Integrand size = 14, antiderivative size = 245 \[ \int \frac {1}{a+b \sin ^3(c+d x)} \, dx=\frac {2 \arctan \left (\frac {\sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 a^{2/3} \sqrt {a^{2/3}-b^{2/3}} d}+\frac {2 \arctan \left (\frac {(-1)^{2/3} \sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 a^{2/3} \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}} d}-\frac {2 \arctan \left (\frac {\sqrt [3]{-1} \left (\sqrt [3]{b}+(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 a^{2/3} \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}} d} \] Output:
2/3*arctan((b^(1/3)+a^(1/3)*tan(1/2*d*x+1/2*c))/(a^(2/3)-b^(2/3))^(1/2))/a ^(2/3)/(a^(2/3)-b^(2/3))^(1/2)/d+2/3*arctan(((-1)^(2/3)*b^(1/3)+a^(1/3)*ta n(1/2*d*x+1/2*c))/(a^(2/3)+(-1)^(1/3)*b^(2/3))^(1/2))/a^(2/3)/(a^(2/3)+(-1 )^(1/3)*b^(2/3))^(1/2)/d-2/3*arctan((-1)^(1/3)*(b^(1/3)+(-1)^(2/3)*a^(1/3) *tan(1/2*d*x+1/2*c))/(a^(2/3)-(-1)^(2/3)*b^(2/3))^(1/2))/a^(2/3)/(a^(2/3)- (-1)^(2/3)*b^(2/3))^(1/2)/d
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.05 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.51 \[ \int \frac {1}{a+b \sin ^3(c+d x)} \, dx=-\frac {2 i \text {RootSum}\left [-i b+3 i b \text {$\#$1}^2+8 a \text {$\#$1}^3-3 i b \text {$\#$1}^4+i b \text {$\#$1}^6\&,\frac {2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}-i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}}{b-4 i a \text {$\#$1}-2 b \text {$\#$1}^2+b \text {$\#$1}^4}\&\right ]}{3 d} \] Input:
Integrate[(a + b*Sin[c + d*x]^3)^(-1),x]
Output:
(((-2*I)/3)*RootSum[(-I)*b + (3*I)*b*#1^2 + 8*a*#1^3 - (3*I)*b*#1^4 + I*b* #1^6 & , (2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1 - I*Log[1 - 2*Cos[ c + d*x]*#1 + #1^2]*#1)/(b - (4*I)*a*#1 - 2*b*#1^2 + b*#1^4) & ])/d
Time = 0.50 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3042, 3692, 2009}
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.
| \(\displaystyle \int \frac {1}{a+b \sin ^3(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{a+b \sin (c+d x)^3}dx\) |
| \(\Big \downarrow \) 3692 |
| \(\displaystyle \int \left (-\frac {1}{3 a^{2/3} \left (-\sqrt [3]{a}-\sqrt [3]{b} \sin (c+d x)\right )}-\frac {1}{3 a^{2/3} \left (\sqrt [3]{-1} \sqrt [3]{b} \sin (c+d x)-\sqrt [3]{a}\right )}-\frac {1}{3 a^{2/3} \left (-\sqrt [3]{a}-(-1)^{2/3} \sqrt [3]{b} \sin (c+d x)\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 a^{2/3} d \sqrt {a^{2/3}-b^{2/3}}}+\frac {2 \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+(-1)^{2/3} \sqrt [3]{b}}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 a^{2/3} d \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}-\frac {2 \arctan \left (\frac {\sqrt [3]{-1} \left ((-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 a^{2/3} d \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\) |
Input:
Int[(a + b*Sin[c + d*x]^3)^(-1),x]
Output:
