Integrand size = 26, antiderivative size = 1106 \[ \int \frac {(e+f x)^2 \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx =\text {Too large to display} \] Output:
-I*a^2*(f*x+e)^2*ln(1-I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)^ (3/2)/d+2*a*f*(f*x+e)*ln(1-I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2- b^2)/d^2-2*I*a*f^2*polylog(2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^ 2-b^2)/d^3+2*I*f^2*polylog(3,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^ 2-b^2)^(1/2)/d^3+2*a*f*(f*x+e)*ln(1-I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)) )/b/(a^2-b^2)/d^2-2*I*f^2*polylog(3,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)) )/b/(a^2-b^2)^(1/2)/d^3-2*I*a^2*f^2*polylog(3,I*b*exp(I*(d*x+c))/(a+(a^2-b ^2)^(1/2)))/b/(a^2-b^2)^(3/2)/d^3+I*a^2*(f*x+e)^2*ln(1-I*b*exp(I*(d*x+c))/ (a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3/2)/d+2*a^2*f*(f*x+e)*polylog(2,I*b*exp (I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3/2)/d^2-2*f*(f*x+e)*polylog (2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(1/2)/d^2+I*(f*x+e) ^2*ln(1-I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(1/2)/d-2*a^2* f*(f*x+e)*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3 /2)/d^2+2*f*(f*x+e)*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a ^2-b^2)^(1/2)/d^2-I*(f*x+e)^2*ln(1-I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2))) /b/(a^2-b^2)^(1/2)/d+2*I*a^2*f^2*polylog(3,I*b*exp(I*(d*x+c))/(a-(a^2-b^2) ^(1/2)))/b/(a^2-b^2)^(3/2)/d^3-I*a*(f*x+e)^2/b/(a^2-b^2)/d-2*I*a*f^2*polyl og(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)/d^3-a*(f*x+e)^2*c os(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. \(3595\) vs. \(2(1106)=2212\).
Time = 25.60 (sec) , antiderivative size = 3595, normalized size of antiderivative = 3.25 \[ \int \frac {(e+f x)^2 \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Result too large to show} \] Input:
Integrate[((e + f*x)^2*Sin[c + d*x])/(a + b*Sin[c + d*x])^2,x]
Output:
(2*b*e*f*((Pi*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + (2*(-c + Pi/2 - d*x)*ArcTanh[((a + b)*Cot[(-c + Pi/2 - d*x)/2])/Sq rt[-a^2 + b^2]] - 2*(-c + ArcCos[-(a/b)])*ArcTanh[((-a + b)*Tan[(-c + Pi/2 - d*x)/2])/Sqrt[-a^2 + b^2]] + (ArcCos[-(a/b)] - (2*I)*(ArcTanh[((a + b)* Cot[(-c + Pi/2 - d*x)/2])/Sqrt[-a^2 + b^2]] - ArcTanh[((-a + b)*Tan[(-c + Pi/2 - d*x)/2])/Sqrt[-a^2 + b^2]]))*Log[Sqrt[-a^2 + b^2]/(Sqrt[2]*Sqrt[b]* E^((I/2)*(-c + Pi/2 - d*x))*Sqrt[a + b*Sin[c + d*x]])] + (ArcCos[-(a/b)] + (2*I)*(ArcTanh[((a + b)*Cot[(-c + Pi/2 - d*x)/2])/Sqrt[-a^2 + b^2]] - Arc Tanh[((-a + b)*Tan[(-c + Pi/2 - d*x)/2])/Sqrt[-a^2 + b^2]]))*Log[(Sqrt[-a^ 2 + b^2]*E^((I/2)*(-c + Pi/2 - d*x)))/(Sqrt[2]*Sqrt[b]*Sqrt[a + b*Sin[c + d*x]])] - (ArcCos[-(a/b)] + (2*I)*ArcTanh[((-a + b)*Tan[(-c + Pi/2 - d*x)/ 2])/Sqrt[-a^2 + b^2]])*Log[1 - ((a - I*Sqrt[-a^2 + b^2])*(a + b - Sqrt[-a^ 2 + b^2]*Tan[(-c + Pi/2 - d*x)/2]))/(b*(a + b + Sqrt[-a^2 + b^2]*Tan[(-c + Pi/2 - d*x)/2]))] + (-ArcCos[-(a/b)] + (2*I)*ArcTanh[((-a + b)*Tan[(-c + Pi/2 - d*x)/2])/Sqrt[-a^2 + b^2]])*Log[1 - ((a + I*Sqrt[-a^2 + b^2])*(a + b - Sqrt[-a^2 + b^2]*Tan[(-c + Pi/2 - d*x)/2]))/(b*(a + b + Sqrt[-a^2 + b^ 2]*Tan[(-c + Pi/2 - d*x)/2]))] + I*(PolyLog[2, ((a - I*Sqrt[-a^2 + b^2])*( a + b - Sqrt[-a^2 + b^2]*Tan[(-c + Pi/2 - d*x)/2]))/(b*(a + b + Sqrt[-a^2 + b^2]*Tan[(-c + Pi/2 - d*x)/2]))] - PolyLog[2, ((a + I*Sqrt[-a^2 + b^2])* (a + b - Sqrt[-a^2 + b^2]*Tan[(-c + Pi/2 - d*x)/2]))/(b*(a + b + Sqrt[-...
