3.322 \(\int \frac {(e+f x) \cos (c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=116 \[ \frac {a f \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b d^2 \left (a^2-b^2\right )^{3/2}}+\frac {f \cos (c+d x)}{2 d^2 \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac {e+f x}{2 b d (a+b \sin (c+d x))^2} \]

[Out]

a*f*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/b/(a^2-b^2)^(3/2)/d^2+1/2*(-f*x-e)/b/d/(a+b*sin(d*x+c))^2
+1/2*f*cos(d*x+c)/(a^2-b^2)/d^2/(a+b*sin(d*x+c))

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Rubi [A]  time = 0.10, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4422, 2664, 12, 2660, 618, 204} \[ \frac {a f \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b d^2 \left (a^2-b^2\right )^{3/2}}+\frac {f \cos (c+d x)}{2 d^2 \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac {e+f x}{2 b d (a+b \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[((e + f*x)*Cos[c + d*x])/(a + b*Sin[c + d*x])^3,x]

[Out]

(a*f*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b*(a^2 - b^2)^(3/2)*d^2) - (e + f*x)/(2*b*d*(a + b*Sin
[c + d*x])^2) + (f*Cos[c + d*x])/(2*(a^2 - b^2)*d^2*(a + b*Sin[c + d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 2664

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n +
1))/(d*(n + 1)*(a^2 - b^2)), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1
) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integer
Q[2*n]

Rule 4422

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

Rubi steps

\begin {align*} \int \frac {(e+f x) \cos (c+d x)}{(a+b \sin (c+d x))^3} \, dx &=-\frac {e+f x}{2 b d (a+b \sin (c+d x))^2}+\frac {f \int \frac {1}{(a+b \sin (c+d x))^2} \, dx}{2 b d}\\ &=-\frac {e+f x}{2 b d (a+b \sin (c+d x))^2}+\frac {f \cos (c+d x)}{2 \left (a^2-b^2\right ) d^2 (a+b \sin (c+d x))}+\frac {f \int \frac {a}{a+b \sin (c+d x)} \, dx}{2 b \left (a^2-b^2\right ) d}\\ &=-\frac {e+f x}{2 b d (a+b \sin (c+d x))^2}+\frac {f \cos (c+d x)}{2 \left (a^2-b^2\right ) d^2 (a+b \sin (c+d x))}+\frac {(a f) \int \frac {1}{a+b \sin (c+d x)} \, dx}{2 b \left (a^2-b^2\right ) d}\\ &=-\frac {e+f x}{2 b d (a+b \sin (c+d x))^2}+\frac {f \cos (c+d x)}{2 \left (a^2-b^2\right ) d^2 (a+b \sin (c+d x))}+\frac {(a f) \operatorname {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{b \left (a^2-b^2\right ) d^2}\\ &=-\frac {e+f x}{2 b d (a+b \sin (c+d x))^2}+\frac {f \cos (c+d x)}{2 \left (a^2-b^2\right ) d^2 (a+b \sin (c+d x))}-\frac {(2 a f) \operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b \left (a^2-b^2\right ) d^2}\\ &=\frac {a f \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d^2}-\frac {e+f x}{2 b d (a+b \sin (c+d x))^2}+\frac {f \cos (c+d x)}{2 \left (a^2-b^2\right ) d^2 (a+b \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 1.17, size = 112, normalized size = 0.97 \[ \frac {\frac {2 a f \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2}}+\frac {\frac {f \cos (c+d x) (a+b \sin (c+d x))}{(a-b) (a+b)}-\frac {d (e+f x)}{b}}{(a+b \sin (c+d x))^2}}{2 d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[((e + f*x)*Cos[c + d*x])/(a + b*Sin[c + d*x])^3,x]

[Out]

((2*a*f*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b*(a^2 - b^2)^(3/2)) + (-((d*(e + f*x))/b) + (f*Cos
[c + d*x]*(a + b*Sin[c + d*x]))/((a - b)*(a + b)))/(a + b*Sin[c + d*x])^2)/(2*d^2)

