3.111 \(\int \frac{1}{(a+b \sin ^2(c+d x))^4} \, dx\)

Optimal. Leaf size=206 \[ \frac{(2 a+b) \left (8 a^2+8 a b+5 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{16 a^{7/2} d (a+b)^{7/2}}+\frac{b \left (44 a^2+44 a b+15 b^2\right ) \sin (c+d x) \cos (c+d x)}{48 a^3 d (a+b)^3 \left (a+b \sin ^2(c+d x)\right )}+\frac{5 b (2 a+b) \sin (c+d x) \cos (c+d x)}{24 a^2 d (a+b)^2 \left (a+b \sin ^2(c+d x)\right )^2}+\frac{b \sin (c+d x) \cos (c+d x)}{6 a d (a+b) \left (a+b \sin ^2(c+d x)\right )^3} \]

[Out]

((2*a + b)*(8*a^2 + 8*a*b + 5*b^2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(16*a^(7/2)*(a + b)^(7/2)*d) +
(b*Cos[c + d*x]*Sin[c + d*x])/(6*a*(a + b)*d*(a + b*Sin[c + d*x]^2)^3) + (5*b*(2*a + b)*Cos[c + d*x]*Sin[c + d
*x])/(24*a^2*(a + b)^2*d*(a + b*Sin[c + d*x]^2)^2) + (b*(44*a^2 + 44*a*b + 15*b^2)*Cos[c + d*x]*Sin[c + d*x])/
(48*a^3*(a + b)^3*d*(a + b*Sin[c + d*x]^2))

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Rubi [A]  time = 0.296316, antiderivative size = 206, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {3184, 3173, 12, 3181, 205} \[ \frac{(2 a+b) \left (8 a^2+8 a b+5 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{16 a^{7/2} d (a+b)^{7/2}}+\frac{b \left (44 a^2+44 a b+15 b^2\right ) \sin (c+d x) \cos (c+d x)}{48 a^3 d (a+b)^3 \left (a+b \sin ^2(c+d x)\right )}+\frac{5 b (2 a+b) \sin (c+d x) \cos (c+d x)}{24 a^2 d (a+b)^2 \left (a+b \sin ^2(c+d x)\right )^2}+\frac{b \sin (c+d x) \cos (c+d x)}{6 a d (a+b) \left (a+b \sin ^2(c+d x)\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x]^2)^(-4),x]

[Out]

((2*a + b)*(8*a^2 + 8*a*b + 5*b^2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(16*a^(7/2)*(a + b)^(7/2)*d) +
(b*Cos[c + d*x]*Sin[c + d*x])/(6*a*(a + b)*d*(a + b*Sin[c + d*x]^2)^3) + (5*b*(2*a + b)*Cos[c + d*x]*Sin[c + d
*x])/(24*a^2*(a + b)^2*d*(a + b*Sin[c + d*x]^2)^2) + (b*(44*a^2 + 44*a*b + 15*b^2)*Cos[c + d*x]*Sin[c + d*x])/
(48*a^3*(a + b)^3*d*(a + b*Sin[c + d*x]^2))

Rule 3184

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> -Simp[(b*Cos[e + f*x]*Sin[e + f*x]*(a + b*Sin[
e + f*x]^2)^(p + 1))/(2*a*f*(p + 1)*(a + b)), x] + Dist[1/(2*a*(p + 1)*(a + b)), Int[(a + b*Sin[e + f*x]^2)^(p
+ 1)*Simp[2*a*(p + 1) + b*(2*p + 3) - 2*b*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f}, x] && NeQ
[a + b, 0] && LtQ[p, -1]

Rule 3173

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Sim
p[((A*b - a*B)*Cos[e + f*x]*Sin[e + f*x]*(a + b*Sin[e + f*x]^2)^(p + 1))/(2*a*f*(a + b)*(p + 1)), x] - Dist[1/
(2*a*(a + b)*(p + 1)), Int[(a + b*Sin[e + f*x]^2)^(p + 1)*Simp[a*B - A*(2*a*(p + 1) + b*(2*p + 3)) + 2*(A*b -
a*B)*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && LtQ[p, -1] && NeQ[a + b, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3181

