Integrand size = 24, antiderivative size = 751 \[ \int \frac {(e+f x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {3 i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{5/2} d}-\frac {3 i a (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d}-\frac {3 i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{5/2} d}+\frac {3 i a (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d}+\frac {3 a^2 f \log (a+b \sin (c+d x))}{2 b \left (a^2-b^2\right )^2 d^2}-\frac {f \log (a+b \sin (c+d x))}{b \left (a^2-b^2\right ) d^2}+\frac {3 a^3 f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{5/2} d^2}-\frac {3 a f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d^2}-\frac {3 a^3 f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{5/2} d^2}+\frac {3 a f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d^2}-\frac {a (e+f x) \cos (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac {a f}{2 b \left (a^2-b^2\right ) d^2 (a+b \sin (c+d x))}-\frac {3 a^2 (e+f x) \cos (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac {(e+f x) \cos (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))} \]
3/2*a^2*f*ln(a+b*sin(d*x+c))/b/(a^2-b^2)^2/d^2-f*ln(a+b*sin(d*x+c))/b/(a^2 -b^2)/d^2+3/2*I*a^3*(f*x+e)*ln(1-I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b /(a^2-b^2)^(5/2)/d-3/2*I*a*(f*x+e)*ln(1-I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1 /2)))/b/(a^2-b^2)^(3/2)/d-3/2*I*a^3*(f*x+e)*ln(1-I*b*exp(I*(d*x+c))/(a+(a^ 2-b^2)^(1/2)))/b/(a^2-b^2)^(5/2)/d+3/2*I*a*(f*x+e)*ln(1-I*b*exp(I*(d*x+c)) /(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3/2)/d+3/2*a^3*f*polylog(2,I*b*exp(I*(d *x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(5/2)/d^2-3/2*a*f*polylog(2,I*b*ex p(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3/2)/d^2-3/2*a^3*f*polylog( 2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(5/2)/d^2+3/2*a*f*po lylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3/2)/d^2-1/2* a*(f*x+e)*cos(d*x+c)/(a^2-b^2)/d/(a+b*sin(d*x+c))^2-1/2*a*f/b/(a^2-b^2)/d^ 2/(a+b*sin(d*x+c))-3/2*a^2*(f*x+e)*cos(d*x+c)/(a^2-b^2)^2/d/(a+b*sin(d*x+c ))+(f*x+e)*cos(d*x+c)/(a^2-b^2)/d/(a+b*sin(d*x+c))
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(2666\) vs. \(2(751)=1502\).
Time = 15.90 (sec) , antiderivative size = 2666, normalized size of antiderivative = 3.55 \[ \int \frac {(e+f x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Result too large to show} \]
(-(a*d*e*Cos[c + d*x]) + a*c*f*Cos[c + d*x] - a*f*(c + d*x)*Cos[c + d*x])/ (2*(a - b)*(a + b)*d^2*(a + b*Sin[c + d*x])^2) + (-(a^3*f) + a*b^2*f - a^2 *b*d*e*Cos[c + d*x] - 2*b^3*d*e*Cos[c + d*x] + a^2*b*c*f*Cos[c + d*x] + 2* b^3*c*f*Cos[c + d*x] - a^2*b*f*(c + d*x)*Cos[c + d*x] - 2*b^3*f*(c + d*x)* Cos[c + d*x])/(2*(a - b)^2*b*(a + b)^2*d^2*(a + b*Sin[c + d*x])) + (((-3*a *b*d*e*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + (3*a*b*c*f*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b ^2] - ((a^2 + 2*b^2)*f*Log[Sec[(c + d*x)/2]^2])/(2*b) + (a^2*f*Log[Sec[(c + d*x)/2]^2*(a + b*Sin[c + d*x])])/(2*b) + b*f*Log[Sec[(c + d*x)/2]^2*(a + b*Sin[c + d*x])] + (((3*I)/2)*a*b*f*Log[1 + I*Tan[(c + d*x)/2]]*Log[(b - Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/(I*a + b - Sqrt[-a^2 + b^2])])/Sqrt [-a^2 + b^2] - (((3*I)/2)*a*b*f*Log[1 - I*Tan[(c + d*x)/2]]*Log[-((b - Sqr t[-a^2 + b^2] + a*Tan[(c + d*x)/2])/(I*a - b + Sqrt[-a^2 + b^2]))])/Sqrt[- a^2 + b^2] + (((3*I)/2)*a*b*f*Log[1 - I*Tan[(c + d*x)/2]]*Log[(b + Sqrt[-a ^2 + b^2] + a*Tan[(c + d*x)/2])/((-I)*a + b + Sqrt[-a^2 + b^2])])/Sqrt[-a^ 2 + b^2] - (((3*I)/2)*a*b*f*Log[1 + I*Tan[(c + d*x)/2]]*Log[(b + Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/(I*a + b + Sqrt[-a^2 + b^2])])/Sqrt[-a^2 + b ^2] + (((3*I)/2)*a*b*f*PolyLog[2, (a*(1 - I*Tan[(c + d*x)/2]))/(a + I*(b + Sqrt[-a^2 + b^2]))])/Sqrt[-a^2 + b^2] - (((3*I)/2)*a*b*f*PolyLog[2, (a*(1 + I*Tan[(c + d*x)/2]))/(a - I*(b + Sqrt[-a^2 + b^2]))])/Sqrt[-a^2 + b^...
