3.100 \(\int \frac {\sin ^6(c+d x)}{(a+b \sin ^2(c+d x))^2} \, dx\)

Optimal. Leaf size=148 \[ \frac {a^{3/2} (4 a+5 b) \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{2 b^3 d (a+b)^{3/2}}-\frac {x (4 a-b)}{2 b^3}-\frac {a (2 a+b) \tan (c+d x)}{2 b^2 d (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )}-\frac {\sin ^2(c+d x) \tan (c+d x)}{2 b d \left ((a+b) \tan ^2(c+d x)+a\right )} \]

[Out]

-1/2*(4*a-b)*x/b^3+1/2*a^(3/2)*(4*a+5*b)*arctan((a+b)^(1/2)*tan(d*x+c)/a^(1/2))/b^3/(a+b)^(3/2)/d-1/2*a*(2*a+b
)*tan(d*x+c)/b^2/(a+b)/d/(a+(a+b)*tan(d*x+c)^2)-1/2*sin(d*x+c)^2*tan(d*x+c)/b/d/(a+(a+b)*tan(d*x+c)^2)

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Rubi [A]  time = 0.29, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3187, 470, 578, 522, 203, 205} \[ \frac {a^{3/2} (4 a+5 b) \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{2 b^3 d (a+b)^{3/2}}-\frac {a (2 a+b) \tan (c+d x)}{2 b^2 d (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )}-\frac {x (4 a-b)}{2 b^3}-\frac {\sin ^2(c+d x) \tan (c+d x)}{2 b d \left ((a+b) \tan ^2(c+d x)+a\right )} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^6/(a + b*Sin[c + d*x]^2)^2,x]

[Out]

-((4*a - b)*x)/(2*b^3) + (a^(3/2)*(4*a + 5*b)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(2*b^3*(a + b)^(3/2)
*d) - (a*(2*a + b)*Tan[c + d*x])/(2*b^2*(a + b)*d*(a + (a + b)*Tan[c + d*x]^2)) - (Sin[c + d*x]^2*Tan[c + d*x]
)/(2*b*d*(a + (a + b)*Tan[c + d*x]^2))

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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]

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2
*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2
*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +
(a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 578

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> Simp[(g^(n - 1)*(b*e - a*f)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c -
a*d)*(p + 1)), x] - Dist[g^n/(b*n*(b*c - a*d)*(p + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*S
imp[c*(b*e - a*f)*(m - n + 1) + (d*(b*e - a*f)*(m + n*q + 1) - b*n*(c*f - d*e)*(p + 1))*x^n, x], x], x] /; Fre
eQ[{a, b, c, d, e, f, g, q}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, 0]

Rule 3187

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

Rubi steps

\begin {align*} \int \frac {\sin ^6(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^6}{\left (1+x^2\right )^2 \left (a+(a+b) x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac {\sin ^2(c+d x) \tan (c+d x)}{2 b d \left (a+(a+b) \tan ^2(c+d x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (3 a+(-a+b) x^2\right )}{\left (1+x^2\right ) \left (a+(a+b) x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{2 b d}\\ &=-\frac {a (2 a+b) \tan (c+d x)}{2 b^2 (a+b) d \left (a+(a+b) \tan ^2(c+d x)\right )}-\frac {\sin ^2(c+d x) \tan (c+d x)}{2 b d \left (a+(a+b) \tan ^2(c+d x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {2 a (2 a+b)-2 \left (2 a^2+2 a b-b^2\right ) x^2}{\left (1+x^2\right ) \left (a+(a+b) x^2\right )} \, dx,x,\tan (c+d x)\right )}{4 b^2 (a+b) d}\\ &=-\frac {a (2 a+b) \tan (c+d x)}{2 b^2 (a+b) d \left (a+(a+b) \tan ^2(c+d x)\right )}-\frac {\sin ^2(c+d x) \tan (c+d x)}{2 b d \left (a+(a+b) \tan ^2(c+d x)\right )}-\frac {(4 a-b) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 b^3 d}+\frac {\left (a^2 (4 a+5 b)\right ) \operatorname {Subst}\left (\int \frac {1}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{2 b^3 (a+b) d}\\ &=-\frac {(4 a-b) x}{2 b^3}+\frac {a^{3/2} (4 a+5 b) \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{2 b^3 (a+b)^{3/2} d}-\frac {a (2 a+b) \tan (c+d x)}{2 b^2 (a+b) d \left (a+(a+b) \tan ^2(c+d x)\right )}-\frac {\sin ^2(c+d x) \tan (c+d x)}{2 b d \left (a+(a+b) \tan ^2(c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 1.53, size = 106, normalized size = 0.72 \[ \frac {\frac {2 a^{3/2} (4 a+5 b) \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{(a+b)^{3/2}}+b \sin (2 (c+d x)) \left (-\frac {2 a^2}{(a+b) (2 a-b \cos (2 (c+d x))+b)}-1\right )-2 (4 a-b) (c+d x)}{4 b^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^6/(a + b*Sin[c + d*x]^2)^2,x]