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fricas [B]  time = 0.50, size = 625, normalized size = 5.39 \[ \left [\frac {2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d f x - 2 \, {\left (a^{2} b^{2} - b^{4}\right )} f \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d e - 2 \, {\left (a^{3} b - a b^{3}\right )} f \cos \left (d x + c\right ) + {\left (a b^{2} f \cos \left (d x + c\right )^{2} - 2 \, a^{2} b f \sin \left (d x + c\right ) - {\left (a^{3} + a b^{2}\right )} f\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right )}{4 \, {\left ({\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} d^{2} \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} d^{2} \sin \left (d x + c\right ) - {\left (a^{6} b - a^{4} b^{3} - a^{2} b^{5} + b^{7}\right )} d^{2}\right )}}, \frac {{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d f x - {\left (a^{2} b^{2} - b^{4}\right )} f \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d e - {\left (a^{3} b - a b^{3}\right )} f \cos \left (d x + c\right ) - {\left (a b^{2} f \cos \left (d x + c\right )^{2} - 2 \, a^{2} b f \sin \left (d x + c\right ) - {\left (a^{3} + a b^{2}\right )} f\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right )}{2 \, {\left ({\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} d^{2} \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} d^{2} \sin \left (d x + c\right ) - {\left (a^{6} b - a^{4} b^{3} - a^{2} b^{5} + b^{7}\right )} d^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*cos(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

[1/4*(2*(a^4 - 2*a^2*b^2 + b^4)*d*f*x - 2*(a^2*b^2 - b^4)*f*cos(d*x + c)*sin(d*x + c) + 2*(a^4 - 2*a^2*b^2 + b
^4)*d*e - 2*(a^3*b - a*b^3)*f*cos(d*x + c) + (a*b^2*f*cos(d*x + c)^2 - 2*a^2*b*f*sin(d*x + c) - (a^3 + a*b^2)*
f)*sqrt(-a^2 + b^2)*log(-((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 - 2*(a*cos(d*x + c)*si
n(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)))/((a^4*b
^3 - 2*a^2*b^5 + b^7)*d^2*cos(d*x + c)^2 - 2*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*d^2*sin(d*x + c) - (a^6*b - a^4*b^3
- a^2*b^5 + b^7)*d^2), 1/2*((a^4 - 2*a^2*b^2 + b^4)*d*f*x - (a^2*b^2 - b^4)*f*cos(d*x + c)*sin(d*x + c) + (a^
4 - 2*a^2*b^2 + b^4)*d*e - (a^3*b - a*b^3)*f*cos(d*x + c) - (a*b^2*f*cos(d*x + c)^2 - 2*a^2*b*f*sin(d*x + c) -
(a^3 + a*b^2)*f)*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))))/((a^4*b^3 - 2*
a^2*b^5 + b^7)*d^2*cos(d*x + c)^2 - 2*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*d^2*sin(d*x + c) - (a^6*b - a^4*b^3 - a^2*
b^5 + b^7)*d^2)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (f x + e\right )} \cos \left (d x + c\right )}{{\left (b \sin \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*cos(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((f*x + e)*cos(d*x + c)/(b*sin(d*x + c) + a)^3, x)

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maple [C]  time = 2.60, size = 349, normalized size = 3.01 \[ \frac {2 a^{2} d f x \,{\mathrm e}^{2 i \left (d x +c \right )}-2 b^{2} d f x \,{\mathrm e}^{2 i \left (d x +c \right )}+2 i a^{2} f \,{\mathrm e}^{2 i \left (d x +c \right )}+i b^{2} f \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a^{2} d e \,{\mathrm e}^{2 i \left (d x +c \right )}+b a f \,{\mathrm e}^{3 i \left (d x +c \right )}-2 b^{2} d e \,{\mathrm e}^{2 i \left (d x +c \right )}-i b^{2} f -3 a b f \,{\mathrm e}^{i \left (d x +c \right )}}{\left (b \,{\mathrm e}^{2 i \left (d x +c \right )}-b +2 i a \,{\mathrm e}^{i \left (d x +c \right )}\right )^{2} d^{2} \left (a^{2}-b^{2}\right ) b}-\frac {f a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d^{2} b}+\frac {f a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d^{2} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)*cos(d*x+c)/(a+b*sin(d*x+c))^3,x)

[Out]

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

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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((f*x+e)*cos(d*x+c)/(a+b*sin(d*x+c))^3,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 = -0.01 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)*(e + f*x))/(a + b*sin(c + d*x))^3,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((f*x+e)*cos(d*x+c)/(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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