3.1.10 \(\int \frac {\sin ^7(x)}{a+b \cos ^2(x)} \, dx\) [10]

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

[Out]

(a^2+3*a*b+3*b^2)*cos(x)/b^3-1/3*(a+3*b)*cos(x)^3/b^2+1/5*cos(x)^5/b-(a+b)^3*arctan(cos(x)*b^(1/2)/a^(1/2))/b^
(7/2)/a^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3269, 398, 211} \begin {gather*} \frac {\left (a^2+3 a b+3 b^2\right ) \cos (x)}{b^3}-\frac {(a+b)^3 \text {ArcTan}\left (\frac {\sqrt {b} \cos (x)}{\sqrt {a}}\right )}{\sqrt {a} b^{7/2}}-\frac {(a+3 b) \cos ^3(x)}{3 b^2}+\frac {\cos ^5(x)}{5 b} \end {gather*}

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 211

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

Rule 398

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 3269

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]

Rubi steps

\begin {align*} \int \frac {\sin ^7(x)}{a+b \cos ^2(x)} \, dx &=-\text {Subst}\left (\int \frac {\left (1-x^2\right )^3}{a+b x^2} \, dx,x,\cos (x)\right )\\ &=-\text {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 \text {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*}

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Mathematica [A]
time = 0.26, size = 143, normalized size = 1.83 \begin {gather*} -\frac {(a+b)^3 \text {ArcTan}\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 \text {ArcTan}\left (\frac {\sqrt {b}+\sqrt {a+b} \tan \left (\frac {x}{2}\right )}{\sqrt {a}}\right )}{\sqrt {a} b^{7/2}}+\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 {\cos (5 x)}{80 b} \end {gather*}

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 [A]
time = 0.23, size = 94, normalized size = 1.21

method	result	size
default	\(\frac {\frac {\left (\cos ^{5}\left (x \right )\right ) b^{2}}{5}-\frac {a b \left (\cos ^{3}\left (x \right )\right )}{3}-b^{2} \left (\cos ^{3}\left (x \right )\right )+a^{2} \cos \left (x \right )+3 a b \cos \left (x \right )+3 b^{2} \cos \left (x \right )}{b^{3}}+\frac {\left (-a^{3}-3 a^{2} b -3 b^{2} a -b^{3}\right ) \arctan \left (\frac {b \cos \left (x \right )}{\sqrt {a b}}\right )}{b^{3} \sqrt {a b}}\)	\(94\)
risch	\(\frac {{\mathrm e}^{i x} a^{2}}{2 b^{3}}+\frac {11 a \,{\mathrm e}^{i x}}{8 b^{2}}+\frac {19 \,{\mathrm e}^{i x}}{16 b}+\frac {{\mathrm e}^{-i x} a^{2}}{2 b^{3}}+\frac {11 \,{\mathrm e}^{-i x} a}{8 b^{2}}+\frac {19 \,{\mathrm e}^{-i x}}{16 b}-\frac {3 i \ln \left ({\mathrm e}^{2 i x}+\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {a b}}+1\right ) a}{2 \sqrt {a b}\, b}-\frac {3 i \ln \left ({\mathrm e}^{2 i x}+\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {a b}}+1\right ) a^{2}}{2 \sqrt {a b}\, b^{2}}-\frac {i \ln \left ({\mathrm e}^{2 i x}+\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {a b}}+1\right )}{2 \sqrt {a b}}+\frac {i \ln \left ({\mathrm e}^{2 i x}-\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {a b}}+1\right ) a^{3}}{2 \sqrt {a b}\, b^{3}}+\frac {i \ln \left ({\mathrm e}^{2 i x}-\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {a b}}+1\right )}{2 \sqrt {a b}}-\frac {i \ln \left ({\mathrm e}^{2 i x}+\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {a b}}+1\right ) a^{3}}{2 \sqrt {a b}\, b^{3}}+\frac {3 i \ln \left ({\mathrm e}^{2 i x}-\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {a b}}+1\right ) a}{2 \sqrt {a b}\, b}+\frac {3 i \ln \left ({\mathrm e}^{2 i x}-\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {a b}}+1\right ) a^{2}}{2 \sqrt {a b}\, b^{2}}+\frac {\cos \left (5 x \right )}{80 b}-\frac {3 \cos \left (3 x \right )}{16 b}-\frac {\cos \left (3 x \right ) a}{12 b^{2}}\)	\(370\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^3*(1/5*cos(x)^5*b^2-1/3*a*b*cos(x)^3-b^2*cos(x)^3+a^2*cos(x)+3*a*b*cos(x)+3*b^2*cos(x))+(-a^3-3*a^2*b-3*a*
b^2-b^3)/b^3/(a*b)^(1/2)*arctan(b*cos(x)/(a*b)^(1/2))

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Maxima [A]
time = 0.47, size = 87, normalized size = 1.12 \begin {gather*} -\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^{2} \cos \left (x\right )^{5} - 5 \, {\left (a b + 3 \, b^{2}\right )} \cos \left (x\right )^{3} + 15 \, {\left (a^{2} + 3 \, a b + 3 \, b^{2}\right )} \cos \left (x\right )}{15 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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^2*cos(x)^5 - 5*(a*b +
3*b^2)*cos(x)^3 + 15*(a^2 + 3*a*b + 3*b^2)*cos(x))/b^3

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Fricas [A]
time = 0.41, size = 225, normalized size = 2.88 \begin {gather*} \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 {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.42, size = 99, normalized size = 1.27 \begin {gather*} -\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 {gather*}

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

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Mupad [B]
time = 2.13, size = 100, normalized size = 1.28 \begin {gather*} \cos \left (x\right )\,\left (\frac {3}{b}+\frac {a\,\left (\frac {a}{b^2}+\frac {3}{b}\right )}{b}\right )-{\cos \left (x\right )}^3\,\left (\frac {a}{3\,b^2}+\frac {1}{b}\right )+\frac {{\cos \left (x\right )}^5}{5\,b}-\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\cos \left (x\right )\,{\left (a+b\right )}^3}{\sqrt {a}\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}\right )\,{\left (a+b\right )}^3}{\sqrt {a}\,b^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^7/(a + b*cos(x)^2),x)

[Out]

cos(x)*(3/b + (a*(a/b^2 + 3/b))/b) - cos(x)^3*(a/(3*b^2) + 1/b) + cos(x)^5/(5*b) - (atan((b^(1/2)*cos(x)*(a +
b)^3)/(a^(1/2)*(3*a*b^2 + 3*a^2*b + a^3 + b^3)))*(a + b)^3)/(a^(1/2)*b^(7/2))

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