[Out]

(-2*(4*a - b)*(c + d*x) + (2*a^(3/2)*(4*a + 5*b)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(a + b)^(3/2) + b
*(-1 - (2*a^2)/((a + b)*(2*a + b - b*Cos[2*(c + d*x)])))*Sin[2*(c + d*x)])/(4*b^3*d)

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fricas [A]  time = 0.49, size = 623, normalized size = 4.21 \[ \left [-\frac {4 \, {\left (4 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} d x \cos \left (d x + c\right )^{2} - 4 \, {\left (4 \, a^{3} + 7 \, a^{2} b + 2 \, a b^{2} - b^{3}\right )} d x + {\left (4 \, a^{3} + 9 \, a^{2} b + 5 \, a b^{2} - {\left (4 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {-\frac {a}{a + b}} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left ({\left (2 \, a^{2} + 3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {-\frac {a}{a + b}} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) + 4 \, {\left ({\left (a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{3} - {\left (2 \, a^{2} b + 2 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, {\left ({\left (a b^{4} + b^{5}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{2} b^{3} + 2 \, a b^{4} + b^{5}\right )} d\right )}}, -\frac {2 \, {\left (4 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} d x \cos \left (d x + c\right )^{2} - 2 \, {\left (4 \, a^{3} + 7 \, a^{2} b + 2 \, a b^{2} - b^{3}\right )} d x - {\left (4 \, a^{3} + 9 \, a^{2} b + 5 \, a b^{2} - {\left (4 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {\frac {a}{a + b}} \arctan \left (\frac {{\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt {\frac {a}{a + b}}}{2 \, a \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) + 2 \, {\left ({\left (a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{3} - {\left (2 \, a^{2} b + 2 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a b^{4} + b^{5}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{2} b^{3} + 2 \, a b^{4} + b^{5}\right )} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^6/(a+b*sin(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

[-1/8*(4*(4*a^2*b + 3*a*b^2 - b^3)*d*x*cos(d*x + c)^2 - 4*(4*a^3 + 7*a^2*b + 2*a*b^2 - b^3)*d*x + (4*a^3 + 9*a
^2*b + 5*a*b^2 - (4*a^2*b + 5*a*b^2)*cos(d*x + c)^2)*sqrt(-a/(a + b))*log(((8*a^2 + 8*a*b + b^2)*cos(d*x + c)^
4 - 2*(4*a^2 + 5*a*b + b^2)*cos(d*x + c)^2 - 4*((2*a^2 + 3*a*b + b^2)*cos(d*x + c)^3 - (a^2 + 2*a*b + b^2)*cos
(d*x + c))*sqrt(-a/(a + b))*sin(d*x + c) + a^2 + 2*a*b + b^2)/(b^2*cos(d*x + c)^4 - 2*(a*b + b^2)*cos(d*x + c)
^2 + a^2 + 2*a*b + b^2)) + 4*((a*b^2 + b^3)*cos(d*x + c)^3 - (2*a^2*b + 2*a*b^2 + b^3)*cos(d*x + c))*sin(d*x +
c))/((a*b^4 + b^5)*d*cos(d*x + c)^2 - (a^2*b^3 + 2*a*b^4 + b^5)*d), -1/4*(2*(4*a^2*b + 3*a*b^2 - b^3)*d*x*cos
(d*x + c)^2 - 2*(4*a^3 + 7*a^2*b + 2*a*b^2 - b^3)*d*x - (4*a^3 + 9*a^2*b + 5*a*b^2 - (4*a^2*b + 5*a*b^2)*cos(d
*x + c)^2)*sqrt(a/(a + b))*arctan(1/2*((2*a + b)*cos(d*x + c)^2 - a - b)*sqrt(a/(a + b))/(a*cos(d*x + c)*sin(d
*x + c))) + 2*((a*b^2 + b^3)*cos(d*x + c)^3 - (2*a^2*b + 2*a*b^2 + b^3)*cos(d*x + c))*sin(d*x + c))/((a*b^4 +
b^5)*d*cos(d*x + c)^2 - (a^2*b^3 + 2*a*b^4 + b^5)*d)]

