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

Optimal. Leaf size=128 \[ \frac{a^3 \cos (c+d x)}{2 b^3 d (a+b) \left (a-b \cos ^2(c+d x)+b\right )}-\frac{a^2 (5 a+6 b) \tanh ^{-1}\left (\frac{\sqrt{b} \cos (c+d x)}{\sqrt{a+b}}\right )}{2 b^{7/2} d (a+b)^{3/2}}+\frac{(2 a-b) \cos (c+d x)}{b^3 d}+\frac{\cos ^3(c+d x)}{3 b^2 d} \]

[Out]

-(a^2*(5*a + 6*b)*ArcTanh[(Sqrt[b]*Cos[c + d*x])/Sqrt[a + b]])/(2*b^(7/2)*(a + b)^(3/2)*d) + ((2*a - b)*Cos[c
+ d*x])/(b^3*d) + Cos[c + d*x]^3/(3*b^2*d) + (a^3*Cos[c + d*x])/(2*b^3*(a + b)*d*(a + b - b*Cos[c + d*x]^2))

________________________________________________________________________________________

Rubi [A]  time = 0.186128, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3186, 390, 385, 208} \[ \frac{a^3 \cos (c+d x)}{2 b^3 d (a+b) \left (a-b \cos ^2(c+d x)+b\right )}-\frac{a^2 (5 a+6 b) \tanh ^{-1}\left (\frac{\sqrt{b} \cos (c+d x)}{\sqrt{a+b}}\right )}{2 b^{7/2} d (a+b)^{3/2}}+\frac{(2 a-b) \cos (c+d x)}{b^3 d}+\frac{\cos ^3(c+d x)}{3 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*(5*a + 6*b)*ArcTanh[(Sqrt[b]*Cos[c + d*x])/Sqrt[a + b]])/(2*b^(7/2)*(a + b)^(3/2)*d) + ((2*a - b)*Cos[c
+ d*x])/(b^3*d) + Cos[c + d*x]^3/(3*b^2*d) + (a^3*Cos[c + d*x])/(2*b^3*(a + b)*d*(a + b - b*Cos[c + d*x]^2))

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 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

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)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^3}{\left (a+b-b x^2\right )^2} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int \left (-\frac{2 a-b}{b^3}-\frac{x^2}{b^2}+\frac{a^2 (2 a+3 b)-3 a^2 b x^2}{b^3 \left (a+b-b x^2\right )^2}\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac{(2 a-b) \cos (c+d x)}{b^3 d}+\frac{\cos ^3(c+d x)}{3 b^2 d}-\frac{\operatorname{Subst}\left (\int \frac{a^2 (2 a+3 b)-3 a^2 b x^2}{\left (a+b-b x^2\right )^2} \, dx,x,\cos (c+d x)\right )}{b^3 d}\\ &=\frac{(2 a-b) \cos (c+d x)}{b^3 d}+\frac{\cos ^3(c+d x)}{3 b^2 d}+\frac{a^3 \cos (c+d x)}{2 b^3 (a+b) d \left (a+b-b \cos ^2(c+d x)\right )}-\frac{\left (a^2 (5 a+6 b)\right ) \operatorname{Subst}\left (\int \frac{1}{a+b-b x^2} \, dx,x,\cos (c+d x)\right )}{2 b^3 (a+b) d}\\ &=-\frac{a^2 (5 a+6 b) \tanh ^{-1}\left (\frac{\sqrt{b} \cos (c+d x)}{\sqrt{a+b}}\right )}{2 b^{7/2} (a+b)^{3/2} d}+\frac{(2 a-b) \cos (c+d x)}{b^3 d}+\frac{\cos ^3(c+d x)}{3 b^2 d}+\frac{a^3 \cos (c+d x)}{2 b^3 (a+b) d \left (a+b-b \cos ^2(c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 1.57376, size = 194, normalized size = 1.52 \[ \frac{\sqrt{b} \left (\cos (c+d x) \left (\frac{12 a^3}{(a+b) (2 a-b \cos (2 (c+d x))+b)}+24 a-9 b\right )+b \cos (3 (c+d x))\right )-\frac{6 a^2 (5 a+6 b) \tan ^{-1}\left (\frac{\sqrt{b}-i \sqrt{a} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a-b}}\right )}{(-a-b)^{3/2}}-\frac{6 a^2 (5 a+6 b) \tan ^{-1}\left (\frac{\sqrt{b}+i \sqrt{a} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a-b}}\right )}{(-a-b)^{3/2}}}{12 b^{7/2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-6*a^2*(5*a + 6*b)*ArcTan[(Sqrt[b] - I*Sqrt[a]*Tan[(c + d*x)/2])/Sqrt[-a - b]])/(-a - b)^(3/2) - (6*a^2*(5*a
+ 6*b)*ArcTan[(Sqrt[b] + I*Sqrt[a]*Tan[(c + d*x)/2])/Sqrt[-a - b]])/(-a - b)^(3/2) + Sqrt[b]*(Cos[c + d*x]*(2
4*a - 9*b + (12*a^3)/((a + b)*(2*a + b - b*Cos[2*(c + d*x)]))) + b*Cos[3*(c + d*x)]))/(12*b^(7/2)*d)

________________________________________________________________________________________

Maple [A]  time = 0.085, size = 118, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({\frac{1}{{b}^{3}} \left ({\frac{b \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{3}}+2\,\cos \left ( dx+c \right ) a-b\cos \left ( dx+c \right ) \right ) }+{\frac{{a}^{2}}{{b}^{3}} \left ( -{\frac{\cos \left ( dx+c \right ) a}{ \left ( 2\,a+2\,b \right ) \left ( b \left ( \cos \left ( dx+c \right ) \right ) ^{2}-a-b \right ) }}-{\frac{5\,a+6\,b}{2\,a+2\,b}{\it Artanh} \left ({b\cos \left ( dx+c \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ) } \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)^2,x)