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(-1), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[1/(a + (a + b)*ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x]

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]

Rubi steps

\begin{align*} \int \frac{1}{\left (a+b \sin ^2(c+d x)\right )^4} \, dx &=\frac{b \cos (c+d x) \sin (c+d x)}{6 a (a+b) d \left (a+b \sin ^2(c+d x)\right )^3}-\frac{\int \frac{-6 a-5 b+4 b \sin ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx}{6 a (a+b)}\\ &=\frac{b \cos (c+d x) \sin (c+d x)}{6 a (a+b) d \left (a+b \sin ^2(c+d x)\right )^3}+\frac{5 b (2 a+b) \cos (c+d x) \sin (c+d x)}{24 a^2 (a+b)^2 d \left (a+b \sin ^2(c+d x)\right )^2}-\frac{\int \frac{-24 a^2-34 a b-15 b^2+10 b (2 a+b) \sin ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx}{24 a^2 (a+b)^2}\\ &=\frac{b \cos (c+d x) \sin (c+d x)}{6 a (a+b) d \left (a+b \sin ^2(c+d x)\right )^3}+\frac{5 b (2 a+b) \cos (c+d x) \sin (c+d x)}{24 a^2 (a+b)^2 d \left (a+b \sin ^2(c+d x)\right )^2}+\frac{b \left (44 a^2+44 a b+15 b^2\right ) \cos (c+d x) \sin (c+d x)}{48 a^3 (a+b)^3 d \left (a+b \sin ^2(c+d x)\right )}-\frac{\int -\frac{3 (2 a+b) \left (8 a^2+8 a b+5 b^2\right )}{a+b \sin ^2(c+d x)} \, dx}{48 a^3 (a+b)^3}\\ &=\frac{b \cos (c+d x) \sin (c+d x)}{6 a (a+b) d \left (a+b \sin ^2(c+d x)\right )^3}+\frac{5 b (2 a+b) \cos (c+d x) \sin (c+d x)}{24 a^2 (a+b)^2 d \left (a+b \sin ^2(c+d x)\right )^2}+\frac{b \left (44 a^2+44 a b+15 b^2\right ) \cos (c+d x) \sin (c+d x)}{48 a^3 (a+b)^3 d \left (a+b \sin ^2(c+d x)\right )}+\frac{\left ((2 a+b) \left (8 a^2+8 a b+5 b^2\right )\right ) \int \frac{1}{a+b \sin ^2(c+d x)} \, dx}{16 a^3 (a+b)^3}\\ &=\frac{b \cos (c+d x) \sin (c+d x)}{6 a (a+b) d \left (a+b \sin ^2(c+d x)\right )^3}+\frac{5 b (2 a+b) \cos (c+d x) \sin (c+d x)}{24 a^2 (a+b)^2 d \left (a+b \sin ^2(c+d x)\right )^2}+\frac{b \left (44 a^2+44 a b+15 b^2\right ) \cos (c+d x) \sin (c+d x)}{48 a^3 (a+b)^3 d \left (a+b \sin ^2(c+d x)\right )}+\frac{\left ((2 a+b) \left (8 a^2+8 a b+5 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{16 a^3 (a+b)^3 d}\\ &=\frac{(2 a+b) \left (8 a^2+8 a b+5 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{16 a^{7/2} (a+b)^{7/2} d}+\frac{b \cos (c+d x) \sin (c+d x)}{6 a (a+b) d \left (a+b \sin ^2(c+d x)\right )^3}+\frac{5 b (2 a+b) \cos (c+d x) \sin (c+d x)}{24 a^2 (a+b)^2 d \left (a+b \sin ^2(c+d x)\right )^2}+\frac{b \left (44 a^2+44 a b+15 b^2\right ) \cos (c+d x) \sin (c+d x)}{48 a^3 (a+b)^3 d \left (a+b \sin ^2(c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 1.39787, size = 201, normalized size = 0.98 \[ \frac{\frac{3 \left (24 a^2 b+16 a^3+18 a b^2+5 b^3\right ) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{(a+b)^{7/2}}+\frac{\sqrt{a} b \left (44 a^2+44 a b+15 b^2\right ) \sin (2 (c+d x))}{(a+b)^3 (2 a-b \cos (2 (c+d x))+b)}+\frac{32 a^{5/2} b \sin (2 (c+d x))}{(a+b) (2 a-b \cos (2 (c+d x))+b)^3}+\frac{20 a^{3/2} b (2 a+b) \sin (2 (c+d x))}{(a+b)^2 (2 a-b \cos (2 (c+d x))+b)^2}}{48 a^{7/2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*(16*a^3 + 24*a^2*b + 18*a*b^2 + 5*b^3)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(a + b)^(7/2) + (32*a^(
5/2)*b*Sin[2*(c + d*x)])/((a + b)*(2*a + b - b*Cos[2*(c + d*x)])^3) + (20*a^(3/2)*b*(2*a + b)*Sin[2*(c + d*x)]
)/((a + b)^2*(2*a + b - b*Cos[2*(c + d*x)])^2) + (Sqrt[a]*b*(44*a^2 + 44*a*b + 15*b^2)*Sin[2*(c + d*x)])/((a +
b)^3*(2*a + b - b*Cos[2*(c + d*x)])))/(48*a^(7/2)*d)

