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

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

[Out]

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

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Rubi [A]  time = 0.0903888, 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} \[ \frac{\left (a^2+3 a b+3 b^2\right ) \cos (x)}{b^3}-\frac{(a+3 b) \cos ^3(x)}{3 b^2}-\frac{(a+b)^3 \tan ^{-1}\left (\frac{\sqrt{b} \cos (x)}{\sqrt{a}}\right )}{\sqrt{a} b^{7/2}}+\frac{\cos ^5(x)}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^7/(a + b*Cos[x]^2),x]

[Out]

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

Mathematica [A]  time = 0.24336, size = 143, normalized size = 1.83 \[ \frac{\left (8 a^2+22 a b+19 b^2\right ) \cos (x)}{8 b^3}-\frac{(4 a+9 b) \cos (3 x)}{48 b^2}-\frac{(a+b)^3 \tan ^{-1}\left (\frac{\sqrt{b}-\sqrt{a+b} \tan \left (\frac{x}{2}\right )}{\sqrt{a}}\right )}{\sqrt{a} b^{7/2}}-\frac{(a+b)^3 \tan ^{-1}\left (\frac{\sqrt{a+b} \tan \left (\frac{x}{2}\right )+\sqrt{b}}{\sqrt{a}}\right )}{\sqrt{a} b^{7/2}}+\frac{\cos (5 x)}{80 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a + b)^3*ArcTan[(Sqrt[b] - Sqrt[a + b]*Tan[x/2])/Sqrt[a]])/(Sqrt[a]*b^(7/2))) - ((a + b)^3*ArcTan[(Sqrt[b]
+ Sqrt[a + b]*Tan[x/2])/Sqrt[a]])/(Sqrt[a]*b^(7/2)) + ((8*a^2 + 22*a*b + 19*b^2)*Cos[x])/(8*b^3) - ((4*a + 9*
b)*Cos[3*x])/(48*b^2) + Cos[5*x]/(80*b)

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Maple [B]  time = 0.024, size = 138, normalized size = 1.8 \begin{align*}{\frac{ \left ( \cos \left ( x \right ) \right ) ^{5}}{5\,b}}-{\frac{ \left ( \cos \left ( x \right ) \right ) ^{3}a}{3\,{b}^{2}}}-{\frac{ \left ( \cos \left ( x \right ) \right ) ^{3}}{b}}+{\frac{{a}^{2}\cos \left ( x \right ) }{{b}^{3}}}+3\,{\frac{a\cos \left ( x \right ) }{{b}^{2}}}+3\,{\frac{\cos \left ( x \right ) }{b}}-{\frac{{a}^{3}}{{b}^{3}}\arctan \left ({b\cos \left ( x \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-3\,{\frac{{a}^{2}}{{b}^{2}\sqrt{ab}}\arctan \left ({\frac{b\cos \left ( x \right ) }{\sqrt{ab}}} \right ) }-3\,{\frac{a}{b\sqrt{ab}}\arctan \left ({\frac{b\cos \left ( x \right ) }{\sqrt{ab}}} \right ) }-{\arctan \left ({b\cos \left ( x \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.08668, size = 547, normalized size = 7.01 \begin{align*} \left [\frac{6 \, a b^{3} \cos \left (x\right )^{5} - 10 \,{\left (a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (x\right )^{3} - 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} \cos \left (x\right ) - a}{b \cos \left (x\right )^{2} + a}\right ) + 30 \,{\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (x\right )}{30 \, a b^{4}}, \frac{3 \, a b^{3} \cos \left (x\right )^{5} - 5 \,{\left (a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (x\right )^{3} - 15 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b} \cos \left (x\right )}{a}\right ) + 15 \,{\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (x\right )}{15 \, a b^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/30*(6*a*b^3*cos(x)^5 - 10*(a^2*b^2 + 3*a*b^3)*cos(x)^3 - 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)*cos(x) - a)/(b*cos(x)^2 + a)) + 30*(a^3*b + 3*a^2*b^2 + 3*a*b^3)*cos(x))/(a*b^4),
1/15*(3*a*b^3*cos(x)^5 - 5*(a^2*b^2 + 3*a*b^3)*cos(x)^3 - 15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sqrt(a*b)*arctan(
sqrt(a*b)*cos(x)/a) + 15*(a^3*b + 3*a^2*b^2 + 3*a*b^3)*cos(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(sin(x)**7/(a+b*cos(x)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.14163, size = 134, normalized size = 1.72 \begin{align*} -\frac{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \arctan \left (\frac{b \cos \left (x\right )}{\sqrt{a b}}\right )}{\sqrt{a b} b^{3}} + \frac{3 \, b^{4} \cos \left (x\right )^{5} - 5 \, a b^{3} \cos \left (x\right )^{3} - 15 \, b^{4} \cos \left (x\right )^{3} + 15 \, a^{2} b^{2} \cos \left (x\right ) + 45 \, a b^{3} \cos \left (x\right ) + 45 \, b^{4} \cos \left (x\right )}{15 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*arctan(b*cos(x)/sqrt(a*b))/(sqrt(a*b)*b^3) + 1/15*(3*b^4*cos(x)^5 - 5*a*b^3*c
os(x)^3 - 15*b^4*cos(x)^3 + 15*a^2*b^2*cos(x) + 45*a*b^3*cos(x) + 45*b^4*cos(x))/b^5