Time = 3.01 (sec) , antiderivative size = 1106, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, 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.
\(\displaystyle \int \frac {(e+f x)^2 \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {(e+f x)^2}{b (a+b \sin (c+d x))}-\frac {a (e+f x)^2}{b (a+b \sin (c+d x))^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {i (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) a^2}{b \left (a^2-b^2\right )^{3/2} d}-\frac {i (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) a^2}{b \left (a^2-b^2\right )^{3/2} d}+\frac {2 f (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) a^2}{b \left (a^2-b^2\right )^{3/2} d^2}-\frac {2 f (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) a^2}{b \left (a^2-b^2\right )^{3/2} d^2}+\frac {2 i f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) a^2}{b \left (a^2-b^2\right )^{3/2} d^3}-\frac {2 i f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) a^2}{b \left (a^2-b^2\right )^{3/2} d^3}-\frac {i (e+f x)^2 a}{b \left (a^2-b^2\right ) d}+\frac {2 f (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) a}{b \left (a^2-b^2\right ) d^2}+\frac {2 f (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) a}{b \left (a^2-b^2\right ) d^2}-\frac {2 i f^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) a}{b \left (a^2-b^2\right ) d^3}-\frac {2 i f^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) a}{b \left (a^2-b^2\right ) d^3}-\frac {(e+f x)^2 \cos (c+d x) a}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {i (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d}+\frac {i (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d}-\frac {2 f (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^2}+\frac {2 f (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^2}-\frac {2 i f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^3}+\frac {2 i f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^3}\) |
Input:
Int[((e + f*x)^2*Sin[c + d*x])/(a + b*Sin[c + d*x])^2,x]
Output:
((-I)*a*(e + f*x)^2)/(b*(a^2 - b^2)*d) + (2*a*f*(e + f*x)*Log[1 - (I*b*E^( I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)*d^2) + (I*a^2*(e + f* x)^2*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^ (3/2)*d) - (I*(e + f*x)^2*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^ 2])])/(b*Sqrt[a^2 - b^2]*d) + (2*a*f*(e + f*x)*Log[1 - (I*b*E^(I*(c + d*x) ))/(a + Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)*d^2) - (I*a^2*(e + f*x)^2*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(3/2)*d) + (I*(e + f*x)^2*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*Sq rt[a^2 - b^2]*d) - ((2*I)*a*f^2*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt [a^2 - b^2])])/(b*(a^2 - b^2)*d^3) + (2*a^2*f*(e + f*x)*PolyLog[2, (I*b*E^ (I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(3/2)*d^2) - (2*f*(e + f*x)*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*Sqrt[a ^2 - b^2]*d^2) - ((2*I)*a*f^2*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a ^2 - b^2])])/(b*(a^2 - b^2)*d^3) - (2*a^2*f*(e + f*x)*PolyLog[2, (I*b*E^(I *(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(3/2)*d^2) + (2*f*(e + f*x)*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*Sqrt[a^2 - b^2]*d^2) + ((2*I)*a^2*f^2*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a ^2 - b^2])])/(b*(a^2 - b^2)^(3/2)*d^3) - ((2*I)*f^2*PolyLog[3, (I*b*E^(I*( c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*Sqrt[a^2 - b^2]*d^3) - ((2*I)*a^2*f^ 2*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*(a^2 - b^...