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Maple [B]  time = 0.099, size = 705, normalized size = 3.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+sin(d*x+c)^2*b)^4,x)

[Out]

3/2/d/(a*tan(d*x+c)^2+tan(d*x+c)^2*b+a)^3/a*b/(a+b)*tan(d*x+c)^5+9/8/d/(a*tan(d*x+c)^2+tan(d*x+c)^2*b+a)^3/a^2
*b^2/(a+b)*tan(d*x+c)^5+5/16/d/(a*tan(d*x+c)^2+tan(d*x+c)^2*b+a)^3/a^3*b^3/(a+b)*tan(d*x+c)^5+3/d/(a*tan(d*x+c
)^2+tan(d*x+c)^2*b+a)^3*b/(a^2+2*a*b+b^2)*tan(d*x+c)^3+3/d/(a*tan(d*x+c)^2+tan(d*x+c)^2*b+a)^3/a*b^2/(a^2+2*a*
b+b^2)*tan(d*x+c)^3+5/6/d/(a*tan(d*x+c)^2+tan(d*x+c)^2*b+a)^3/a^2*b^3/(a^2+2*a*b+b^2)*tan(d*x+c)^3+3/2/d/(a*ta
n(d*x+c)^2+tan(d*x+c)^2*b+a)^3*b*a/(a^3+3*a^2*b+3*a*b^2+b^3)*tan(d*x+c)+15/8/d/(a*tan(d*x+c)^2+tan(d*x+c)^2*b+
a)^3*b^2/(a^3+3*a^2*b+3*a*b^2+b^3)*tan(d*x+c)+11/16/d/(a*tan(d*x+c)^2+tan(d*x+c)^2*b+a)^3*b^3/a/(a^3+3*a^2*b+3
*a*b^2+b^3)*tan(d*x+c)+1/d/(a^3+3*a^2*b+3*a*b^2+b^3)/(a*(a+b))^(1/2)*arctan((a+b)*tan(d*x+c)/(a*(a+b))^(1/2))+
3/2/d/a/(a^3+3*a^2*b+3*a*b^2+b^3)/(a*(a+b))^(1/2)*arctan((a+b)*tan(d*x+c)/(a*(a+b))^(1/2))*b+9/8/d/a^2/(a^3+3*
a^2*b+3*a*b^2+b^3)/(a*(a+b))^(1/2)*arctan((a+b)*tan(d*x+c)/(a*(a+b))^(1/2))*b^2+5/16/d/a^3/(a^3+3*a^2*b+3*a*b^
2+b^3)/(a*(a+b))^(1/2)*arctan((a+b)*tan(d*x+c)/(a*(a+b))^(1/2))*b^3

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.42058, size = 3070, normalized size = 14.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(d*x+c)^2)^4,x, algorithm="fricas")

[Out]

