3.2.1 \(\int \frac {\sin ^4(c+d x)}{(a+b \sin ^2(c+d x))^2} \, dx\) [101]

Optimal. Leaf size=93 \[ \frac {x}{b^2}-\frac {\sqrt {a} (2 a+3 b) \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{2 b^2 (a+b)^{3/2} d}+\frac {a \tan (c+d x)}{2 b (a+b) d \left (a+(a+b) \tan ^2(c+d x)\right )} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3266, 481, 536, 209, 211} \begin {gather*} -\frac {\sqrt {a} (2 a+3 b) \text {ArcTan}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{2 b^2 d (a+b)^{3/2}}+\frac {a \tan (c+d x)}{2 b d (a+b) \left ((a+b) \tan ^2(c+d x)+a\right )}+\frac {x}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^4/(a + b*Sin[c + d*x]^2)^2,x]

[Out]

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

Rule 209

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

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 481

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 536

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 3266

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

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Mathematica [A]
time = 0.59, size = 93, normalized size = 1.00 \begin {gather*} \frac {2 (c+d x)-\frac {\sqrt {a} (2 a+3 b) \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{(a+b)^{3/2}}+\frac {a b \sin (2 (c+d x))}{(a+b) (2 a+b-b \cos (2 (c+d x)))}}{2 b^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^4/(a + b*Sin[c + d*x]^2)^2,x]

[Out]

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

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Maple [A]
time = 0.35, size = 101, normalized size = 1.09

method	result	size
derivativedivides	\(\frac {\frac {\arctan \left (\tan \left (d x +c \right )\right )}{b^{2}}-\frac {a \left (-\frac {b \tan \left (d x +c \right )}{2 \left (a +b \right ) \left (a \left (\tan ^{2}\left (d x +c \right )\right )+b \left (\tan ^{2}\left (d x +c \right )\right )+a \right )}+\frac {\left (2 a +3 b \right ) \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{2 \left (a +b \right ) \sqrt {a \left (a +b \right )}}\right )}{b^{2}}}{d}\)	\(101\)
default	\(\frac {\frac {\arctan \left (\tan \left (d x +c \right )\right )}{b^{2}}-\frac {a \left (-\frac {b \tan \left (d x +c \right )}{2 \left (a +b \right ) \left (a \left (\tan ^{2}\left (d x +c \right )\right )+b \left (\tan ^{2}\left (d x +c \right )\right )+a \right )}+\frac {\left (2 a +3 b \right ) \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{2 \left (a +b \right ) \sqrt {a \left (a +b \right )}}\right )}{b^{2}}}{d}\)	\(101\)
risch	\(\frac {x}{b^{2}}-\frac {i a \left (2 a \,{\mathrm e}^{2 i \left (d x +c \right )}+b \,{\mathrm e}^{2 i \left (d x +c \right )}-b \right )}{b^{2} \left (a +b \right ) d \left (-b \,{\mathrm e}^{4 i \left (d x +c \right )}+4 a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-b \right )}+\frac {\sqrt {-a \left (a +b \right )}\, a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a \left (a +b \right )}+2 a +b}{b}\right )}{2 \left (a +b \right )^{2} d \,b^{2}}+\frac {3 \sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a \left (a +b \right )}+2 a +b}{b}\right )}{4 \left (a +b \right )^{2} d b}-\frac {\sqrt {-a \left (a +b \right )}\, a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a \left (a +b \right )}-2 a -b}{b}\right )}{2 \left (a +b \right )^{2} d \,b^{2}}-\frac {3 \sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a \left (a +b \right )}-2 a -b}{b}\right )}{4 \left (a +b \right )^{2} d b}\)	\(307\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^4/(a+sin(d*x+c)^2*b)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.53, size = 109, normalized size = 1.17 \begin {gather*} \frac {\frac {a \tan \left (d x + c\right )}{a^{2} b + a b^{2} + {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \tan \left (d x + c\right )^{2}} - \frac {{\left (2 \, a^{2} + 3 \, a 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^{2} + b^{3}\right )} \sqrt {{\left (a + b\right )} a}} + \frac {2 \, {\left (d x + c\right )}}{b^{2}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^4/(a+b*sin(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 191 vs. \(2 (81) = 162\).
time = 0.46, size = 492, normalized size = 5.29 \begin {gather*} \left [\frac {8 \, {\left (a b + b^{2}\right )} d x \cos \left (d x + c\right )^{2} - 4 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 8 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + {\left ({\left (2 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - 5 \, a b - 3 \, b^{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 )}{8 \, {\left ({\left (a b^{3} + b^{4}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} d\right )}}, \frac {4 \, {\left (a b + b^{2}\right )} d x \cos \left (d x + c\right )^{2} - 2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + {\left ({\left (2 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - 5 \, a b - 3 \, b^{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 )}{4 \, {\left ({\left (a b^{3} + b^{4}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} d\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^4/(a+b*sin(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

