3.301 \(\int \frac{\cos ^7(x)}{a+b \sin ^2(x)} \, dx\)

Optimal. Leaf size=78 \[ -\frac{\left (a^2+3 a b+3 b^2\right ) \sin (x)}{b^3}+\frac{(a+3 b) \sin ^3(x)}{3 b^2}+\frac{(a+b)^3 \tan ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a}}\right )}{\sqrt{a} b^{7/2}}-\frac{\sin ^5(x)}{5 b} \]

[Out]

((a + b)^3*ArcTan[(Sqrt[b]*Sin[x])/Sqrt[a]])/(Sqrt[a]*b^(7/2)) - ((a^2 + 3*a*b + 3*b^2)*Sin[x])/b^3 + ((a + 3*
b)*Sin[x]^3)/(3*b^2) - Sin[x]^5/(5*b)

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Rubi [A]  time = 0.0895978, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3190, 390, 205} \[ -\frac{\left (a^2+3 a b+3 b^2\right ) \sin (x)}{b^3}+\frac{(a+3 b) \sin ^3(x)}{3 b^2}+\frac{(a+b)^3 \tan ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a}}\right )}{\sqrt{a} b^{7/2}}-\frac{\sin ^5(x)}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]^7/(a + b*Sin[x]^2),x]

[Out]

((a + b)^3*ArcTan[(Sqrt[b]*Sin[x])/Sqrt[a]])/(Sqrt[a]*b^(7/2)) - ((a^2 + 3*a*b + 3*b^2)*Sin[x])/b^3 + ((a + 3*
b)*Sin[x]^3)/(3*b^2) - Sin[x]^5/(5*b)

Rule 3190

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e +
f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 205

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

Rubi steps

\begin{align*} \int \frac{\cos ^7(x)}{a+b \sin ^2(x)} \, dx &=\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^3}{a+b x^2} \, dx,x,\sin (x)\right )\\ &=\operatorname{Subst}\left (\int \left (-\frac{a^2+3 a b+3 b^2}{b^3}+\frac{(a+3 b) x^2}{b^2}-\frac{x^4}{b}+\frac{a^3+3 a^2 b+3 a b^2+b^3}{b^3 \left (a+b x^2\right )}\right ) \, dx,x,\sin (x)\right )\\ &=-\frac{\left (a^2+3 a b+3 b^2\right ) \sin (x)}{b^3}+\frac{(a+3 b) \sin ^3(x)}{3 b^2}-\frac{\sin ^5(x)}{5 b}+\frac{(a+b)^3 \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\sin (x)\right )}{b^3}\\ &=\frac{(a+b)^3 \tan ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a}}\right )}{\sqrt{a} b^{7/2}}-\frac{\left (a^2+3 a b+3 b^2\right ) \sin (x)}{b^3}+\frac{(a+3 b) \sin ^3(x)}{3 b^2}-\frac{\sin ^5(x)}{5 b}\\ \end{align*}

Mathematica [A]  time = 0.275232, size = 109, normalized size = 1.4 \[ \frac{-2 \sqrt{a} \sqrt{b} \sin (x) \left (120 a^2+4 b (5 a+12 b) \cos (2 x)+340 a b+3 b^2 \cos (4 x)+309 b^2\right )+120 (a+b)^3 \tan ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a}}\right )-120 (a+b)^3 \tan ^{-1}\left (\frac{\sqrt{a} \csc (x)}{\sqrt{b}}\right )}{240 \sqrt{a} b^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]^7/(a + b*Sin[x]^2),x]

[Out]

(-120*(a + b)^3*ArcTan[(Sqrt[a]*Csc[x])/Sqrt[b]] + 120*(a + b)^3*ArcTan[(Sqrt[b]*Sin[x])/Sqrt[a]] - 2*Sqrt[a]*
Sqrt[b]*(120*a^2 + 340*a*b + 309*b^2 + 4*b*(5*a + 12*b)*Cos[2*x] + 3*b^2*Cos[4*x])*Sin[x])/(240*Sqrt[a]*b^(7/2
))

