3.78 \(\int \frac{\sin ^7(c+d x)}{a+b \sin ^2(c+d x)} \, dx\)

Optimal. Leaf size=106 \[ -\frac{\left (a^2-a b+b^2\right ) \cos (c+d x)}{b^3 d}+\frac{a^3 \tanh ^{-1}\left (\frac{\sqrt{b} \cos (c+d x)}{\sqrt{a+b}}\right )}{b^{7/2} d \sqrt{a+b}}-\frac{(a-2 b) \cos ^3(c+d x)}{3 b^2 d}-\frac{\cos ^5(c+d x)}{5 b d} \]

[Out]

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

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Rubi [A]  time = 0.11261, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3186, 390, 208} \[ -\frac{\left (a^2-a b+b^2\right ) \cos (c+d x)}{b^3 d}+\frac{a^3 \tanh ^{-1}\left (\frac{\sqrt{b} \cos (c+d x)}{\sqrt{a+b}}\right )}{b^{7/2} d \sqrt{a+b}}-\frac{(a-2 b) \cos ^3(c+d x)}{3 b^2 d}-\frac{\cos ^5(c+d x)}{5 b d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^7/(a + b*Sin[c + d*x]^2),x]

[Out]

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

Rule 3186

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Cos[e + f*x], x]}, -Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos
[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 208

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

Rubi steps

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

Mathematica [C]  time = 1.39507, size = 180, normalized size = 1.7 \[ \frac{-2 \sqrt{b} \sqrt{-a-b} \cos (c+d x) \left (120 a^2+4 b (5 a-7 b) \cos (2 (c+d x))-100 a b+3 b^2 \cos (4 (c+d x))+89 b^2\right )-240 a^3 \tan ^{-1}\left (\frac{\sqrt{b}-i \sqrt{a} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a-b}}\right )-240 a^3 \tan ^{-1}\left (\frac{\sqrt{b}+i \sqrt{a} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a-b}}\right )}{240 b^{7/2} d \sqrt{-a-b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-240*a^3*ArcTan[(Sqrt[b] - I*Sqrt[a]*Tan[(c + d*x)/2])/Sqrt[-a - b]] - 240*a^3*ArcTan[(Sqrt[b] + I*Sqrt[a]*Ta
n[(c + d*x)/2])/Sqrt[-a - b]] - 2*Sqrt[-a - b]*Sqrt[b]*Cos[c + d*x]*(120*a^2 - 100*a*b + 89*b^2 + 4*(5*a - 7*b
)*b*Cos[2*(c + d*x)] + 3*b^2*Cos[4*(c + d*x)]))/(240*Sqrt[-a - b]*b^(7/2)*d)

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Maple [A]  time = 0.085, size = 110, normalized size = 1. \begin{align*}{\frac{1}{d} \left ( -{\frac{1}{{b}^{3}} \left ({\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{5}{b}^{2}}{5}}+{\frac{ab \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{3}}-{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}{b}^{2}}{3}}+\cos \left ( dx+c \right ){a}^{2}-ab\cos \left ( dx+c \right ) +\cos \left ( dx+c \right ){b}^{2} \right ) }+{\frac{{a}^{3}}{{b}^{3}}{\it Artanh} \left ({\cos \left ( dx+c \right ) b{\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-1/b^3*(1/5*cos(d*x+c)^5*b^2+1/3*a*b*cos(d*x+c)^3-2/3*cos(d*x+c)^3*b^2+cos(d*x+c)*a^2-a*b*cos(d*x+c)+cos(
d*x+c)*b^2)+a^3/b^3/((a+b)*b)^(1/2)*arctanh(cos(d*x+c)*b/((a+b)*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(d*x+c)^7/(a+b*sin(d*x+c)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.85589, size = 626, normalized size = 5.91 \begin{align*} \left [-\frac{6 \,{\left (a b^{3} + b^{4}\right )} \cos \left (d x + c\right )^{5} - 15 \, \sqrt{a b + b^{2}} a^{3} \log \left (\frac{b \cos \left (d x + c\right )^{2} + 2 \, \sqrt{a b + b^{2}} \cos \left (d x + c\right ) + a + b}{b \cos \left (d x + c\right )^{2} - a - b}\right ) + 10 \,{\left (a^{2} b^{2} - a b^{3} - 2 \, b^{4}\right )} \cos \left (d x + c\right )^{3} + 30 \,{\left (a^{3} b + b^{4}\right )} \cos \left (d x + c\right )}{30 \,{\left (a b^{4} + b^{5}\right )} d}, -\frac{3 \,{\left (a b^{3} + b^{4}\right )} \cos \left (d x + c\right )^{5} + 15 \, \sqrt{-a b - b^{2}} a^{3} \arctan \left (\frac{\sqrt{-a b - b^{2}} \cos \left (d x + c\right )}{a + b}\right ) + 5 \,{\left (a^{2} b^{2} - a b^{3} - 2 \, b^{4}\right )} \cos \left (d x + c\right )^{3} + 15 \,{\left (a^{3} b + b^{4}\right )} \cos \left (d x + c\right )}{15 \,{\left (a b^{4} + b^{5}\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.15983, size = 448, normalized size = 4.23 \begin{align*} -\frac{\frac{15 \, a^{3} \arctan \left (\frac{b \cos \left (d x + c\right ) + a + b}{\sqrt{-a b - b^{2}} \cos \left (d x + c\right ) + \sqrt{-a b - b^{2}}}\right )}{\sqrt{-a b - b^{2}} b^{3}} - \frac{2 \,{\left (15 \, a^{2} - 10 \, a b + 8 \, b^{2} - \frac{60 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{50 \, a b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{40 \, b^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{90 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{70 \, a b{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{80 \, b^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{60 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{30 \, a b{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{15 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}\right )}}{b^{3}{\left (\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}^{5}}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(15*a^3*arctan((b*cos(d*x + c) + a + b)/(sqrt(-a*b - b^2)*cos(d*x + c) + sqrt(-a*b - b^2)))/(sqrt(-a*b -
b^2)*b^3) - 2*(15*a^2 - 10*a*b + 8*b^2 - 60*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 50*a*b*(cos(d*x + c)
- 1)/(cos(d*x + c) + 1) - 40*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 90*a^2*(cos(d*x + c) - 1)^2/(cos(d*x
+ c) + 1)^2 - 70*a*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 80*b^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1
)^2 - 60*a^2*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 30*a*b*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 15
*a^2*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4)/(b^3*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)^5))/d