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giac [A]  time = 0.18, size = 223, normalized size = 1.51 \[ \frac {\frac {{\left (4 \, a^{3} + 5 \, a^{2} b\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )\right )}}{{\left (a b^{3} + b^{4}\right )} \sqrt {a^{2} + a b}} - \frac {2 \, a^{2} \tan \left (d x + c\right )^{3} + 2 \, a b \tan \left (d x + c\right )^{3} + b^{2} \tan \left (d x + c\right )^{3} + 2 \, a^{2} \tan \left (d x + c\right ) + a b \tan \left (d x + c\right )}{{\left (a \tan \left (d x + c\right )^{4} + b \tan \left (d x + c\right )^{4} + 2 \, a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right )^{2} + a\right )} {\left (a b^{2} + b^{3}\right )}} - \frac {{\left (d x + c\right )} {\left (4 \, a - b\right )}}{b^{3}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^6/(a+b*sin(d*x+c)^2)^2,x, algorithm="giac")

[Out]

1/2*((4*a^3 + 5*a^2*b)*(pi*floor((d*x + c)/pi + 1/2)*sgn(2*a + 2*b) + arctan((a*tan(d*x + c) + b*tan(d*x + c))
/sqrt(a^2 + a*b)))/((a*b^3 + b^4)*sqrt(a^2 + a*b)) - (2*a^2*tan(d*x + c)^3 + 2*a*b*tan(d*x + c)^3 + b^2*tan(d*
x + c)^3 + 2*a^2*tan(d*x + c) + a*b*tan(d*x + c))/((a*tan(d*x + c)^4 + b*tan(d*x + c)^4 + 2*a*tan(d*x + c)^2 +
b*tan(d*x + c)^2 + a)*(a*b^2 + b^3)) - (d*x + c)*(4*a - b)/b^3)/d

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maple [A]  time = 0.34, size = 187, normalized size = 1.26 \[ -\frac {a^{2} \tan \left (d x +c \right )}{2 d \,b^{2} \left (a +b \right ) \left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )}+\frac {2 a^{3} \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{d \,b^{3} \left (a +b \right ) \sqrt {a \left (a +b \right )}}+\frac {5 a^{2} \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{2 d \,b^{2} \left (a +b \right ) \sqrt {a \left (a +b \right )}}-\frac {\tan \left (d x +c \right )}{2 d \,b^{2} \left (\tan ^{2}\left (d x +c \right )+1\right )}+\frac {\arctan \left (\tan \left (d x +c \right )\right )}{2 d \,b^{2}}-\frac {2 \arctan \left (\tan \left (d x +c \right )\right ) a}{d \,b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^6/(a+b*sin(d*x+c)^2)^2,x)

[Out]

-1/2/d*a^2/b^2/(a+b)*tan(d*x+c)/(a*tan(d*x+c)^2+tan(d*x+c)^2*b+a)+2/d*a^3/b^3/(a+b)/(a*(a+b))^(1/2)*arctan((a+
b)*tan(d*x+c)/(a*(a+b))^(1/2))+5/2/d*a^2/b^2/(a+b)/(a*(a+b))^(1/2)*arctan((a+b)*tan(d*x+c)/(a*(a+b))^(1/2))-1/
2/d/b^2*tan(d*x+c)/(tan(d*x+c)^2+1)+1/2/d/b^2*arctan(tan(d*x+c))-2/d/b^3*arctan(tan(d*x+c))*a

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maxima [A]  time = 0.45, size = 181, normalized size = 1.22 \[ \frac {\frac {{\left (4 \, a^{3} + 5 \, a^{2} b\right )} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{{\left (a b^{3} + b^{4}\right )} \sqrt {{\left (a + b\right )} a}} - \frac {{\left (2 \, a^{2} + 2 \, a b + b^{2}\right )} \tan \left (d x + c\right )^{3} + {\left (2 \, a^{2} + a b\right )} \tan \left (d x + c\right )}{{\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} \tan \left (d x + c\right )^{4} + a^{2} b^{2} + a b^{3} + {\left (2 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} \tan \left (d x + c\right )^{2}} - \frac {{\left (d x + c\right )} {\left (4 \, a - b\right )}}{b^{3}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^6/(a+b*sin(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 16.06, size = 2295, normalized size = 15.51 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^6/(a + b*sin(c + d*x)^2)^2,x)

