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[(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]))

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Rubi [A]  time = 0.0967485, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, 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 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]

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 12

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

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 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 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])

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*}

Mathematica [A]  time = 1.11504, 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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Maple [C]  time = 1.747, size = 349, normalized size = 3. \begin{align*}{\frac{2\,{a}^{2}dfx{{\rm e}^{2\,i \left ( dx+c \right ) }}-2\,{b}^{2}dfx{{\rm e}^{2\,i \left ( dx+c \right ) }}+2\,i{a}^{2}f{{\rm e}^{2\,i \left ( dx+c \right ) }}+i{b}^{2}f{{\rm e}^{2\,i \left ( dx+c \right ) }}+2\,{a}^{2}de{{\rm e}^{2\,i \left ( dx+c \right ) }}+baf{{\rm e}^{3\,i \left ( dx+c \right ) }}-2\,{b}^{2}de{{\rm e}^{2\,i \left ( dx+c \right ) }}-i{b}^{2}f-3\,abf{{\rm e}^{i \left ( dx+c \right ) }}}{ \left ( b{{\rm e}^{2\,i \left ( dx+c \right ) }}-b+2\,ia{{\rm e}^{i \left ( dx+c \right ) }} \right ) ^{2}{d}^{2} \left ({a}^{2}-{b}^{2} \right ) b}}-{\frac{af}{ \left ( 2\,a+2\,b \right ) \left ( a-b \right ){d}^{2}b}\ln \left ({{\rm e}^{i \left ( dx+c \right ) }}+{\frac{1}{b} \left ( ia\sqrt{-{a}^{2}+{b}^{2}}-{a}^{2}+{b}^{2} \right ){\frac{1}{\sqrt{-{a}^{2}+{b}^{2}}}}} \right ){\frac{1}{\sqrt{-{a}^{2}+{b}^{2}}}}}+{\frac{af}{ \left ( 2\,a+2\,b \right ) \left ( a-b \right ){d}^{2}b}\ln \left ({{\rm e}^{i \left ( dx+c \right ) }}+{\frac{1}{b} \left ( ia\sqrt{-{a}^{2}+{b}^{2}}+{a}^{2}-{b}^{2} \right ){\frac{1}{\sqrt{-{a}^{2}+{b}^{2}}}}} \right ){\frac{1}{\sqrt{-{a}^{2}+{b}^{2}}}}} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

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Fricas [B]  time = 2.03261, size = 1364, normalized size = 11.76 \begin{align*} \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 ] \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)