(2*ArcTan[(b^(1/3) + a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) - b^(2/3)]])/( 3*a^(2/3)*Sqrt[a^(2/3) - b^(2/3)]*d) + (2*ArcTan[((-1)^(2/3)*b^(1/3) + a^( 1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) + (-1)^(1/3)*b^(2/3)]])/(3*a^(2/3)*Sqr t[a^(2/3) + (-1)^(1/3)*b^(2/3)]*d) - (2*ArcTan[((-1)^(1/3)*(b^(1/3) + (-1) ^(2/3)*a^(1/3)*Tan[(c + d*x)/2]))/Sqrt[a^(2/3) - (-1)^(2/3)*b^(2/3)]])/(3* a^(2/3)*Sqrt[a^(2/3) - (-1)^(2/3)*b^(2/3)]*d)
Int[((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Int[ExpandTrig[(a + b*(c*sin[e + f*x])^n)^p, x], x] /; FreeQ[{a, b, c, e, f , n}, x] && (IGtQ[p, 0] || (EqQ[p, -1] && IntegerQ[n]))
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.42 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.34
| method | result | size |
| derivativedivides | \(\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{4}+2 \textit {\_R}^{2}+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +a \textit {\_R}}}{3 d}\) | \(83\) |
| default | \(\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{4}+2 \textit {\_R}^{2}+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +a \textit {\_R}}}{3 d}\) | \(83\) |
| risch | \(\munderset {\textit {\_R} =\operatorname {RootOf}\left (1+\left (729 a^{6} d^{6}-729 a^{4} b^{2} d^{6}\right ) \textit {\_Z}^{6}+243 a^{4} d^{4} \textit {\_Z}^{4}+27 a^{2} d^{2} \textit {\_Z}^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\left (-\frac {486 d^{5} a^{6}}{b}+486 b \,d^{5} a^{4}\right ) \textit {\_R}^{5}+\left (-\frac {81 i a^{5} d^{4}}{b}+81 i b \,a^{3} d^{4}\right ) \textit {\_R}^{4}+\left (-\frac {135 d^{3} a^{4}}{b}-27 a^{2} b \,d^{3}\right ) \textit {\_R}^{3}-\frac {27 i a^{3} d^{2} \textit {\_R}^{2}}{b}-\frac {9 d \,a^{2} \textit {\_R}}{b}-\frac {2 i a}{b}\right )\) | \(175\) |
Input:
int(1/(a+b*sin(d*x+c)^3),x,method=_RETURNVERBOSE)
Output:
1/3/d*sum((_R^4+2*_R^2+1)/(_R^5*a+2*_R^3*a+4*_R^2*b+_R*a)*ln(tan(1/2*d*x+1 /2*c)-_R),_R=RootOf(_Z^6*a+3*_Z^4*a+8*_Z^3*b+3*_Z^2*a+a))
Result contains complex when optimal does not.
Time = 1.05 (sec) , antiderivative size = 25429, normalized size of antiderivative = 103.79 \[ \int \frac {1}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate(1/(a+b*sin(d*x+c)^3),x, algorithm="fricas")
Output:
Too large to include
\[ \int \frac {1}{a+b \sin ^3(c+d x)} \, dx=\int \frac {1}{a + b \sin ^{3}{\left (c + d x \right )}}\, dx \] Input:
integrate(1/(a+b*sin(d*x+c)**3),x)
Output:
Integral(1/(a + b*sin(c + d*x)**3), x)
\[ \int \frac {1}{a+b \sin ^3(c+d x)} \, dx=\int { \frac {1}{b \sin \left (d x + c\right )^{3} + a} \,d x } \] Input:
integrate(1/(a+b*sin(d*x+c)^3),x, algorithm="maxima")
Output:
integrate(1/(b*sin(d*x + c)^3 + a), x)
\[ \int \frac {1}{a+b \sin ^3(c+d x)} \, dx=\int { \frac {1}{b \sin \left (d x + c\right )^{3} + a} \,d x } \] Input:
integrate(1/(a+b*sin(d*x+c)^3),x, algorithm="giac")
Output:
integrate(1/(b*sin(d*x + c)^3 + a), x)