[Out]

(atan((((a + (5*b)/4)*(-a^3*(a + b)^3)^(1/2)*((tan(c + d*x)*(96*a^5*b - 4*a*b^5 + 32*a^6 + b^6 - 10*a^2*b^4 +
20*a^3*b^3 + 90*a^4*b^2))/(2*(a*b^4 + b^5)) + (((2*a*b^9 + 8*a^2*b^8 + 10*a^3*b^7 + 4*a^4*b^6)/(a*b^6 + b^7) +
(tan(c + d*x)*(a + (5*b)/4)*(-a^3*(a + b)^3)^(1/2)*(80*a*b^9 + 16*b^10 + 144*a^2*b^8 + 112*a^3*b^7 + 32*a^4*b
^6))/(2*(a*b^4 + b^5)*(3*a*b^5 + b^6 + 3*a^2*b^4 + a^3*b^3)))*(a + (5*b)/4)*(-a^3*(a + b)^3)^(1/2))/(3*a*b^5 +
b^6 + 3*a^2*b^4 + a^3*b^3))*1i)/(3*a*b^5 + b^6 + 3*a^2*b^4 + a^3*b^3) + ((a + (5*b)/4)*(-a^3*(a + b)^3)^(1/2)
*((tan(c + d*x)*(96*a^5*b - 4*a*b^5 + 32*a^6 + b^6 - 10*a^2*b^4 + 20*a^3*b^3 + 90*a^4*b^2))/(2*(a*b^4 + b^5))
- (((2*a*b^9 + 8*a^2*b^8 + 10*a^3*b^7 + 4*a^4*b^6)/(a*b^6 + b^7) - (tan(c + d*x)*(a + (5*b)/4)*(-a^3*(a + b)^3
)^(1/2)*(80*a*b^9 + 16*b^10 + 144*a^2*b^8 + 112*a^3*b^7 + 32*a^4*b^6))/(2*(a*b^4 + b^5)*(3*a*b^5 + b^6 + 3*a^2
*b^4 + a^3*b^3)))*(a + (5*b)/4)*(-a^3*(a + b)^3)^(1/2))/(3*a*b^5 + b^6 + 3*a^2*b^4 + a^3*b^3))*1i)/(3*a*b^5 +
b^6 + 3*a^2*b^4 + a^3*b^3))/((16*a^5*b + 8*a^6 + (5*a^2*b^4)/4 - (13*a^3*b^3)/2 + (3*a^4*b^2)/2)/(a*b^6 + b^7)
+ ((a + (5*b)/4)*(-a^3*(a + b)^3)^(1/2)*((tan(c + d*x)*(96*a^5*b - 4*a*b^5 + 32*a^6 + b^6 - 10*a^2*b^4 + 20*a
^3*b^3 + 90*a^4*b^2))/(2*(a*b^4 + b^5)) + (((2*a*b^9 + 8*a^2*b^8 + 10*a^3*b^7 + 4*a^4*b^6)/(a*b^6 + b^7) + (ta
n(c + d*x)*(a + (5*b)/4)*(-a^3*(a + b)^3)^(1/2)*(80*a*b^9 + 16*b^10 + 144*a^2*b^8 + 112*a^3*b^7 + 32*a^4*b^6))
/(2*(a*b^4 + b^5)*(3*a*b^5 + b^6 + 3*a^2*b^4 + a^3*b^3)))*(a + (5*b)/4)*(-a^3*(a + b)^3)^(1/2))/(3*a*b^5 + b^6
+ 3*a^2*b^4 + a^3*b^3)))/(3*a*b^5 + b^6 + 3*a^2*b^4 + a^3*b^3) - ((a + (5*b)/4)*(-a^3*(a + b)^3)^(1/2)*((tan(
c + d*x)*(96*a^5*b - 4*a*b^5 + 32*a^6 + b^6 - 10*a^2*b^4 + 20*a^3*b^3 + 90*a^4*b^2))/(2*(a*b^4 + b^5)) - (((2*
a*b^9 + 8*a^2*b^8 + 10*a^3*b^7 + 4*a^4*b^6)/(a*b^6 + b^7) - (tan(c + d*x)*(a + (5*b)/4)*(-a^3*(a + b)^3)^(1/2)
*(80*a*b^9 + 16*b^10 + 144*a^2*b^8 + 112*a^3*b^7 + 32*a^4*b^6))/(2*(a*b^4 + b^5)*(3*a*b^5 + b^6 + 3*a^2*b^4 +
