+b x)}d\sqrt {d \cos (a+b x)}}{2 d}+\frac {\arctan \left (\sqrt {d} \cos (a+b x)\right )}{2 d^{3/2}}\right )-2 \sqrt {d \cos (a+b x)}\right )\right )}{b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {d^3 \left (\frac {(d \cos (a+b x))^{5/2}}{2 \left (d^2-d^2 \cos ^2(a+b x)\right )}-\frac {5}{4} \left (2 d^2 \left (\frac {\arctan \left (\sqrt {d} \cos (a+b x)\right )}{2 d^{3/2}}+\frac {\text {arctanh}\left (\sqrt {d} \cos (a+b x)\right )}{2 d^{3/2}}\right )-2 \sqrt {d \cos (a+b x)}\right )\right )}{b}\) |
-((d^3*((d*Cos[a + b*x])^(5/2)/(2*(d^2 - d^2*Cos[a + b*x]^2)) - (5*(2*d^2* (ArcTan[Sqrt[d]*Cos[a + b*x]]/(2*d^(3/2)) + ArcTanh[Sqrt[d]*Cos[a + b*x]]/ (2*d^(3/2))) - 2*Sqrt[d*Cos[a + b*x]]))/4))/b)
3.3.44.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_ Symbol] :> Simp[-(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])
Leaf count of result is larger than twice the leaf count of optimal. \(309\) vs. \(2(89)=178\).
Time = 6.02 (sec) , antiderivative size = 310, normalized size of antiderivative = 2.74
method | result | size |
default | \(\frac {-\frac {d^{3} \sqrt {2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d -d}}{8 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}-\frac {5 d^{4} \ln \left (\frac {-2 d +2 \sqrt {-d}\, \sqrt {2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d -d}}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right )}{4 \sqrt {-d}}-2 d^{3} \sqrt {d \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right )}+\frac {5 d^{\frac {7}{2}} \ln \left (\frac {4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+2 \sqrt {d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right )}{8}+\frac {5 d^{\frac {7}{2}} \ln \left (\frac {-4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+2 \sqrt {d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right )}{8}+\frac {d^{3} \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}}{16 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-16}-\frac {d^{3} \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}}{16 \left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}}{b}\) | \(310\) |
(-1/8*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)-5/4*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))-2*d^3*(d*(2*cos(1/2*b*x+1/2*a)^2-1))^(1/2)+5/8*d^(7/2)*ln(( 4*d*cos(1/2*b*x+1/2*a)+2*d^(1/2)*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2)-2*d)/ (cos(1/2*b*x+1/2*a)-1))+5/8*d^(7/2)*ln((-4*d*cos(1/2*b*x+1/2*a)+2*d^(1/2)* (-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2)-2*d)/(cos(1/2*b*x+1/2*a)+1))+1/16*d^3/ (cos(1/2*b*x+1/2*a)-1)*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2)-1/16*d^3/(cos(1 /2*b*x+1/2*a)+1)*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2))/b
Leaf count of result is larger than twice the leaf count of optimal. 193 vs. \(2 (89) = 178\).
Time = 0.42 (sec) , antiderivative size = 393, normalized size of antiderivative = 3.48 \[ \int (d \cos (a+b x))^{7/2} \csc ^3(a+b x) \, dx=\left [-\frac {10 \, {\left (d^{3} \cos \left (b x + a\right )^{2} - d^{3}\right )} \sqrt {-d} \arctan \left (\frac {2 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d}}{d \cos \left (b x + a\right ) + d}\right ) - 5 \, {\left (d^{3} \cos \left (b x + a\right )^{2} - d^{3}\right )} \sqrt {-d} \log \left (-\frac {d \cos \left (b x + a\right )^{2} + 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) - 1\right )} - 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, {\left (4 \, d^{3} \cos \left (b x + a\right )^{2} - 5 \, d^{3}\right )} \sqrt {d \cos \left (b x + a\right )}}{16 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )}}, -\frac {10 \, {\left (d^{3} \cos \left (b x + a\right )^{2} - d^{3}\right )} \sqrt {d} \arctan \left (\frac {2 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d}}{d \cos \left (b x + a\right ) - d}\right ) - 5 \, {\left (d^{3} \cos \left (b x + a\right )^{2} - d^{3}\right )} \sqrt {d} \log \left (-\frac {d \cos \left (b x + a\right )^{2} + 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d} {\left (\cos \left (b x + a\right ) + 1\right )} + 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, {\left (4 \, d^{3} \cos \left (b x + a\right )^{2} - 5 \, d^{3}\right )} \sqrt {d \cos \left (b x + a\right )}}{16 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )}}\right ] \]
[-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), -1 /16*(10*(d^3*cos(b*x + a)^2 - d^3)*sqrt(d)*arctan(2*sqrt(d*cos(b*x + a))*s qrt(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)]
Timed out. \[ \int (d \cos (a+b x))^{7/2} \csc ^3(a+b x) \, dx=\text {Timed out} \]
Time = 0.28 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.04 \[ \int (d \cos (a+b x))^{7/2} \csc ^3(a+b x) \, dx=\frac {\frac {4 \, \sqrt {d \cos \left (b x + a\right )} d^{6}}{d^{2} \cos \left (b x + a\right )^{2} - d^{2}} + 10 \, d^{\frac {9}{2}} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right ) - 5 \, d^{\frac {9}{2}} \log \left (\frac {\sqrt {d \cos \left (b x + a\right )} - \sqrt {d}}{\sqrt {d \cos \left (b x + a\right )} + \sqrt {d}}\right ) - 16 \, \sqrt {d \cos \left (b x + a\right )} d^{4}}{8 \, b d} \]
1/8*(4*sqrt(d*cos(b*x + a))*d^6/(d^2*cos(b*x + a)^2 - d^2) + 10*d^(9/2)*ar ctan(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))) - 16*sqrt(d*cos(b*x + a))*d^4) /(b*d)
\[ \int (d \cos (a+b x))^{7/2} \csc ^3(a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {7}{2}} \csc \left (b x + a\right )^{3} \,d x } \]
Timed out. \[ \int (d \cos (a+b x))^{7/2} \csc ^3(a+b x) \, dx=\int \frac {{\left (d\,\cos \left (a+b\,x\right )\right )}^{7/2}}{{\sin \left (a+b\,x\right )}^3} \,d x \]