[Out]

1/d*(1/b^3*(1/3*b*cos(d*x+c)^3+2*cos(d*x+c)*a-b*cos(d*x+c))+1/b^3*a^2*(-1/2/(a+b)*a*cos(d*x+c)/(b*cos(d*x+c)^2
-a-b)-1/2*(5*a+6*b)/(a+b)/((a+b)*b)^(1/2)*arctanh(cos(d*x+c)*b/((a+b)*b)^(1/2))))

________________________________________________________________________________________

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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.07301, size = 1162, normalized size = 9.08 \begin{align*} \left [\frac{4 \,{\left (a^{2} b^{3} + 2 \, a b^{4} + b^{5}\right )} \cos \left (d x + c\right )^{5} + 4 \,{\left (5 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - 3 \, a b^{4} - 4 \, b^{5}\right )} \cos \left (d x + c\right )^{3} - 3 \,{\left (5 \, a^{4} + 11 \, a^{3} b + 6 \, a^{2} b^{2} -{\left (5 \, a^{3} b + 6 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt{a b + b^{2}} \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 ) - 6 \,{\left (5 \, a^{4} b + 11 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - 2 \, a b^{4} - 2 \, b^{5}\right )} \cos \left (d x + c\right )}{12 \,{\left ({\left (a^{2} b^{5} + 2 \, a b^{6} + b^{7}\right )} d \cos \left (d x + c\right )^{2} -{\left (a^{3} b^{4} + 3 \, a^{2} b^{5} + 3 \, a b^{6} + b^{7}\right )} d\right )}}, \frac{2 \,{\left (a^{2} b^{3} + 2 \, a b^{4} + b^{5}\right )} \cos \left (d x + c\right )^{5} + 2 \,{\left (5 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - 3 \, a b^{4} - 4 \, b^{5}\right )} \cos \left (d x + c\right )^{3} - 3 \,{\left (5 \, a^{4} + 11 \, a^{3} b + 6 \, a^{2} b^{2} -{\left (5 \, a^{3} b + 6 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt{-a b - b^{2}} \arctan \left (\frac{\sqrt{-a b - b^{2}} \cos \left (d x + c\right )}{a + b}\right ) - 3 \,{\left (5 \, a^{4} b + 11 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - 2 \, a b^{4} - 2 \, b^{5}\right )} \cos \left (d x + c\right )}{6 \,{\left ({\left (a^{2} b^{5} + 2 \, a b^{6} + b^{7}\right )} d \cos \left (d x + c\right )^{2} -{\left (a^{3} b^{4} + 3 \, a^{2} b^{5} + 3 \, a b^{6} + b^{7}\right )} d\right )}}\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)^2,x, algorithm="fricas")

[Out]

[1/12*(4*(a^2*b^3 + 2*a*b^4 + b^5)*cos(d*x + c)^5 + 4*(5*a^3*b^2 + 6*a^2*b^3 - 3*a*b^4 - 4*b^5)*cos(d*x + c)^3
- 3*(5*a^4 + 11*a^3*b + 6*a^2*b^2 - (5*a^3*b + 6*a^2*b^2)*cos(d*x + c)^2)*sqrt(a*b + b^2)*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)) - 6*(5*a^4*b + 11*a^3*b^2 + 6*a^2*b^
3 - 2*a*b^4 - 2*b^5)*cos(d*x + c))/((a^2*b^5 + 2*a*b^6 + b^7)*d*cos(d*x + c)^2 - (a^3*b^4 + 3*a^2*b^5 + 3*a*b^
6 + b^7)*d), 1/6*(2*(a^2*b^3 + 2*a*b^4 + b^5)*cos(d*x + c)^5 + 2*(5*a^3*b^2 + 6*a^2*b^3 - 3*a*b^4 - 4*b^5)*cos
(d*x + c)^3 - 3*(5*a^4 + 11*a^3*b + 6*a^2*b^2 - (5*a^3*b + 6*a^2*b^2)*cos(d*x + c)^2)*sqrt(-a*b - b^2)*arctan(
sqrt(-a*b - b^2)*cos(d*x + c)/(a + b)) - 3*(5*a^4*b + 11*a^3*b^2 + 6*a^2*b^3 - 2*a*b^4 - 2*b^5)*cos(d*x + c))/
((a^2*b^5 + 2*a*b^6 + b^7)*d*cos(d*x + c)^2 - (a^3*b^4 + 3*a^2*b^5 + 3*a*b^6 + b^7)*d)]

________________________________________________________________________________________

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)**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.18171, size = 435, normalized size = 3.4 \begin{align*} \frac{\frac{3 \,{\left (5 \, a^{3} + 6 \, a^{2} b\right )} \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 )}{{\left (a b^{3} + b^{4}\right )} \sqrt{-a b - b^{2}}} + \frac{6 \,{\left (a^{3} - \frac{a^{3}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{2 \, a^{2} b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}}{{\left (a b^{3} + b^{4}\right )}{\left (a - \frac{2 \, a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{4 \, b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{a{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}} - \frac{8 \,{\left (3 \, a - b - \frac{6 \, a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{3 \, b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{3 \, a{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}}{b^{3}{\left (\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}^{3}}}{6 \, 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)^2,x, algorithm="giac")

[Out]

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