Time = 37.41 (sec) , antiderivative size = 609, normalized size of antiderivative = 2.49 \[ \int \frac {1}{a+b \sin ^3(c+d x)} \, dx=\frac {\sum _{k=1}^6\ln \left (-\frac {8192\,a\,b^3\,\left (-729\,a^5+243\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,b-324\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,\mathrm {root}\left (d^6+27\,a^2\,d^4+243\,a^4\,d^2+729\,a^4\,\left (a^2-b^2\right ),d,k\right )+972\,a^3\,b^2+a^3\,b\,\mathrm {root}\left (d^6+27\,a^2\,d^4+243\,a^4\,d^2+729\,a^4\,\left (a^2-b^2\right ),d,k\right )\,243-162\,a^3\,{\mathrm {root}\left (d^6+27\,a^2\,d^4+243\,a^4\,d^2+729\,a^4\,\left (a^2-b^2\right ),d,k\right )}^2+648\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^2\,\mathrm {root}\left (d^6+27\,a^2\,d^4+243\,a^4\,d^2+729\,a^4\,\left (a^2-b^2\right ),d,k\right )+216\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b\,{\mathrm {root}\left (d^6+27\,a^2\,d^4+243\,a^4\,d^2+729\,a^4\,\left (a^2-b^2\right ),d,k\right )}^2-72\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,{\mathrm {root}\left (d^6+27\,a^2\,d^4+243\,a^4\,d^2+729\,a^4\,\left (a^2-b^2\right ),d,k\right )}^3+36\,a\,b\,{\mathrm {root}\left (d^6+27\,a^2\,d^4+243\,a^4\,d^2+729\,a^4\,\left (a^2-b^2\right ),d,k\right )}^3-9\,a\,{\mathrm {root}\left (d^6+27\,a^2\,d^4+243\,a^4\,d^2+729\,a^4\,\left (a^2-b^2\right ),d,k\right )}^4+24\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b\,{\mathrm {root}\left (d^6+27\,a^2\,d^4+243\,a^4\,d^2+729\,a^4\,\left (a^2-b^2\right ),d,k\right )}^4-4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\mathrm {root}\left (d^6+27\,a^2\,d^4+243\,a^4\,d^2+729\,a^4\,\left (a^2-b^2\right ),d,k\right )}^5\right )}{{\mathrm {root}\left (d^6+27\,a^2\,d^4+243\,a^4\,d^2+729\,a^4\,\left (a^2-b^2\right ),d,k\right )}^5}\right )\,\mathrm {root}\left (729\,a^4\,b^2\,d^6-729\,a^6\,d^6-243\,a^4\,d^4-27\,a^2\,d^2-1,d,k\right )}{d} \] Input:
int(1/(a + b*sin(c + d*x)^3),x)
Output:
symsum(log(-(8192*a*b^3*(972*a^3*b^2 - 729*a^5 - 9*a*root(d^6 + 27*a^2*d^4 + 243*a^4*d^2 + 729*a^4*(a^2 - b^2), d, k)^4 - 162*a^3*root(d^6 + 27*a^2* d^4 + 243*a^4*d^2 + 729*a^4*(a^2 - b^2), d, k)^2 - 4*tan(c/2 + (d*x)/2)*ro ot(d^6 + 27*a^2*d^4 + 243*a^4*d^2 + 729*a^4*(a^2 - b^2), d, k)^5 + 243*a^4 *b*tan(c/2 + (d*x)/2) - 324*tan(c/2 + (d*x)/2)*a^4*root(d^6 + 27*a^2*d^4 + 243*a^4*d^2 + 729*a^4*(a^2 - b^2), d, k) + 24*b*tan(c/2 + (d*x)/2)*root(d ^6 + 27*a^2*d^4 + 243*a^4*d^2 + 729*a^4*(a^2 - b^2), d, k)^4 - 72*a^2*tan( c/2 + (d*x)/2)*root(d^6 + 27*a^2*d^4 + 243*a^4*d^2 + 729*a^4*(a^2 - b^2), d, k)^3 + 36*a*b*root(d^6 + 27*a^2*d^4 + 243*a^4*d^2 + 729*a^4*(a^2 - b^2) , d, k)^3 + 243*b*a^3*root(d^6 + 27*a^2*d^4 + 243*a^4*d^2 + 729*a^4*(a^2 - b^2), d, k) + 648*tan(c/2 + (d*x)/2)*a^2*b^2*root(d^6 + 27*a^2*d^4 + 243* a^4*d^2 + 729*a^4*(a^2 - b^2), d, k) + 216*a^2*b*tan(c/2 + (d*x)/2)*root(d ^6 + 27*a^2*d^4 + 243*a^4*d^2 + 729*a^4*(a^2 - b^2), d, k)^2))/root(d^6 + 27*a^2*d^4 + 243*a^4*d^2 + 729*a^4*(a^2 - b^2), d, k)^5)*root(729*a^4*b^2* d^6 - 729*a^6*d^6 - 243*a^4*d^4 - 27*a^2*d^2 - 1, d, k), k, 1, 6)/d
\[ \int \frac {1}{a+b \sin ^3(c+d x)} \, dx=\int \frac {1}{\sin \left (d x +c \right )^{3} b +a}d x \] Input:
int(1/(a+b*sin(d*x+c)^3),x)
Output:
int(1/(sin(c + d*x)**3*b + a),x)