3.133 \(\int \frac{\cos ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=119 \[ \frac{\left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )-b}{\sqrt{a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{5/2}}-\frac{b \left (\left (4 a^2+b^2\right ) \cos (c+d x)+3 a b \sin (c+d x)\right )}{2 d \left (a^2+b^2\right )^2 (a \cos (c+d x)+b \sin (c+d x))^2} \]

[Out]

((2*a^2 - b^2)*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]])/((a^2 + b^2)^(5/2)*d) - (b*((4*a^2 + b^2)*C
os[c + d*x] + 3*a*b*Sin[c + d*x]))/(2*(a^2 + b^2)^2*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)

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Rubi [A]  time = 0.587705, antiderivative size = 225, normalized size of antiderivative = 1.89, number of steps used = 6, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {1660, 12, 618, 206} \[ \frac{2 b^2 \left (\left (a^2+2 b^2\right ) \tan \left (\frac{1}{2} (c+d x)\right )+a b\right )}{a^3 d \left (a^2+b^2\right ) \left (-a \tan ^2\left (\frac{1}{2} (c+d x)\right )+a+2 b \tan \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{b \left (a b \left (5 a^2+2 b^2\right ) \tan \left (\frac{1}{2} (c+d x)\right )+3 a^2 b^2+4 a^4+2 b^4\right )}{a^3 d \left (a^2+b^2\right )^2 \left (-a \tan ^2\left (\frac{1}{2} (c+d x)\right )+a+2 b \tan \left (\frac{1}{2} (c+d x)\right )\right )}-\frac{\left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac{b-a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2/(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]

[Out]

-(((2*a^2 - b^2)*ArcTanh[(b - a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]])/((a^2 + b^2)^(5/2)*d)) + (2*b^2*(a*b + (a^
2 + 2*b^2)*Tan[(c + d*x)/2]))/(a^3*(a^2 + b^2)*d*(a + 2*b*Tan[(c + d*x)/2] - a*Tan[(c + d*x)/2]^2)^2) - (b*(4*
a^4 + 3*a^2*b^2 + 2*b^4 + a*b*(5*a^2 + 2*b^2)*Tan[(c + d*x)/2]))/(a^3*(a^2 + b^2)^2*d*(a + 2*b*Tan[(c + d*x)/2
] - a*Tan[(c + d*x)/2]^2))

Rule 1660

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x + c*
x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b
*x + c*x^2, x], x, 1]}, Simp[((b*f - 2*a*g + (2*c*f - b*g)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c
)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (
2*p + 3)*(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