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Maple [B]  time = 0.043, size = 136, normalized size = 1.7 \begin{align*} -{\frac{ \left ( \sin \left ( x \right ) \right ) ^{5}}{5\,b}}+{\frac{a \left ( \sin \left ( x \right ) \right ) ^{3}}{3\,{b}^{2}}}+{\frac{ \left ( \sin \left ( x \right ) \right ) ^{3}}{b}}-{\frac{{a}^{2}\sin \left ( x \right ) }{{b}^{3}}}-3\,{\frac{a\sin \left ( x \right ) }{{b}^{2}}}-3\,{\frac{\sin \left ( x \right ) }{b}}+{\frac{{a}^{3}}{{b}^{3}}\arctan \left ({\sin \left ( x \right ) b{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+3\,{\frac{{a}^{2}}{{b}^{2}\sqrt{ab}}\arctan \left ({\frac{\sin \left ( x \right ) b}{\sqrt{ab}}} \right ) }+3\,{\frac{a}{\sqrt{ab}b}\arctan \left ({\frac{\sin \left ( x \right ) b}{\sqrt{ab}}} \right ) }+{\arctan \left ({\sin \left ( x \right ) b{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^7/(a+b*sin(x)^2),x)

[Out]

-1/5*sin(x)^5/b+1/3/b^2*sin(x)^3*a+sin(x)^3/b-1/b^3*a^2*sin(x)-3/b^2*sin(x)*a-3*sin(x)/b+1/b^3/(a*b)^(1/2)*arc
tan(sin(x)*b/(a*b)^(1/2))*a^3+3/b^2/(a*b)^(1/2)*arctan(sin(x)*b/(a*b)^(1/2))*a^2+3/b/(a*b)^(1/2)*arctan(sin(x)
*b/(a*b)^(1/2))*a+1/(a*b)^(1/2)*arctan(sin(x)*b/(a*b)^(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(x)^7/(a+b*sin(x)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.4446, size = 566, normalized size = 7.26 \begin{align*} \left [-\frac{15 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt{-a b} \log \left (-\frac{b \cos \left (x\right )^{2} + 2 \, \sqrt{-a b} \sin \left (x\right ) + a - b}{b \cos \left (x\right )^{2} - a - b}\right ) + 2 \,{\left (3 \, a b^{3} \cos \left (x\right )^{4} + 15 \, a^{3} b + 40 \, a^{2} b^{2} + 33 \, a b^{3} +{\left (5 \, a^{2} b^{2} + 9 \, a b^{3}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{30 \, a b^{4}}, \frac{15 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b} \sin \left (x\right )}{a}\right ) -{\left (3 \, a b^{3} \cos \left (x\right )^{4} + 15 \, a^{3} b + 40 \, a^{2} b^{2} + 33 \, a b^{3} +{\left (5 \, a^{2} b^{2} + 9 \, a b^{3}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{15 \, a b^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^7/(a+b*sin(x)^2),x, algorithm="fricas")

[Out]

[-1/30*(15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sqrt(-a*b)*log(-(b*cos(x)^2 + 2*sqrt(-a*b)*sin(x) + a - b)/(b*cos(x
)^2 - a - b)) + 2*(3*a*b^3*cos(x)^4 + 15*a^3*b + 40*a^2*b^2 + 33*a*b^3 + (5*a^2*b^2 + 9*a*b^3)*cos(x)^2)*sin(x
))/(a*b^4), 1/15*(15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sqrt(a*b)*arctan(sqrt(a*b)*sin(x)/a) - (3*a*b^3*cos(x)^4
+ 15*a^3*b + 40*a^2*b^2 + 33*a*b^3 + (5*a^2*b^2 + 9*a*b^3)*cos(x)^2)*sin(x))/(a*b^4)]

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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(x)**7/(a+b*sin(x)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.11695, size = 132, normalized size = 1.69 \begin{align*} \frac{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \arctan \left (\frac{b \sin \left (x\right )}{\sqrt{a b}}\right )}{\sqrt{a b} b^{3}} - \frac{3 \, b^{4} \sin \left (x\right )^{5} - 5 \, a b^{3} \sin \left (x\right )^{3} - 15 \, b^{4} \sin \left (x\right )^{3} + 15 \, a^{2} b^{2} \sin \left (x\right ) + 45 \, a b^{3} \sin \left (x\right ) + 45 \, b^{4} \sin \left (x\right )}{15 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^7/(a+b*sin(x)^2),x, algorithm="giac")

[Out]

(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*arctan(b*sin(x)/sqrt(a*b))/(sqrt(a*b)*b^3) - 1/15*(3*b^4*sin(x)^5 - 5*a*b^3*si
n(x)^3 - 15*b^4*sin(x)^3 + 15*a^2*b^2*sin(x) + 45*a*b^3*sin(x) + 45*b^4*sin(x))/b^5