a^3*b^3)))*(a + (5*b)/4)*(-a^3*(a + b)^3)^(1/2))/(3*a*b^5 + b^6 + 3*a^2*b^4 + a^3*b^3)))/(3*a*b^5 + b^6 + 3*a^
2*b^4 + a^3*b^3)))*(a + (5*b)/4)*(-a^3*(a + b)^3)^(1/2)*2i)/(d*(3*a*b^5 + b^6 + 3*a^2*b^4 + a^3*b^3)) - (atan(
((((((2*a*b^9 + 8*a^2*b^8 + 10*a^3*b^7 + 4*a^4*b^6)/(a*b^6 + b^7) - (tan(c + d*x)*(a*1i - (b*1i)/4)*(80*a*b^9
+ 16*b^10 + 144*a^2*b^8 + 112*a^3*b^7 + 32*a^4*b^6))/(2*b^3*(a*b^4 + b^5)))*(a*1i - (b*1i)/4))/b^3 - (tan(c +
d*x)*(96*a^5*b - 4*a*b^5 + 32*a^6 + b^6 - 10*a^2*b^4 + 20*a^3*b^3 + 90*a^4*b^2))/(2*(a*b^4 + b^5)))*(a*1i - (b
*1i)/4)*1i)/b^3 - (((((2*a*b^9 + 8*a^2*b^8 + 10*a^3*b^7 + 4*a^4*b^6)/(a*b^6 + b^7) + (tan(c + d*x)*(a*1i - (b*
1i)/4)*(80*a*b^9 + 16*b^10 + 144*a^2*b^8 + 112*a^3*b^7 + 32*a^4*b^6))/(2*b^3*(a*b^4 + b^5)))*(a*1i - (b*1i)/4)
)/b^3 + (tan(c + d*x)*(96*a^5*b - 4*a*b^5 + 32*a^6 + b^6 - 10*a^2*b^4 + 20*a^3*b^3 + 90*a^4*b^2))/(2*(a*b^4 +
b^5)))*(a*1i - (b*1i)/4)*1i)/b^3)/((16*a^5*b + 8*a^6 + (5*a^2*b^4)/4 - (13*a^3*b^3)/2 + (3*a^4*b^2)/2)/(a*b^6
+ b^7) + (((((2*a*b^9 + 8*a^2*b^8 + 10*a^3*b^7 + 4*a^4*b^6)/(a*b^6 + b^7) - (tan(c + d*x)*(a*1i - (b*1i)/4)*(8
0*a*b^9 + 16*b^10 + 144*a^2*b^8 + 112*a^3*b^7 + 32*a^4*b^6))/(2*b^3*(a*b^4 + b^5)))*(a*1i - (b*1i)/4))/b^3 - (
tan(c + d*x)*(96*a^5*b - 4*a*b^5 + 32*a^6 + b^6 - 10*a^2*b^4 + 20*a^3*b^3 + 90*a^4*b^2))/(2*(a*b^4 + b^5)))*(a
*1i - (b*1i)/4))/b^3 + (((((2*a*b^9 + 8*a^2*b^8 + 10*a^3*b^7 + 4*a^4*b^6)/(a*b^6 + b^7) + (tan(c + d*x)*(a*1i
- (b*1i)/4)*(80*a*b^9 + 16*b^10 + 144*a^2*b^8 + 112*a^3*b^7 + 32*a^4*b^6))/(2*b^3*(a*b^4 + b^5)))*(a*1i - (b*1
i)/4))/b^3 + (tan(c + d*x)*(96*a^5*b - 4*a*b^5 + 32*a^6 + b^6 - 10*a^2*b^4 + 20*a^3*b^3 + 90*a^4*b^2))/(2*(a*b
^4 + b^5)))*(a*1i - (b*1i)/4))/b^3))*(a*1i - (b*1i)/4)*2i)/(b^3*d) - ((tan(c + d*x)^3*(2*a*b + 2*a^2 + b^2))/(
2*b^2*(a + b)) + (a*tan(c + d*x)*(2*a + b))/(2*b^2*(a + b)))/(d*(a + tan(c + d*x)^4*(a + b) + tan(c + d*x)^2*(
2*a + b)))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)**6/(a+b*sin(d*x+c)**2)**2,x)

[Out]

Timed out

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