\(\int \frac {c+d x}{a+a \sin (e+f x)} \, dx\) [109]

Optimal result

Integrand size = 18, antiderivative size = 60 \[ \int \frac {c+d x}{a+a \sin (e+f x)} \, dx=-\frac {(c+d x) \cot \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{a f}+\frac {2 d \log \left (\sin \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )\right )}{a f^2} \]

[Out]

-(d*x+c)*cot(1/2*e+1/4*Pi+1/2*f*x)/a/f+2*d*ln(sin(1/2*e+1/4*Pi+1/2*f*x))/a/f^2

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3399, 4269, 3556} \[ \int \frac {c+d x}{a+a \sin (e+f x)} \, dx=\frac {2 d \log \left (\sin \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )\right )}{a f^2}-\frac {(c+d x) \cot \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{a f} \]

[In]

Int[(c + d*x)/(a + a*Sin[e + f*x]),x]

[Out]

-(((c + d*x)*Cot[e/2 + Pi/4 + (f*x)/2])/(a*f)) + (2*d*Log[Sin[e/2 + Pi/4 + (f*x)/2]])/(a*f^2)

Rule 3399

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(2*a)^n, Int[(c
+ d*x)^m*Sin[(1/2)*(e + Pi*(a/(2*b))) + f*(x/2)]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (GtQ[n, 0] || IGtQ[m, 0])

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4269

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(c + d*x)^m)*(Cot[e + f*x]/f), x
] + Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\int (c+d x) \csc ^2\left (\frac {1}{2} \left (e+\frac {\pi }{2}\right )+\frac {f x}{2}\right ) \, dx}{2 a} \\ & = -\frac {(c+d x) \cot \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{a f}+\frac {d \int \cot \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right ) \, dx}{a f} \\ & = -\frac {(c+d x) \cot \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{a f}+\frac {2 d \log \left (\sin \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )\right )}{a f^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.85 \[ \int \frac {c+d x}{a+a \sin (e+f x)} \, dx=\frac {2 d \log \left (\cos \left (\frac {1}{4} (2 e-\pi +2 f x)\right )\right )+f (c+d x) \tan \left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{a f^2} \]

[In]

Integrate[(c + d*x)/(a + a*Sin[e + f*x]),x]

[Out]

(2*d*Log[Cos[(2*e - Pi + 2*f*x)/4]] + f*(c + d*x)*Tan[(2*e - Pi + 2*f*x)/4])/(a*f^2)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.21 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.22

method	result	size
risch	\(-\frac {2 i d x}{a f}-\frac {2 i d e}{a \,f^{2}}-\frac {2 \left (d x +c \right )}{f a \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}+\frac {2 d \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{a \,f^{2}}\)	\(73\)
parallelrisch	\(\frac {-d \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \ln \left (\sec ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+2 d \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )-2 \left (-\frac {d x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}+\frac {d x}{2}+c \right ) f}{a \,f^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}\)	\(96\)
norman	\(\frac {\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f a}+\frac {d x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f a}-\frac {d x}{f a}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}+\frac {2 d \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{a \,f^{2}}-\frac {d \ln \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a \,f^{2}}\)	\(107\)

[In]

int((d*x+c)/(a+a*sin(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

-2*I*d/a/f*x-2*I*d/a/f^2*e-2*(d*x+c)/f/a/(exp(I*(f*x+e))+I)+2*d/a/f^2*ln(exp(I*(f*x+e))+I)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (48) = 96\).

Time = 0.29 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.67 \[ \int \frac {c+d x}{a+a \sin (e+f x)} \, dx=-\frac {d f x + c f + {\left (d f x + c f\right )} \cos \left (f x + e\right ) - {\left (d \cos \left (f x + e\right ) + d \sin \left (f x + e\right ) + d\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - {\left (d f x + c f\right )} \sin \left (f x + e\right )}{a f^{2} \cos \left (f x + e\right ) + a f^{2} \sin \left (f x + e\right ) + a f^{2}} \]

[In]

integrate((d*x+c)/(a+a*sin(f*x+e)),x, algorithm="fricas")

[Out]

-(d*f*x + c*f + (d*f*x + c*f)*cos(f*x + e) - (d*cos(f*x + e) + d*sin(f*x + e) + d)*log(sin(f*x + e) + 1) - (d*
f*x + c*f)*sin(f*x + e))/(a*f^2*cos(f*x + e) + a*f^2*sin(f*x + e) + a*f^2)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 272 vs. \(2 (46) = 92\).