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

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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(d*x+c)**4/(a+b*sin(d*x+c)**2)**2,x)

[Out]

Timed out

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Giac [A]
time = 0.45, size = 140, normalized size = 1.51 \begin {gather*} -\frac {\frac {{\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 (2 \, a^{2} + 3 \, a b\right )}}{{\left (a b^{2} + b^{3}\right )} \sqrt {a^{2} + a b}} - \frac {a \tan \left (d x + c\right )}{{\left (a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right )^{2} + a\right )} {\left (a b + b^{2}\right )}} - \frac {2 \, {\left (d x + c\right )}}{b^{2}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^4/(a+b*sin(d*x+c)^2)^2,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 15.23, size = 1959, normalized size = 21.06 \begin {gather*} \frac {a\,\mathrm {tan}\left (c+d\,x\right )}{2\,d\,\left (\left (a+b\right )\,{\mathrm {tan}\left (c+d\,x\right )}^2+a\right )\,\left (b^2+a\,b\right )}-\frac {\mathrm {atan}\left (\frac {\frac {\frac {-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (32\,a^4\,b^4+112\,a^3\,b^5+144\,a^2\,b^6+80\,a\,b^7+16\,b^8\right )}{8\,b^2\,\left (b^3+a\,b^2\right )}+\frac {\left (2\,a^3\,b^4+4\,a^2\,b^5+2\,a\,b^6\right )\,1{}\mathrm {i}}{2\,\left (b^4+a\,b^3\right )}}{2\,b^2}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (8\,a^4+28\,a^3\,b+33\,a^2\,b^2+16\,a\,b^3+4\,b^4\right )}{4\,\left (b^3+a\,b^2\right )}}{b^2}-\frac {\frac {\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (32\,a^4\,b^4+112\,a^3\,b^5+144\,a^2\,b^6+80\,a\,b^7+16\,b^8\right )}{8\,b^2\,\left (b^3+a\,b^2\right )}+\frac {\left (2\,a^3\,b^4+4\,a^2\,b^5+2\,a\,b^6\right )\,1{}\mathrm {i}}{2\,\left (b^4+a\,b^3\right )}}{2\,b^2}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (8\,a^4+28\,a^3\,b+33\,a^2\,b^2+16\,a\,b^3+4\,b^4\right )}{4\,\left (b^3+a\,b^2\right )}}{b^2}}{\frac {a^3+\frac {7\,a^2\,b}{2}+3\,a\,b^2}{b^4+a\,b^3}+\frac {\frac {\left (-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (32\,a^4\,b^4+112\,a^3\,b^5+144\,a^2\,b^6+80\,a\,b^7+16\,b^8\right )}{8\,b^2\,\left (b^3+a\,b^2\right )}+\frac {\left (2\,a^3\,b^4+4\,a^2\,b^5+2\,a\,b^6\right )\,1{}\mathrm {i}}{2\,\left (b^4+a\,b^3\right )}\right )\,1{}\mathrm {i}}{2\,b^2}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (8\,a^4+28\,a^3\,b+33\,a^2\,b^2+16\,a\,b^3+4\,b^4\right )\,1{}\mathrm {i}}{4\,\left (b^3+a\,b^2\right )}}{b^2}+\frac {\frac {\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (32\,a^4\,b^4+112\,a^3\,b^5+144\,a^2\,b^6+80\,a\,b^7+16\,b^8\right )}{8\,b^2\,\left (b^3+a\,b^2\right )}+\frac {\left (2\,a^3\,b^4+4\,a^2\,b^5+2\,a\,b^6\right )\,1{}\mathrm {i}}{2\,\left (b^4+a\,b^3\right )}\right )\,1{}\mathrm {i}}{2\,b^2}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (8\,a^4+28\,a^3\,b+33\,a^2\,b^2+16\,a\,b^3+4\,b^4\right )\,1{}\mathrm {i}}{4\,\left (b^3+a\,b^2\right )}}{b^2}}\right )}{b^2\,d}-\frac {\mathrm {atan}\left (\frac {\frac {\sqrt {-a\,{\left (a+b\right )}^3}\,\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (8\,a^4+28\,a^3\,b+33\,a^2\,b^2+16\,a\,b^3+4\,b^4\right )}{2\,\left (b^3+a\,b^2\right )}-\frac {\left (\frac {2\,a^3\,b^4+4\,a^2\,b^5+2\,a\,b^6}{b^4+a\,b^3}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-a\,{\left (a+b\right )}^3}\,\left (2\,a+3\,b\right )\,\left (32\,a^4\,b^4+112\,a^3\,b^5+144\,a^2\,b^6+80\,a\,b^7+16\,b^8\right )}{8\,\left (b^3+a\,b^2\right )\,\left (a^3\,b^2+3\,a^2\,b^3+3\,a\,b^4+b^5\right )}\right )\,\sqrt {-a\,{\left (a+b\right )}^3}\,\left (2\,a+3\,b\right )}{4\,\left (a^3\,b^2+3\,a^2\,b^3+3\,a\,b^4+b^5\right )}\right )\,\left (2\,a+3\,b\right )\,1{}\mathrm {i}}{4\,\left (a^3\,b^2+3\,a^2\,b^3+3\,a\,b^4+b^5\right )}+\frac {\sqrt {-a\,{\left (a+b\right )}^3}\,\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (8\,a^4+28\,a^3\,b+33\,a^2\,b^2+16\,a\,b^3+4\,b^4\right )}{2\,\left (b^3+a\,b^2\right )}+\frac {\left (\frac {2\,a^3\,b^4+4\,a^2\,b^5+2\,a\,b^6}{b^4+a\,b^3}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-a\,{\left (a+b\right )}^3}\,\left (2\,a+3\,b\right )\,\left (32\,a^4\,b^4+112\,a^3\,b^5+144\,a^2\,b^6+80\,a\,b^7+16\,b^8\right )}{8\,\left (b^3+a\,b^2\right )\,\left (a^3\,b^2+3\,a^2\,b^3+3\,a\,b^4+b^5\right )}\right )\,\sqrt {-a\,{\left (a+b\right )}^3}\,\left (2\,a+3\,b\right )}{4\,\left (a^3\,b^2+3\,a^2\,b^3+3\,a\,b^4+b^5\right )}\right )\,\left (2\,a+3\,b\right )\,1{}\mathrm {i}}{4\,\left (a^3\,b^2+3\,a^2\,b^3+3\,a\,b^4+b^5\right )}}{\frac {a^3+\frac {7\,a^2\,b}{2}+3\,a\,b^2}{b^4+a\,b^3}-\frac {\sqrt {-a\,{\left (a+b\right )}^3}\,\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (8\,a^4+28\,a^3\,b+33\,a^2\,b^2+16\,a\,b^3+4\,b^4\right )}{2\,\left (b^3+a\,b^2\right )}-\frac {\left (\frac {2\,a^3\,b^4+4\,a^2\,b^5+2\,a\,b^6}{b^4+a\,b^3}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-a\,{\left (a+b\right )}^3}\,\left (2\,a+3\,b\right )\,\left (32\,a^4\,b^4+112\,a^3\,b^5+144\,a^2\,b^6+80\,a\,b^7+16\,b^8\right )}{8\,\left (b^3+a\,b^2\right )\,\left (a^3\,b^2+3\,a^2\,b^3+3\,a\,b^4+b^5\right )}\right )\,\sqrt {-a\,{\left (a+b\right )}^3}\,\left (2\,a+3\,b\right )}{4\,\left (a^3\,b^2+3\,a^2\,b^3+3\,a\,b^4+b^5\right )}\right )\,\left (2\,a+3\,b\right )}{4\,\left (a^3\,b^2+3\,a^2\,b^3+3\,a\,b^4+b^5\right )}+\frac {\sqrt {-a\,{\left (a+b\right )}^3}\,\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (8\,a^4+28\,a^3\,b+33\,a^2\,b^2+16\,a\,b^3+4\,b^4\right )}{2\,\left (b^3+a\,b^2\right )}+\frac {\left (\frac {2\,a^3\,b^4+4\,a^2\,b^5+2\,a\,b^6}{b^4+a\,b^3}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-a\,{\left (a+b\right )}^3}\,\left (2\,a+3\,b\right )\,\left (32\,a^4\,b^4+112\,a^3\,b^5+144\,a^2\,b^6+80\,a\,b^7+16\,b^8\right )}{8\,\left (b^3+a\,b^2\right )\,\left (a^3\,b^2+3\,a^2\,b^3+3\,a\,b^4+b^5\right )}\right )\,\sqrt {-a\,{\left (a+b\right )}^3}\,\left (2\,a+3\,b\right )}{4\,\left (a^3\,b^2+3\,a^2\,b^3+3\,a\,b^4+b^5\right )}\right )\,\left (2\,a+3\,b\right )}{4\,\left (a^3\,b^2+3\,a^2\,b^3+3\,a\,b^4+b^5\right )}}\right )\,\sqrt {-a\,{\left (a+b\right )}^3}\,\left (2\,a+3\,b\right )\,1{}\mathrm {i}}{2\,d\,\left (a^3\,b^2+3\,a^2\,b^3+3\,a\,b^4+b^5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^4/(a + b*sin(c + d*x)^2)^2,x)

