Integrand size = 14, antiderivative size = 79 \[ \int \frac {\sin ^4\left (a+b x^n\right )}{x} \, dx=-\frac {\cos (2 a) \operatorname {CosIntegral}\left (2 b x^n\right )}{2 n}+\frac {\cos (4 a) \operatorname {CosIntegral}\left (4 b x^n\right )}{8 n}+\frac {3 \log (x)}{8}+\frac {\sin (2 a) \text {Si}\left (2 b x^n\right )}{2 n}-\frac {\sin (4 a) \text {Si}\left (4 b x^n\right )}{8 n} \] Output:
-1/2*cos(2*a)*Ci(2*b*x^n)/n+1/8*cos(4*a)*Ci(4*b*x^n)/n+3/8*ln(x)+1/2*sin(2 *a)*Si(2*b*x^n)/n-1/8*sin(4*a)*Si(4*b*x^n)/n
Time = 0.12 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.84 \[ \int \frac {\sin ^4\left (a+b x^n\right )}{x} \, dx=\frac {3 \log (x)}{8}+\frac {-4 \cos (2 a) \operatorname {CosIntegral}\left (2 b x^n\right )+\cos (4 a) \operatorname {CosIntegral}\left (4 b x^n\right )+4 \sin (2 a) \text {Si}\left (2 b x^n\right )-\sin (4 a) \text {Si}\left (4 b x^n\right )}{8 n} \] Input:
Integrate[Sin[a + b*x^n]^4/x,x]
Output:
(3*Log[x])/8 + (-4*Cos[2*a]*CosIntegral[2*b*x^n] + Cos[4*a]*CosIntegral[4* b*x^n] + 4*Sin[2*a]*SinIntegral[2*b*x^n] - Sin[4*a]*SinIntegral[4*b*x^n])/ (8*n)
Time = 0.29 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3906, 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.
\(\displaystyle \int \frac {\sin ^4\left (a+b x^n\right )}{x} \, dx\) |
\(\Big \downarrow \) 3906 |
\(\displaystyle \int \left (-\frac {\cos \left (2 a+2 b x^n\right )}{2 x}+\frac {\cos \left (4 a+4 b x^n\right )}{8 x}+\frac {3}{8 x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\cos (2 a) \operatorname {CosIntegral}\left (2 b x^n\right )}{2 n}+\frac {\cos (4 a) \operatorname {CosIntegral}\left (4 b x^n\right )}{8 n}+\frac {\sin (2 a) \text {Si}\left (2 b x^n\right )}{2 n}-\frac {\sin (4 a) \text {Si}\left (4 b x^n\right )}{8 n}+\frac {3 \log (x)}{8}\) |
Input:
Int[Sin[a + b*x^n]^4/x,x]
Output:
-1/2*(Cos[2*a]*CosIntegral[2*b*x^n])/n + (Cos[4*a]*CosIntegral[4*b*x^n])/( 8*n) + (3*Log[x])/8 + (Sin[2*a]*SinIntegral[2*b*x^n])/(2*n) - (Sin[4*a]*Si nIntegral[4*b*x^n])/(8*n)
Int[((e_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_), x _Symbol] :> Int[ExpandTrigReduce[(e*x)^m, (a + b*Sin[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
Time = 2.47 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.84
method | result | size |
derivativedivides | \(\frac {\frac {3 \ln \left (b \,x^{n}\right )}{8}-\frac {\operatorname {Si}\left (4 b \,x^{n}\right ) \sin \left (4 a \right )}{8}+\frac {\operatorname {Ci}\left (4 b \,x^{n}\right ) \cos \left (4 a \right )}{8}+\frac {\operatorname {Si}\left (2 b \,x^{n}\right ) \sin \left (2 a \right )}{2}-\frac {\operatorname {Ci}\left (2 b \,x^{n}\right ) \cos \left (2 a \right )}{2}}{n}\) | \(66\) |