\[\int \frac {\left (f x +e \right )^{2} \sin \left (d x +c \right )}{\left (a +b \sin \left (d x +c \right )\right )^{2}}d x\]
Input:
int((f*x+e)^2*sin(d*x+c)/(a+b*sin(d*x+c))^2,x)
Output:
int((f*x+e)^2*sin(d*x+c)/(a+b*sin(d*x+c))^2,x)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 3122 vs. \(2 (958) = 1916\).
Time = 0.34 (sec) , antiderivative size = 3122, normalized size of antiderivative = 2.82 \[ \int \frac {(e+f x)^2 \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Too large to display} \] Input:
integrate((f*x+e)^2*sin(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="fricas")
Output:
Too large to include
Timed out. \[ \int \frac {(e+f x)^2 \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Timed out} \] Input:
integrate((f*x+e)**2*sin(d*x+c)/(a+b*sin(d*x+c))**2,x)
Output:
Timed out
Exception generated. \[ \int \frac {(e+f x)^2 \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((f*x+e)^2*sin(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="maxima")
Output:
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
\[ \int \frac {(e+f x)^2 \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\int { \frac {{\left (f x + e\right )}^{2} \sin \left (d x + c\right )}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:
integrate((f*x+e)^2*sin(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="giac")
Output:
integrate((f*x + e)^2*sin(d*x + c)/(b*sin(d*x + c) + a)^2, x)
Timed out. \[ \int \frac {(e+f x)^2 \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Hanged} \] Input:
int((sin(c + d*x)*(e + f*x)^2)/(a + b*sin(c + d*x))^2,x)
Output:
\text{Hanged}
\[ \int \frac {(e+f x)^2 \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {too large to display} \] Input:
int((f*x+e)^2*sin(d*x+c)/(a+b*sin(d*x+c))^2,x)
Output:
(12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin (c + d*x)*a**6*b*f**2 - 6*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/ sqrt(a**2 - b**2))*sin(c + d*x)*a**4*b**3*d**2*e**2 + 12*sqrt(a**2 - b**2) *atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x)*a**4*b**3*f **2 + 24*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2) )*sin(c + d*x)*a**3*b**4*d*e*f - 96*sqrt(a**2 - b**2)*atan((tan((c + d*x)/ 2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x)*a**2*b**5*f**2 - 24*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x)*a*b**6 *d*e*f + 72*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b* *2))*sin(c + d*x)*b**7*f**2 + 12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)* a + b)/sqrt(a**2 - b**2))*a**7*f**2 - 6*sqrt(a**2 - b**2)*atan((tan((c + d *x)/2)*a + b)/sqrt(a**2 - b**2))*a**5*b**2*d**2*e**2 + 12*sqrt(a**2 - b**2 )*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*a**5*b**2*f**2 + 24*sqr t(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*a**4*b**3* d*e*f - 96*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b** 2))*a**3*b**4*f**2 - 24*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sq rt(a**2 - b**2))*a**2*b**5*d*e*f + 72*sqrt(a**2 - b**2)*atan((tan((c + d*x )/2)*a + b)/sqrt(a**2 - b**2))*a*b**6*f**2 - 3*cos(c + d*x)*a**7*b*d**2*e* *2 - 6*cos(c + d*x)*a**7*b*d**2*e*f*x - 3*cos(c + d*x)*a**7*b*d**2*f**2*x* *2 + 6*cos(c + d*x)*a**6*b**2*d*f**2*x + 3*cos(c + d*x)*a**5*b**3*d**2*...