[Out]

(a*tan(c + d*x))/(2*d*(a + tan(c + d*x)^2*(a + b))*(a*b + b^2)) - atan((((((2*a*b^6 + 4*a^2*b^5 + 2*a^3*b^4)*1
i)/(2*(a*b^3 + b^4)) - (tan(c + d*x)*(80*a*b^7 + 16*b^8 + 144*a^2*b^6 + 112*a^3*b^5 + 32*a^4*b^4))/(8*b^2*(a*b
^2 + b^3)))/(2*b^2) + (tan(c + d*x)*(16*a*b^3 + 28*a^3*b + 8*a^4 + 4*b^4 + 33*a^2*b^2))/(4*(a*b^2 + b^3)))/b^2
- ((((2*a*b^6 + 4*a^2*b^5 + 2*a^3*b^4)*1i)/(2*(a*b^3 + b^4)) + (tan(c + d*x)*(80*a*b^7 + 16*b^8 + 144*a^2*b^6
+ 112*a^3*b^5 + 32*a^4*b^4))/(8*b^2*(a*b^2 + b^3)))/(2*b^2) - (tan(c + d*x)*(16*a*b^3 + 28*a^3*b + 8*a^4 + 4*
b^4 + 33*a^2*b^2))/(4*(a*b^2 + b^3)))/b^2)/((3*a*b^2 + (7*a^2*b)/2 + a^3)/(a*b^3 + b^4) + (((((2*a*b^6 + 4*a^2
*b^5 + 2*a^3*b^4)*1i)/(2*(a*b^3 + b^4)) - (tan(c + d*x)*(80*a*b^7 + 16*b^8 + 144*a^2*b^6 + 112*a^3*b^5 + 32*a^
4*b^4))/(8*b^2*(a*b^2 + b^3)))*1i)/(2*b^2) + (tan(c + d*x)*(16*a*b^3 + 28*a^3*b + 8*a^4 + 4*b^4 + 33*a^2*b^2)*
1i)/(4*(a*b^2 + b^3)))/b^2 + (((((2*a*b^6 + 4*a^2*b^5 + 2*a^3*b^4)*1i)/(2*(a*b^3 + b^4)) + (tan(c + d*x)*(80*a
*b^7 + 16*b^8 + 144*a^2*b^6 + 112*a^3*b^5 + 32*a^4*b^4))/(8*b^2*(a*b^2 + b^3)))*1i)/(2*b^2) - (tan(c + d*x)*(1
6*a*b^3 + 28*a^3*b + 8*a^4 + 4*b^4 + 33*a^2*b^2)*1i)/(4*(a*b^2 + b^3)))/b^2))/(b^2*d) - (atan((((-a*(a + b)^3)
^(1/2)*((tan(c + d*x)*(16*a*b^3 + 28*a^3*b + 8*a^4 + 4*b^4 + 33*a^2*b^2))/(2*(a*b^2 + b^3)) - (((2*a*b^6 + 4*a
^2*b^5 + 2*a^3*b^4)/(a*b^3 + b^4) - (tan(c + d*x)*(-a*(a + b)^3)^(1/2)*(2*a + 3*b)*(80*a*b^7 + 16*b^8 + 144*a^
2*b^6 + 112*a^3*b^5 + 32*a^4*b^4))/(8*(a*b^2 + b^3)*(3*a*b^4 + b^5 + 3*a^2*b^3 + a^3*b^2)))*(-a*(a + b)^3)^(1/
2)*(2*a + 3*b))/(4*(3*a*b^4 + b^5 + 3*a^2*b^3 + a^3*b^2)))*(2*a + 3*b)*1i)/(4*(3*a*b^4 + b^5 + 3*a^2*b^3 + a^3