Time = 3.06 (sec) , antiderivative size = 751, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {7293, 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.
\(\displaystyle \int \frac {(e+f x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e+f x}{b (a+b \sin (c+d x))^2}-\frac {a (e+f x)}{b (a+b \sin (c+d x))^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {3 a f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b d^2 \left (a^2-b^2\right )^{3/2}}+\frac {3 a f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b d^2 \left (a^2-b^2\right )^{3/2}}-\frac {a f}{2 b d^2 \left (a^2-b^2\right ) (a+b \sin (c+d x))}+\frac {3 a^2 f \log (a+b \sin (c+d x))}{2 b d^2 \left (a^2-b^2\right )^2}-\frac {f \log (a+b \sin (c+d x))}{b d^2 \left (a^2-b^2\right )}-\frac {3 i a (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b d \left (a^2-b^2\right )^{3/2}}+\frac {3 i a (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{2 b d \left (a^2-b^2\right )^{3/2}}-\frac {3 a^2 (e+f x) \cos (c+d x)}{2 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}-\frac {a (e+f x) \cos (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}+\frac {(e+f x) \cos (c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}+\frac {3 a^3 f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b d^2 \left (a^2-b^2\right )^{5/2}}-\frac {3 a^3 f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b d^2 \left (a^2-b^2\right )^{5/2}}+\frac {3 i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b d \left (a^2-b^2\right )^{5/2}}-\frac {3 i a^3 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{2 b d \left (a^2-b^2\right )^{5/2}}\) |
(((3*I)/2)*a^3*(e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2 ])])/(b*(a^2 - b^2)^(5/2)*d) - (((3*I)/2)*a*(e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(3/2)*d) - (((3*I)/2)*a^3 *(e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(5/2)*d) + (((3*I)/2)*a*(e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(3/2)*d) + (3*a^2*f*Log[a + b*Sin[c + d *x]])/(2*b*(a^2 - b^2)^2*d^2) - (f*Log[a + b*Sin[c + d*x]])/(b*(a^2 - b^2) *d^2) + (3*a^3*f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/ (2*b*(a^2 - b^2)^(5/2)*d^2) - (3*a*f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(2*b*(a^2 - b^2)^(3/2)*d^2) - (3*a^3*f*PolyLog[2, (I*b *E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(2*b*(a^2 - b^2)^(5/2)*d^2) + (3 *a*f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(2*b*(a^2 - b^2)^(3/2)*d^2) - (a*(e + f*x)*Cos[c + d*x])/(2*(a^2 - b^2)*d*(a + b*Sin[c + d*x])^2) - (a*f)/(2*b*(a^2 - b^2)*d^2*(a + b*Sin[c + d*x])) - (3*a^2*(e + f*x)*Cos[c + d*x])/(2*(a^2 - b^2)^2*d*(a + b*Sin[c + d*x])) + ((e + f*x )*Cos[c + d*x])/((a^2 - b^2)*d*(a + b*Sin[c + d*x]))
3.3.48.3.1 Defintions of rubi rules used
Time = 2.33 (sec) , antiderivative size = 1084, normalized size of antiderivative = 1.44