Rule 12

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

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 206

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

Rubi steps

\begin{align*} \int \frac{\cos ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx &=\frac{2 \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2}{\left (a+2 b x-a x^2\right )^3} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{d}\\ &=\frac{2 b^2 \left (a b+\left (a^2+2 b^2\right ) \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^3 \left (a^2+b^2\right ) d \left (a+2 b \tan \left (\frac{1}{2} (c+d x)\right )-a \tan ^2\left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{\operatorname{Subst}\left (\int \frac{-\frac{8 \left (a^4+2 b^4\right )}{a^3}+16 b \left (1+\frac{b^2}{a^2}\right ) x+8 \left (a+\frac{b^2}{a}\right ) x^2}{\left (a+2 b x-a x^2\right )^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{4 \left (a^2+b^2\right ) d}\\ &=\frac{2 b^2 \left (a b+\left (a^2+2 b^2\right ) \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^3 \left (a^2+b^2\right ) d \left (a+2 b \tan \left (\frac{1}{2} (c+d x)\right )-a \tan ^2\left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{b \left (4 a^4+3 a^2 b^2+2 b^4+a b \left (5 a^2+2 b^2\right ) \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^3 \left (a^2+b^2\right )^2 d \left (a+2 b \tan \left (\frac{1}{2} (c+d x)\right )-a \tan ^2\left (\frac{1}{2} (c+d x)\right )\right )}+\frac{\operatorname{Subst}\left (\int \frac{16 \left (2 a^2-b^2\right )}{a+2 b x-a x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{16 \left (a^2+b^2\right )^2 d}\\ &=\frac{2 b^2 \left (a b+\left (a^2+2 b^2\right ) \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^3 \left (a^2+b^2\right ) d \left (a+2 b \tan \left (\frac{1}{2} (c+d x)\right )-a \tan ^2\left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{b \left (4 a^4+3 a^2 b^2+2 b^4+a b \left (5 a^2+2 b^2\right ) \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^3 \left (a^2+b^2\right )^2 d \left (a+2 b \tan \left (\frac{1}{2} (c+d x)\right )-a \tan ^2\left (\frac{1}{2} (c+d x)\right )\right )}+\frac{\left (2 a^2-b^2\right ) \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 )}{\left (a^2+b^2\right )^2 d}\\ &=\frac{2 b^2 \left (a b+\left (a^2+2 b^2\right ) \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^3 \left (a^2+b^2\right ) d \left (a+2 b \tan \left (\frac{1}{2} (c+d x)\right )-a \tan ^2\left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{b \left (4 a^4+3 a^2 b^2+2 b^4+a b \left (5 a^2+2 b^2\right ) \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^3 \left (a^2+b^2\right )^2 d \left (a+2 b \tan \left (\frac{1}{2} (c+d x)\right )-a \tan ^2\left (\frac{1}{2} (c+d x)\right )\right )}-\frac{\left (2 \left (2 a^2-b^2\right )\right ) \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 )}{\left (a^2+b^2\right )^2 d}\\ &=-\frac{\left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac{b-a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2} d}+\frac{2 b^2 \left (a b+\left (a^2+2 b^2\right ) \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^3 \left (a^2+b^2\right ) d \left (a+2 b \tan \left (\frac{1}{2} (c+d x)\right )-a \tan ^2\left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{b \left (4 a^4+3 a^2 b^2+2 b^4+a b \left (5 a^2+2 b^2\right ) \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^3 \left (a^2+b^2\right )^2 d \left (a+2 b \tan \left (\frac{1}{2} (c+d x)\right )-a \tan ^2\left (\frac{1}{2} (c+d x)\right )\right )}\\ \end{align*}

Mathematica [A]  time = 0.655404, size = 119, normalized size = 1. \[ \frac{\frac{2 \left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )-b}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac{b \left (\left (4 a^2+b^2\right ) \cos (c+d x)+3 a b \sin (c+d x)\right )}{\left (a^2+b^2\right )^2 (a \cos (c+d x)+b \sin (c+d x))^2}}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2/(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]

[Out]

((2*(2*a^2 - b^2)*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(5/2) - (b*((4*a^2 + b^2)*Co
s[c + d*x] + 3*a*b*Sin[c + d*x]))/((a^2 + b^2)^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2))/(2*d)

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Maple [B]  time = 0.206, size = 280, normalized size = 2.4 \begin{align*}{\frac{1}{d} \left ( -2\,{\frac{1}{ \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\,\tan \left ( 1/2\,dx+c/2 \right ) b-a \right ) ^{2}} \left ( -1/2\,{\frac{{b}^{2} \left ( 5\,{a}^{2}+2\,{b}^{2} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{a \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) }}-1/2\,{\frac{b \left ( 4\,{a}^{4}-7\,{a}^{2}{b}^{2}-2\,{b}^{4} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{ \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ){a}^{2}}}+1/2\,{\frac{{b}^{2} \left ( 11\,{a}^{2}+2\,{b}^{2} \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{a \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) }}+1/2\,{\frac{b \left ( 4\,{a}^{2}+{b}^{2} \right ) }{{a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4}}} \right ) }+{\frac{2\,{a}^{2}-{b}^{2}}{{a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4}}{\it Artanh} \left ({\frac{1}{2} \left ( 2\,a\tan \left ( 1/2\,dx+c/2 \right ) -2\,b \right ){\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}}} \right ){\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2/(a*cos(d*x+c)+b*sin(d*x+c))^3,x)