default | \(\frac {\frac {3 \ln \left (b \,x^{n}\right )}{8}-\frac {\operatorname {Si}\left (4 b \,x^{n}\right ) \sin \left (4 a \right )}{8}+\frac {\operatorname {Ci}\left (4 b \,x^{n}\right ) \cos \left (4 a \right )}{8}+\frac {\operatorname {Si}\left (2 b \,x^{n}\right ) \sin \left (2 a \right )}{2}-\frac {\operatorname {Ci}\left (2 b \,x^{n}\right ) \cos \left (2 a \right )}{2}}{n}\) | \(66\) |
risch | \(\frac {i {\mathrm e}^{-4 i a} \pi \,\operatorname {csgn}\left (b \,x^{n}\right )}{16 n}-\frac {i {\mathrm e}^{-4 i a} \operatorname {Si}\left (4 b \,x^{n}\right )}{8 n}-\frac {{\mathrm e}^{-4 i a} \operatorname {expIntegral}_{1}\left (-4 i b \,x^{n}\right )}{16 n}-\frac {{\mathrm e}^{4 i a} \operatorname {expIntegral}_{1}\left (-4 i b \,x^{n}\right )}{16 n}+\frac {3 \ln \left (x \right )}{8}-\frac {i {\mathrm e}^{-2 i a} \pi \,\operatorname {csgn}\left (b \,x^{n}\right )}{4 n}+\frac {i {\mathrm e}^{-2 i a} \operatorname {Si}\left (2 b \,x^{n}\right )}{2 n}+\frac {{\mathrm e}^{-2 i a} \operatorname {expIntegral}_{1}\left (-2 i b \,x^{n}\right )}{4 n}+\frac {{\mathrm e}^{2 i a} \operatorname {expIntegral}_{1}\left (-2 i b \,x^{n}\right )}{4 n}\) | \(154\) |
Input:
int(sin(a+b*x^n)^4/x,x,method=_RETURNVERBOSE)
Output:
1/n*(3/8*ln(b*x^n)-1/8*Si(4*b*x^n)*sin(4*a)+1/8*Ci(4*b*x^n)*cos(4*a)+1/2*S i(2*b*x^n)*sin(2*a)-1/2*Ci(2*b*x^n)*cos(2*a))
Time = 0.08 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.78 \[ \int \frac {\sin ^4\left (a+b x^n\right )}{x} \, dx=\frac {\cos \left (4 \, a\right ) \operatorname {Ci}\left (4 \, b x^{n}\right ) - 4 \, \cos \left (2 \, a\right ) \operatorname {Ci}\left (2 \, b x^{n}\right ) + 3 \, n \log \left (x\right ) - \sin \left (4 \, a\right ) \operatorname {Si}\left (4 \, b x^{n}\right ) + 4 \, \sin \left (2 \, a\right ) \operatorname {Si}\left (2 \, b x^{n}\right )}{8 \, n} \] Input:
integrate(sin(a+b*x^n)^4/x,x, algorithm="fricas")
Output:
1/8*(cos(4*a)*cos_integral(4*b*x^n) - 4*cos(2*a)*cos_integral(2*b*x^n) + 3 *n*log(x) - sin(4*a)*sin_integral(4*b*x^n) + 4*sin(2*a)*sin_integral(2*b*x ^n))/n
\[ \int \frac {\sin ^4\left (a+b x^n\right )}{x} \, dx=\int \frac {\sin ^{4}{\left (a + b x^{n} \right )}}{x}\, dx \] Input:
integrate(sin(a+b*x**n)**4/x,x)
Output:
