3.170 \(\int \frac {c+d x}{(a+b \sin (e+f x))^2} \, dx\)

Optimal. Leaf size=305 \[ -\frac {i a (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{3/2}}+\frac {i a (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{f \left (a^2-b^2\right )^{3/2}}+\frac {b (c+d x) \cos (e+f x)}{f \left (a^2-b^2\right ) (a+b \sin (e+f x))}-\frac {a d \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f^2 \left (a^2-b^2\right )^{3/2}}+\frac {a d \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{f^2 \left (a^2-b^2\right )^{3/2}}-\frac {d \log (a+b \sin (e+f x))}{f^2 \left (a^2-b^2\right )} \]

[Out]

-d*ln(a+b*sin(f*x+e))/(a^2-b^2)/f^2-I*a*(d*x+c)*ln(1-I*b*exp(I*(f*x+e))/(a-(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/f
+I*a*(d*x+c)*ln(1-I*b*exp(I*(f*x+e))/(a+(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/f-a*d*polylog(2,I*b*exp(I*(f*x+e))/(
a-(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/f^2+a*d*polylog(2,I*b*exp(I*(f*x+e))/(a+(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/
f^2+b*(d*x+c)*cos(f*x+e)/(a^2-b^2)/f/(a+b*sin(f*x+e))

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Rubi [A]  time = 0.55, antiderivative size = 305, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3324, 3323, 2264, 2190, 2279, 2391, 2668, 31} \[ -\frac {a d \text {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f^2 \left (a^2-b^2\right )^{3/2}}+\frac {a d \text {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{f^2 \left (a^2-b^2\right )^{3/2}}-\frac {i a (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{3/2}}+\frac {i a (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{f \left (a^2-b^2\right )^{3/2}}+\frac {b (c+d x) \cos (e+f x)}{f \left (a^2-b^2\right ) (a+b \sin (e+f x))}-\frac {d \log (a+b \sin (e+f x))}{f^2 \left (a^2-b^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)/(a + b*Sin[e + f*x])^2,x]

[Out]

((-I)*a*(c + d*x)*Log[1 - (I*b*E^(I*(e + f*x)))/(a - Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*f) + (I*a*(c + d*x)
*Log[1 - (I*b*E^(I*(e + f*x)))/(a + Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*f) - (d*Log[a + b*Sin[e + f*x]])/((a
^2 - b^2)*f^2) - (a*d*PolyLog[2, (I*b*E^(I*(e + f*x)))/(a - Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*f^2) + (a*d*
PolyLog[2, (I*b*E^(I*(e + f*x)))/(a + Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*f^2) + (b*(c + d*x)*Cos[e + f*x])/
((a^2 - b^2)*f*(a + b*Sin[e + f*x]))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 3323

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[2, Int[((c + d*x)^m*E
^(I*(e + f*x)))/(I*b + 2*a*E^(I*(e + f*x)) - I*b*E^(2*I*(e + f*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 3324