Time = 0.47 (sec) , antiderivative size = 272, normalized size of antiderivative = 4.53 \[ \int \frac {c+d x}{a+a \sin (e+f x)} \, dx=\begin {cases} - \frac {2 c f}{a f^{2} \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + a f^{2}} + \frac {d f x \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{a f^{2} \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + a f^{2}} - \frac {d f x}{a f^{2} \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + a f^{2}} + \frac {2 d \log {\left (\tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 1 \right )} \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{a f^{2} \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + a f^{2}} + \frac {2 d \log {\left (\tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 1 \right )}}{a f^{2} \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + a f^{2}} - \frac {d \log {\left (\tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 1 \right )} \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{a f^{2} \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + a f^{2}} - \frac {d \log {\left (\tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 1 \right )}}{a f^{2} \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + a f^{2}} & \text {for}\: f \neq 0 \\\frac {c x + \frac {d x^{2}}{2}}{a \sin {\left (e \right )} + a} & \text {otherwise} \end {cases} \]

[In]

integrate((d*x+c)/(a+a*sin(f*x+e)),x)

[Out]

Piecewise((-2*c*f/(a*f**2*tan(e/2 + f*x/2) + a*f**2) + d*f*x*tan(e/2 + f*x/2)/(a*f**2*tan(e/2 + f*x/2) + a*f**
2) - d*f*x/(a*f**2*tan(e/2 + f*x/2) + a*f**2) + 2*d*log(tan(e/2 + f*x/2) + 1)*tan(e/2 + f*x/2)/(a*f**2*tan(e/2
+ f*x/2) + a*f**2) + 2*d*log(tan(e/2 + f*x/2) + 1)/(a*f**2*tan(e/2 + f*x/2) + a*f**2) - d*log(tan(e/2 + f*x/2
)**2 + 1)*tan(e/2 + f*x/2)/(a*f**2*tan(e/2 + f*x/2) + a*f**2) - d*log(tan(e/2 + f*x/2)**2 + 1)/(a*f**2*tan(e/2
+ f*x/2) + a*f**2), Ne(f, 0)), ((c*x + d*x**2/2)/(a*sin(e) + a), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 169 vs. \(2 (48) = 96\).

Time = 0.20 (sec) , antiderivative size = 169, normalized size of antiderivative = 2.82 \[ \int \frac {c+d x}{a+a \sin (e+f x)} \, dx=-\frac {\frac {{\left (2 \, {\left (f x + e\right )} \cos \left (f x + e\right ) - {\left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right )} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right )\right )} d}{a f \cos \left (f x + e\right )^{2} + a f \sin \left (f x + e\right )^{2} + 2 \, a f \sin \left (f x + e\right ) + a f} - \frac {2 \, d e}{a f + \frac {a f \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}} + \frac {2 \, c}{a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}}{f} \]

[In]

integrate((d*x+c)/(a+a*sin(f*x+e)),x, algorithm="maxima")

[Out]

-((2*(f*x + e)*cos(f*x + e) - (cos(f*x + e)^2 + sin(f*x + e)^2 + 2*sin(f*x + e) + 1)*log(cos(f*x + e)^2 + sin(
f*x + e)^2 + 2*sin(f*x + e) + 1))*d/(a*f*cos(f*x + e)^2 + a*f*sin(f*x + e)^2 + 2*a*f*sin(f*x + e) + a*f) - 2*d
*e/(a*f + a*f*sin(f*x + e)/(cos(f*x + e) + 1)) + 2*c/(a + a*sin(f*x + e)/(cos(f*x + e) + 1)))/f

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 548 vs. \(2 (48) = 96\).