Integral(sin(a + b*x**n)**4/x, x)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.27 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.39 \[ \int \frac {\sin ^4\left (a+b x^n\right )}{x} \, dx=\frac {{\left ({\rm Ei}\left (4 i \, b x^{n}\right ) + {\rm Ei}\left (-4 i \, b x^{n}\right ) + {\rm Ei}\left (4 i \, b e^{\left (n \overline {\log \left (x\right )}\right )}\right ) + {\rm Ei}\left (-4 i \, b e^{\left (n \overline {\log \left (x\right )}\right )}\right )\right )} \cos \left (4 \, a\right ) - 4 \, {\left ({\rm Ei}\left (2 i \, b x^{n}\right ) + {\rm Ei}\left (-2 i \, b x^{n}\right ) + {\rm Ei}\left (2 i \, b e^{\left (n \overline {\log \left (x\right )}\right )}\right ) + {\rm Ei}\left (-2 i \, b e^{\left (n \overline {\log \left (x\right )}\right )}\right )\right )} \cos \left (2 \, a\right ) + 12 \, n \log \left (x\right ) + {\left (i \, {\rm Ei}\left (4 i \, b x^{n}\right ) - i \, {\rm Ei}\left (-4 i \, b x^{n}\right ) + i \, {\rm Ei}\left (4 i \, b e^{\left (n \overline {\log \left (x\right )}\right )}\right ) - i \, {\rm Ei}\left (-4 i \, b e^{\left (n \overline {\log \left (x\right )}\right )}\right )\right )} \sin \left (4 \, a\right ) - 4 \, {\left (i \, {\rm Ei}\left (2 i \, b x^{n}\right ) - i \, {\rm Ei}\left (-2 i \, b x^{n}\right ) + i \, {\rm Ei}\left (2 i \, b e^{\left (n \overline {\log \left (x\right )}\right )}\right ) - i \, {\rm Ei}\left (-2 i \, b e^{\left (n \overline {\log \left (x\right )}\right )}\right )\right )} \sin \left (2 \, a\right )}{32 \, n} \] Input:
integrate(sin(a+b*x^n)^4/x,x, algorithm="maxima")
Output:
1/32*((Ei(4*I*b*x^n) + Ei(-4*I*b*x^n) + Ei(4*I*b*e^(n*conjugate(log(x)))) + Ei(-4*I*b*e^(n*conjugate(log(x)))))*cos(4*a) - 4*(Ei(2*I*b*x^n) + Ei(-2* I*b*x^n) + Ei(2*I*b*e^(n*conjugate(log(x)))) + Ei(-2*I*b*e^(n*conjugate(lo g(x)))))*cos(2*a) + 12*n*log(x) + (I*Ei(4*I*b*x^n) - I*Ei(-4*I*b*x^n) + I* Ei(4*I*b*e^(n*conjugate(log(x)))) - I*Ei(-4*I*b*e^(n*conjugate(log(x)))))* sin(4*a) - 4*(I*Ei(2*I*b*x^n) - I*Ei(-2*I*b*x^n) + I*Ei(2*I*b*e^(n*conjuga te(log(x)))) - I*Ei(-2*I*b*e^(n*conjugate(log(x)))))*sin(2*a))/n
\[ \int \frac {\sin ^4\left (a+b x^n\right )}{x} \, dx=\int { \frac {\sin \left (b x^{n} + a\right )^{4}}{x} \,d x } \] Input:
integrate(sin(a+b*x^n)^4/x,x, algorithm="giac")
Output:
integrate(sin(b*x^n + a)^4/x, x)
Timed out. \[ \int \frac {\sin ^4\left (a+b x^n\right )}{x} \, dx=\int \frac {{\sin \left (a+b\,x^n\right )}^4}{x} \,d x \] Input:
int(sin(a + b*x^n)^4/x,x)
Output:
int(sin(a + b*x^n)^4/x, x)
\[ \int \frac {\sin ^4\left (a+b x^n\right )}{x} \, dx=\int \frac {\sin \left (x^{n} b +a \right )^{4}}{x}d x \] Input:
int(sin(a+b*x^n)^4/x,x)
Output:
int(sin(x**n*b + a)**4/x,x)