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

Rubi steps

\begin {align*} \int \frac {c+d x}{(a+b \sin (e+f x))^2} \, dx &=\frac {b (c+d x) \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}+\frac {a \int \frac {c+d x}{a+b \sin (e+f x)} \, dx}{a^2-b^2}-\frac {(b d) \int \frac {\cos (e+f x)}{a+b \sin (e+f x)} \, dx}{\left (a^2-b^2\right ) f}\\ &=\frac {b (c+d x) \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}+\frac {(2 a) \int \frac {e^{i (e+f x)} (c+d x)}{i b+2 a e^{i (e+f x)}-i b e^{2 i (e+f x)}} \, dx}{a^2-b^2}-\frac {d \operatorname {Subst}\left (\int \frac {1}{a+x} \, dx,x,b \sin (e+f x)\right )}{\left (a^2-b^2\right ) f^2}\\ &=-\frac {d \log (a+b \sin (e+f x))}{\left (a^2-b^2\right ) f^2}+\frac {b (c+d x) \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}-\frac {(2 i a b) \int \frac {e^{i (e+f x)} (c+d x)}{2 a-2 \sqrt {a^2-b^2}-2 i b e^{i (e+f x)}} \, dx}{\left (a^2-b^2\right )^{3/2}}+\frac {(2 i a b) \int \frac {e^{i (e+f x)} (c+d x)}{2 a+2 \sqrt {a^2-b^2}-2 i b e^{i (e+f x)}} \, dx}{\left (a^2-b^2\right )^{3/2}}\\ &=-\frac {i a (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}+\frac {i a (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}-\frac {d \log (a+b \sin (e+f x))}{\left (a^2-b^2\right ) f^2}+\frac {b (c+d x) \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}+\frac {(i a d) \int \log \left (1-\frac {2 i b e^{i (e+f x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right )^{3/2} f}-\frac {(i a d) \int \log \left (1-\frac {2 i b e^{i (e+f x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right )^{3/2} f}\\ &=-\frac {i a (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}+\frac {i a (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}-\frac {d \log (a+b \sin (e+f x))}{\left (a^2-b^2\right ) f^2}+\frac {b (c+d x) \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}+\frac {(a d) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {2 i b x}{2 a-2 \sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (e+f x)}\right )}{\left (a^2-b^2\right )^{3/2} f^2}-\frac {(a d) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {2 i b x}{2 a+2 \sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (e+f x)}\right )}{\left (a^2-b^2\right )^{3/2} f^2}\\ &=-\frac {i a (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}+\frac {i a (c+d x) \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}-\frac {d \log (a+b \sin (e+f x))}{\left (a^2-b^2\right ) f^2}-\frac {a d \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f^2}+\frac {a d \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f^2}+\frac {b (c+d x) \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}\\ \end {align*}

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Mathematica [A]  time = 1.08, size = 236, normalized size = 0.77 \[ \frac {\frac {a \left (-i f (c+d x) \left (\log \left (1+\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}-a}\right )-\log \left (1-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )\right )-d \text {Li}_2\left (-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}-a}\right )+d \text {Li}_2\left (\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )\right )}{\sqrt {a^2-b^2}}+\frac {b f (c+d x) \cos (e+f x)}{a+b \sin (e+f x)}-d \log (a+b \sin (e+f x))}{f^2 \left (a^2-b^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)/(a + b*Sin[e + f*x])^2,x]

[Out]

(-(d*Log[a + b*Sin[e + f*x]]) + (a*((-I)*f*(c + d*x)*(Log[1 + (I*b*E^(I*(e + f*x)))/(-a + Sqrt[a^2 - b^2])] -
Log[1 - (I*b*E^(I*(e + f*x)))/(a + Sqrt[a^2 - b^2])]) - d*PolyLog[2, ((-I)*b*E^(I*(e + f*x)))/(-a + Sqrt[a^2 -
b^2])] + d*PolyLog[2, (I*b*E^(I*(e + f*x)))/(a + Sqrt[a^2 - b^2])]))/Sqrt[a^2 - b^2] + (b*f*(c + d*x)*Cos[e +
f*x])/(a + b*Sin[e + f*x]))/((a^2 - b^2)*f^2)

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fricas [B]  time = 0.72, size = 1520, normalized size = 4.98 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(a+b*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