[Out]

1/d*(-2*(-1/2*b^2*(5*a^2+2*b^2)/a/(a^4+2*a^2*b^2+b^4)*tan(1/2*d*x+1/2*c)^3-1/2*b*(4*a^4-7*a^2*b^2-2*b^4)/(a^4+
2*a^2*b^2+b^4)/a^2*tan(1/2*d*x+1/2*c)^2+1/2*b^2*(11*a^2+2*b^2)/(a^4+2*a^2*b^2+b^4)/a*tan(1/2*d*x+1/2*c)+1/2*b*
(4*a^2+b^2)/(a^4+2*a^2*b^2+b^4))/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)^2+(2*a^2-b^2)/(a^4+2*a^2*b^
2+b^4)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tan(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2)))

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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(cos(d*x+c)^2/(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.586519, size = 795, normalized size = 6.68 \begin{align*} -\frac{{\left (2 \, a^{2} b^{2} - b^{4} +{\left (2 \, a^{4} - 3 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (2 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \sqrt{a^{2} + b^{2}} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt{a^{2} + b^{2}}{\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) + 2 \,{\left (4 \, a^{4} b + 5 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right ) + 6 \,{\left (a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )}{4 \,{\left ({\left (a^{8} + 2 \, a^{6} b^{2} - 2 \, a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\right )} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/4*((2*a^2*b^2 - b^4 + (2*a^4 - 3*a^2*b^2 + b^4)*cos(d*x + c)^2 + 2*(2*a^3*b - a*b^3)*cos(d*x + c)*sin(d*x +
c))*sqrt(a^2 + b^2)*log((2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 - 2*a^2 - b^2 - 2*sqrt(
a^2 + b^2)*(b*cos(d*x + c) - a*sin(d*x + c)))/(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 +
b^2)) + 2*(4*a^4*b + 5*a^2*b^3 + b^5)*cos(d*x + c) + 6*(a^3*b^2 + a*b^4)*sin(d*x + c))/((a^8 + 2*a^6*b^2 - 2*a
^2*b^6 - b^8)*d*cos(d*x + c)^2 + 2*(a^7*b + 3*a^5*b^3 + 3*a^3*b^5 + a*b^7)*d*cos(d*x + c)*sin(d*x + c) + (a^6*
b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8)*d)

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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(cos(d*x+c)**2/(a*cos(d*x+c)+b*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.35814, size = 396, normalized size = 3.33 \begin{align*} -\frac{\frac{{\left (2 \, a^{2} - b^{2}\right )} \log \left (\frac{{\left | 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{a^{2} + b^{2}}} - \frac{2 \,{\left (5 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 7 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 2 \, b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 11 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, a^{4} b - a^{2} b^{3}\right )}}{{\left (a^{6} + 2 \, a^{4} b^{2} + a^{2} b^{4}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/2*((2*a^2 - b^2)*log(abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b - 2*sqrt(a^2 + b^2))/abs(2*a*tan(1/2*d*x + 1/2*c) -
2*b + 2*sqrt(a^2 + b^2)))/((a^4 + 2*a^2*b^2 + b^4)*sqrt(a^2 + b^2)) - 2*(5*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 + 2
*a*b^4*tan(1/2*d*x + 1/2*c)^3 + 4*a^4*b*tan(1/2*d*x + 1/2*c)^2 - 7*a^2*b^3*tan(1/2*d*x + 1/2*c)^2 - 2*b^5*tan(
1/2*d*x + 1/2*c)^2 - 11*a^3*b^2*tan(1/2*d*x + 1/2*c) - 2*a*b^4*tan(1/2*d*x + 1/2*c) - 4*a^4*b - a^2*b^3)/((a^6
+ 2*a^4*b^2 + a^2*b^4)*(a*tan(1/2*d*x + 1/2*c)^2 - 2*b*tan(1/2*d*x + 1/2*c) - a)^2))/d