I*(-2*I*b^2*f*exp(2*I*(d*x+c))*a^2+5*I*b^3*a*d*e*exp(I*(d*x+c))+4*I*b*a^3* d*f*x*exp(I*(d*x+c))-3*I*b^3*a*d*f*x*exp(3*I*(d*x+c))+2*a^4*d*f*x*exp(2*I* (d*x+c))+5*exp(2*I*(d*x+c))*a^2*b^2*d*f*x+2*b^4*d*f*x*exp(2*I*(d*x+c))+2*I *a^4*f*exp(2*I*(d*x+c))+5*I*b^3*a*d*f*x*exp(I*(d*x+c))-3*I*b^3*a*d*e*exp(3 *I*(d*x+c))+4*I*b*a^3*d*e*exp(I*(d*x+c))+2*a^4*d*e*exp(2*I*(d*x+c))+b*f*a^ 3*exp(3*I*(d*x+c))+5*exp(2*I*(d*x+c))*a^2*b^2*d*e-b^3*a*f*exp(3*I*(d*x+c)) +2*b^4*d*e*exp(2*I*(d*x+c))-a^2*b^2*d*f*x-2*b^4*d*f*x-b*a^3*f*exp(I*(d*x+c ))-a^2*b^2*d*e+b^3*a*f*exp(I*(d*x+c))-2*b^4*d*e)/(-I*b*exp(2*I*(d*x+c))+2* a*exp(I*(d*x+c))+I*b)^2/(a^2-b^2)^2/d^2/b+1/2/b/d^2/(-a^2+b^2)^2*a^2*f*ln( I*b*exp(2*I*(d*x+c))-I*b-2*a*exp(I*(d*x+c)))-1/b/d^2/(-a^2+b^2)^2*a^2*f*ln (exp(I*(d*x+c)))+b/d^2/(-a^2+b^2)^2*f*ln(I*b*exp(2*I*(d*x+c))-I*b-2*a*exp( I*(d*x+c)))-2*b/d^2/(-a^2+b^2)^2*f*ln(exp(I*(d*x+c)))-3*I*b/d/(-a^2+b^2)^( 5/2)*a*e*arctan(1/2*(2*I*b*exp(I*(d*x+c))-2*a)/(-a^2+b^2)^(1/2))+3*I*b/d^2 /(-a^2+b^2)^(5/2)*a*f*c*arctan(1/2*(2*I*b*exp(I*(d*x+c))-2*a)/(-a^2+b^2)^( 1/2))-3/2*b/d/(-a^2+b^2)^(5/2)*a*f*ln((I*a+b*exp(I*(d*x+c))-(-a^2+b^2)^(1/ 2))/(I*a-(-a^2+b^2)^(1/2)))*x+3/2*b/d/(-a^2+b^2)^(5/2)*a*f*ln((I*a+b*exp(I *(d*x+c))+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))*x-3/2*b/d^2/(-a^2+b^2) ^(5/2)*a*f*ln((I*a+b*exp(I*(d*x+c))-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2 )))*c+3/2*b/d^2/(-a^2+b^2)^(5/2)*a*f*ln((I*a+b*exp(I*(d*x+c))+(-a^2+b^2)^( 1/2))/(I*a+(-a^2+b^2)^(1/2)))*c+3/2*I*b/d^2/(-a^2+b^2)^(5/2)*a*f*dilog(...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2429 vs. \(2 (651) = 1302\).
Time = 0.55 (sec) , antiderivative size = 2429, normalized size of antiderivative = 3.23 \[ \int \frac {(e+f x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Too large to display} \]
-1/4*(3*(I*a*b^5*f*cos(d*x + c)^2 - 2*I*a^2*b^4*f*sin(d*x + c) - I*(a^3*b^ 3 + a*b^5)*f)*sqrt(-(a^2 - b^2)/b^2)*dilog((I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) + 3*(-I*a*b^5*f*cos(d*x + c)^2 + 2*I*a^2*b^4*f*sin(d*x + c) + I*(a^3*b^ 3 + a*b^5)*f)*sqrt(-(a^2 - b^2)/b^2)*dilog((I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) + 3*(-I*a*b^5*f*cos(d*x + c)^2 + 2*I*a^2*b^4*f*sin(d*x + c) + I*(a^3*b^ 3 + a*b^5)*f)*sqrt(-(a^2 - b^2)/b^2)*dilog((-I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) + 3*(I*a*b^5*f*cos(d*x + c)^2 - 2*I*a^2*b^4*f*sin(d*x + c) - I*(a^3*b^ 3 + a*b^5)*f)*sqrt(-(a^2 - b^2)/b^2)*dilog((-I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) + 3*((a^3*b^3 + a*b^5)*d*f*x + (a^3*b^3 + a*b^5)*c*f - (a*b^5*d*f*x + a*b^5*c*f)*cos(d*x + c)^2 + 2*(a^2*b^4*d*f*x + a^2*b^4*c*f)*sin(d*x + c))* sqrt(-(a^2 - b^2)/b^2)*log(-(I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d* x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b) - 3*((a^3*b^3 + a*b^5)*d*f*x + (a^3*b^3 + a*b^5)*c*f - (a*b^5*d*f*x + a*b^5*c*f)*cos(d*x + c)^2 + 2*(a^2*b^4*d*f*x + a^2*b^4*c*f)*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^ 2)*log(-(I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b) + 3*((a^3*b^3 + a*b^5)*d*f*x + (a...
Timed out. \[ \int \frac {(e+f x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(e+f x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f or more de
\[ \int \frac {(e+f x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\int { \frac {{\left (f x + e\right )} \sin \left (d x + c\right )}{{\left (b \sin \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {(e+f x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Hanged} \]