1/2*((I*a*b^2*d*sin(f*x + e) + I*a^2*b*d)*sqrt(-(a^2 - b^2)/b^2)*dilog(-1/2*(2*I*a*cos(f*x + e) + 2*a*sin(f*x
+ e) + 2*(b*cos(f*x + e) - I*b*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) + (-I*a*b^2*d*sin(f*x + e) -
I*a^2*b*d)*sqrt(-(a^2 - b^2)/b^2)*dilog(-1/2*(2*I*a*cos(f*x + e) + 2*a*sin(f*x + e) - 2*(b*cos(f*x + e) - I*b
*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) + (-I*a*b^2*d*sin(f*x + e) - I*a^2*b*d)*sqrt(-(a^2 - b^2)/
b^2)*dilog(-1/2*(-2*I*a*cos(f*x + e) + 2*a*sin(f*x + e) + 2*(b*cos(f*x + e) + I*b*sin(f*x + e))*sqrt(-(a^2 - b
^2)/b^2) + 2*b)/b + 1) + (I*a*b^2*d*sin(f*x + e) + I*a^2*b*d)*sqrt(-(a^2 - b^2)/b^2)*dilog(-1/2*(-2*I*a*cos(f*
x + e) + 2*a*sin(f*x + e) - 2*(b*cos(f*x + e) + I*b*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) + (a^2*
b*d*f*x + a^2*b*d*e + (a*b^2*d*f*x + a*b^2*d*e)*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2)*log(1/2*(2*I*a*cos(f*x +
e) + 2*a*sin(f*x + e) + 2*(b*cos(f*x + e) - I*b*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) - (a^2*b*d*f*x
+ a^2*b*d*e + (a*b^2*d*f*x + a*b^2*d*e)*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2)*log(1/2*(2*I*a*cos(f*x + e) + 2*a
*sin(f*x + e) - 2*(b*cos(f*x + e) - I*b*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) + (a^2*b*d*f*x + a^2*b*
d*e + (a*b^2*d*f*x + a*b^2*d*e)*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2)*log(1/2*(-2*I*a*cos(f*x + e) + 2*a*sin(f*
x + e) + 2*(b*cos(f*x + e) + I*b*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) - (a^2*b*d*f*x + a^2*b*d*e + (
a*b^2*d*f*x + a*b^2*d*e)*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2)*log(1/2*(-2*I*a*cos(f*x + e) + 2*a*sin(f*x + e)
- 2*(b*cos(f*x + e) + I*b*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) + 2*((a^2*b - b^3)*d*f*x + (a^2*b - b
^3)*c*f)*cos(f*x + e) - ((a^2*b - b^3)*d*sin(f*x + e) + (a^3 - a*b^2)*d + (a^2*b*d*e - a^2*b*c*f + (a*b^2*d*e
- a*b^2*c*f)*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2))*log(2*b*cos(f*x + e) + 2*I*b*sin(f*x + e) + 2*b*sqrt(-(a^2
- b^2)/b^2) + 2*I*a) - ((a^2*b - b^3)*d*sin(f*x + e) + (a^3 - a*b^2)*d + (a^2*b*d*e - a^2*b*c*f + (a*b^2*d*e -
a*b^2*c*f)*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2))*log(2*b*cos(f*x + e) - 2*I*b*sin(f*x + e) + 2*b*sqrt(-(a^2 -
b^2)/b^2) - 2*I*a) - ((a^2*b - b^3)*d*sin(f*x + e) + (a^3 - a*b^2)*d - (a^2*b*d*e - a^2*b*c*f + (a*b^2*d*e -
a*b^2*c*f)*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2))*log(-2*b*cos(f*x + e) + 2*I*b*sin(f*x + e) + 2*b*sqrt(-(a^2 -
b^2)/b^2) + 2*I*a) - ((a^2*b - b^3)*d*sin(f*x + e) + (a^3 - a*b^2)*d - (a^2*b*d*e - a^2*b*c*f + (a*b^2*d*e -
a*b^2*c*f)*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2))*log(-2*b*cos(f*x + e) - 2*I*b*sin(f*x + e) + 2*b*sqrt(-(a^2 -
b^2)/b^2) - 2*I*a))/((a^4*b - 2*a^2*b^3 + b^5)*f^2*sin(f*x + e) + (a^5 - 2*a^3*b^2 + a*b^4)*f^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {d x + c}{{\left (b \sin \left (f x + e\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(a+b*sin(f*x+e))^2,x, algorithm="giac")

[Out]

integrate((d*x + c)/(b*sin(f*x + e) + a)^2, x)