Time = 0.36 (sec) , antiderivative size = 548, normalized size of antiderivative = 9.13 \[ \int \frac {c+d x}{a+a \sin (e+f x)} \, dx=\text {Too large to display} \]

[In]

integrate((d*x+c)/(a+a*sin(f*x+e)),x, algorithm="giac")

[Out]

-(d*f*x*tan(1/2*f*x)*tan(1/2*e) + d*f*x*tan(1/2*f*x) + d*f*x*tan(1/2*e) + c*f*tan(1/2*f*x)*tan(1/2*e) - d*log(
2*(tan(1/2*f*x)^2*tan(1/2*e)^2 - 2*tan(1/2*f*x)^2*tan(1/2*e) - 2*tan(1/2*f*x)*tan(1/2*e)^2 + tan(1/2*f*x)^2 +
tan(1/2*e)^2 + 2*tan(1/2*f*x) + 2*tan(1/2*e) + 1)/(tan(1/2*f*x)^2*tan(1/2*e)^2 + tan(1/2*f*x)^2 + tan(1/2*e)^2
+ 1))*tan(1/2*f*x)*tan(1/2*e) - d*f*x + c*f*tan(1/2*f*x) + d*log(2*(tan(1/2*f*x)^2*tan(1/2*e)^2 - 2*tan(1/2*f
*x)^2*tan(1/2*e) - 2*tan(1/2*f*x)*tan(1/2*e)^2 + tan(1/2*f*x)^2 + tan(1/2*e)^2 + 2*tan(1/2*f*x) + 2*tan(1/2*e)
+ 1)/(tan(1/2*f*x)^2*tan(1/2*e)^2 + tan(1/2*f*x)^2 + tan(1/2*e)^2 + 1))*tan(1/2*f*x) + c*f*tan(1/2*e) + d*log
(2*(tan(1/2*f*x)^2*tan(1/2*e)^2 - 2*tan(1/2*f*x)^2*tan(1/2*e) - 2*tan(1/2*f*x)*tan(1/2*e)^2 + tan(1/2*f*x)^2 +
tan(1/2*e)^2 + 2*tan(1/2*f*x) + 2*tan(1/2*e) + 1)/(tan(1/2*f*x)^2*tan(1/2*e)^2 + tan(1/2*f*x)^2 + tan(1/2*e)^
2 + 1))*tan(1/2*e) - c*f + d*log(2*(tan(1/2*f*x)^2*tan(1/2*e)^2 - 2*tan(1/2*f*x)^2*tan(1/2*e) - 2*tan(1/2*f*x)
*tan(1/2*e)^2 + tan(1/2*f*x)^2 + tan(1/2*e)^2 + 2*tan(1/2*f*x) + 2*tan(1/2*e) + 1)/(tan(1/2*f*x)^2*tan(1/2*e)^
2 + tan(1/2*f*x)^2 + tan(1/2*e)^2 + 1)))/(a*f^2*tan(1/2*f*x)*tan(1/2*e) - a*f^2*tan(1/2*f*x) - a*f^2*tan(1/2*e
) - a*f^2)

Mupad [B] (verification not implemented)

Time = 0.78 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.10 \[ \int \frac {c+d x}{a+a \sin (e+f x)} \, dx=\frac {2\,d\,\ln \left ({\mathrm {e}}^{e\,1{}\mathrm {i}}\,{\mathrm {e}}^{f\,x\,1{}\mathrm {i}}+1{}\mathrm {i}\right )}{a\,f^2}-\frac {2\,\left (c+d\,x\right )}{a\,f\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+1{}\mathrm {i}\right )}-\frac {d\,x\,2{}\mathrm {i}}{a\,f} \]

[In]

int((c + d*x)/(a + a*sin(e + f*x)),x)

[Out]

(2*d*log(exp(e*1i)*exp(f*x*1i) + 1i))/(a*f^2) - (2*(c + d*x))/(a*f*(exp(e*1i + f*x*1i) + 1i)) - (d*x*2i)/(a*f)