*b^2)) + ((-a*(a + b)^3)^(1/2)*((tan(c + d*x)*(16*a*b^3 + 28*a^3*b + 8*a^4 + 4*b^4 + 33*a^2*b^2))/(2*(a*b^2 +
b^3)) + (((2*a*b^6 + 4*a^2*b^5 + 2*a^3*b^4)/(a*b^3 + b^4) + (tan(c + d*x)*(-a*(a + b)^3)^(1/2)*(2*a + 3*b)*(80
*a*b^7 + 16*b^8 + 144*a^2*b^6 + 112*a^3*b^5 + 32*a^4*b^4))/(8*(a*b^2 + b^3)*(3*a*b^4 + b^5 + 3*a^2*b^3 + a^3*b
^2)))*(-a*(a + b)^3)^(1/2)*(2*a + 3*b))/(4*(3*a*b^4 + b^5 + 3*a^2*b^3 + a^3*b^2)))*(2*a + 3*b)*1i)/(4*(3*a*b^4
+ b^5 + 3*a^2*b^3 + a^3*b^2)))/((3*a*b^2 + (7*a^2*b)/2 + a^3)/(a*b^3 + b^4) - ((-a*(a + b)^3)^(1/2)*((tan(c +
d*x)*(16*a*b^3 + 28*a^3*b + 8*a^4 + 4*b^4 + 33*a^2*b^2))/(2*(a*b^2 + b^3)) - (((2*a*b^6 + 4*a^2*b^5 + 2*a^3*b
^4)/(a*b^3 + b^4) - (tan(c + d*x)*(-a*(a + b)^3)^(1/2)*(2*a + 3*b)*(80*a*b^7 + 16*b^8 + 144*a^2*b^6 + 112*a^3*
b^5 + 32*a^4*b^4))/(8*(a*b^2 + b^3)*(3*a*b^4 + b^5 + 3*a^2*b^3 + a^3*b^2)))*(-a*(a + b)^3)^(1/2)*(2*a + 3*b))/
(4*(3*a*b^4 + b^5 + 3*a^2*b^3 + a^3*b^2)))*(2*a + 3*b))/(4*(3*a*b^4 + b^5 + 3*a^2*b^3 + a^3*b^2)) + ((-a*(a +
b)^3)^(1/2)*((tan(c + d*x)*(16*a*b^3 + 28*a^3*b + 8*a^4 + 4*b^4 + 33*a^2*b^2))/(2*(a*b^2 + b^3)) + (((2*a*b^6
+ 4*a^2*b^5 + 2*a^3*b^4)/(a*b^3 + b^4) + (tan(c + d*x)*(-a*(a + b)^3)^(1/2)*(2*a + 3*b)*(80*a*b^7 + 16*b^8 + 1
44*a^2*b^6 + 112*a^3*b^5 + 32*a^4*b^4))/(8*(a*b^2 + b^3)*(3*a*b^4 + b^5 + 3*a^2*b^3 + a^3*b^2)))*(-a*(a + b)^3
)^(1/2)*(2*a + 3*b))/(4*(3*a*b^4 + b^5 + 3*a^2*b^3 + a^3*b^2)))*(2*a + 3*b))/(4*(3*a*b^4 + b^5 + 3*a^2*b^3 + a
^3*b^2))))*(-a*(a + b)^3)^(1/2)*(2*a + 3*b)*1i)/(2*d*(3*a*b^4 + b^5 + 3*a^2*b^3 + a^3*b^2))

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