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maple [B]  time = 1.24, size = 650, normalized size = 2.13 \[ \frac {2 \left (d x +c \right ) \left (i b +a \,{\mathrm e}^{i \left (f x +e \right )}\right )}{f \left (a^{2}-b^{2}\right ) \left (b \,{\mathrm e}^{2 i \left (f x +e \right )}-b +2 i a \,{\mathrm e}^{i \left (f x +e \right )}\right )}-\frac {2 d \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{\left (-a^{2}+b^{2}\right ) f^{2}}+\frac {d \ln \left (i {\mathrm e}^{2 i \left (f x +e \right )} b -i b -2 a \,{\mathrm e}^{i \left (f x +e \right )}\right )}{\left (-a^{2}+b^{2}\right ) f^{2}}+\frac {i d a \dilog \left (\frac {i a +b \,{\mathrm e}^{i \left (f x +e \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right )}{\left (-a^{2}+b^{2}\right )^{\frac {3}{2}} f^{2}}+\frac {d a \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) x}{\left (-a^{2}+b^{2}\right )^{\frac {3}{2}} f}+\frac {d a \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) e}{\left (-a^{2}+b^{2}\right )^{\frac {3}{2}} f^{2}}+\frac {2 i a d e \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (f x +e \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{\left (-a^{2}+b^{2}\right )^{\frac {3}{2}} f^{2}}-\frac {i d a \dilog \left (\frac {i a +b \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right )}{\left (-a^{2}+b^{2}\right )^{\frac {3}{2}} f^{2}}-\frac {d a \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (f x +e \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right ) x}{\left (-a^{2}+b^{2}\right )^{\frac {3}{2}} f}-\frac {d a \ln \left (\frac {i a +b \,{\mathrm e}^{i \left (f x +e \right )}-\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right ) e}{\left (-a^{2}+b^{2}\right )^{\frac {3}{2}} f^{2}}-\frac {2 i a c \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (f x +e \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{\left (-a^{2}+b^{2}\right )^{\frac {3}{2}} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)/(a+b*sin(f*x+e))^2,x)

[Out]

2*(d*x+c)*(I*b+a*exp(I*(f*x+e)))/f/(a^2-b^2)/(b*exp(2*I*(f*x+e))-b+2*I*a*exp(I*(f*x+e)))-2/(-a^2+b^2)/f^2*d*ln
(exp(I*(f*x+e)))+1/(-a^2+b^2)/f^2*d*ln(I*exp(2*I*(f*x+e))*b-I*b-2*a*exp(I*(f*x+e)))+I/(-a^2+b^2)^(3/2)/f^2*d*a
*dilog((I*a+b*exp(I*(f*x+e))-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))+1/(-a^2+b^2)^(3/2)/f*d*a*ln((I*a+b*exp(
I*(f*x+e))+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))*x+1/(-a^2+b^2)^(3/2)/f^2*d*a*ln((I*a+b*exp(I*(f*x+e))+(-a
^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))*e+2*I/(-a^2+b^2)^(3/2)/f^2*a*d*e*arctan(1/2*(2*I*b*exp(I*(f*x+e))-2*a)/
(-a^2+b^2)^(1/2))-I/(-a^2+b^2)^(3/2)/f^2*d*a*dilog((I*a+b*exp(I*(f*x+e))+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/
2)))-1/(-a^2+b^2)^(3/2)/f*d*a*ln((I*a+b*exp(I*(f*x+e))-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))*x-1/(-a^2+b^2
)^(3/2)/f^2*d*a*ln((I*a+b*exp(I*(f*x+e))-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))*e-2*I/(-a^2+b^2)^(3/2)/f*a*
c*arctan(1/2*(2*I*b*exp(I*(f*x+e))-2*a)/(-a^2+b^2)^(1/2))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(a+b*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)/(a + b*sin(e + f*x))^2,x)

[Out]

\text{Hanged}

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(a+b*sin(f*x+e))**2,x)

[Out]

Timed out

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