[-1/192*(3*((16*a^3*b^3 + 24*a^2*b^4 + 18*a*b^5 + 5*b^6)*cos(d*x + c)^6 - 16*a^6 - 72*a^5*b - 138*a^4*b^2 - 14
7*a^3*b^3 - 93*a^2*b^4 - 33*a*b^5 - 5*b^6 - 3*(16*a^4*b^2 + 40*a^3*b^3 + 42*a^2*b^4 + 23*a*b^5 + 5*b^6)*cos(d*
x + c)^4 + 3*(16*a^5*b + 56*a^4*b^2 + 82*a^3*b^3 + 65*a^2*b^4 + 28*a*b^5 + 5*b^6)*cos(d*x + c)^2)*sqrt(-a^2 -
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 + b)*cos(d*x
+ c)^3 - (a + b)*cos(d*x + c))*sqrt(-a^2 - 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*((44*a^4*b^3 + 88*a^3*b^4 + 59*a^2*b^5 + 15*a*b^6)*cos(d*x +
c)^5 - 2*(54*a^5*b^2 + 157*a^4*b^3 + 167*a^3*b^4 + 79*a^2*b^5 + 15*a*b^6)*cos(d*x + c)^3 + 3*(24*a^6*b + 90*a
^5*b^2 + 131*a^4*b^3 + 93*a^3*b^4 + 33*a^2*b^5 + 5*a*b^6)*cos(d*x + c))*sin(d*x + c))/((a^8*b^3 + 4*a^7*b^4 +
6*a^6*b^5 + 4*a^5*b^6 + a^4*b^7)*d*cos(d*x + c)^6 - 3*(a^9*b^2 + 5*a^8*b^3 + 10*a^7*b^4 + 10*a^6*b^5 + 5*a^5*b
^6 + a^4*b^7)*d*cos(d*x + c)^4 + 3*(a^10*b + 6*a^9*b^2 + 15*a^8*b^3 + 20*a^7*b^4 + 15*a^6*b^5 + 6*a^5*b^6 + a^
4*b^7)*d*cos(d*x + c)^2 - (a^11 + 7*a^10*b + 21*a^9*b^2 + 35*a^8*b^3 + 35*a^7*b^4 + 21*a^6*b^5 + 7*a^5*b^6 + a
^4*b^7)*d), -1/96*(3*((16*a^3*b^3 + 24*a^2*b^4 + 18*a*b^5 + 5*b^6)*cos(d*x + c)^6 - 16*a^6 - 72*a^5*b - 138*a^
4*b^2 - 147*a^3*b^3 - 93*a^2*b^4 - 33*a*b^5 - 5*b^6 - 3*(16*a^4*b^2 + 40*a^3*b^3 + 42*a^2*b^4 + 23*a*b^5 + 5*b
^6)*cos(d*x + c)^4 + 3*(16*a^5*b + 56*a^4*b^2 + 82*a^3*b^3 + 65*a^2*b^4 + 28*a*b^5 + 5*b^6)*cos(d*x + c)^2)*sq
rt(a^2 + a*b)*arctan(1/2*((2*a + b)*cos(d*x + c)^2 - a - b)/(sqrt(a^2 + a*b)*cos(d*x + c)*sin(d*x + c))) + 2*(
(44*a^4*b^3 + 88*a^3*b^4 + 59*a^2*b^5 + 15*a*b^6)*cos(d*x + c)^5 - 2*(54*a^5*b^2 + 157*a^4*b^3 + 167*a^3*b^4 +
79*a^2*b^5 + 15*a*b^6)*cos(d*x + c)^3 + 3*(24*a^6*b + 90*a^5*b^2 + 131*a^4*b^3 + 93*a^3*b^4 + 33*a^2*b^5 + 5*
a*b^6)*cos(d*x + c))*sin(d*x + c))/((a^8*b^3 + 4*a^7*b^4 + 6*a^6*b^5 + 4*a^5*b^6 + a^4*b^7)*d*cos(d*x + c)^6 -
3*(a^9*b^2 + 5*a^8*b^3 + 10*a^7*b^4 + 10*a^6*b^5 + 5*a^5*b^6 + a^4*b^7)*d*cos(d*x + c)^4 + 3*(a^10*b + 6*a^9*
b^2 + 15*a^8*b^3 + 20*a^7*b^4 + 15*a^6*b^5 + 6*a^5*b^6 + a^4*b^7)*d*cos(d*x + c)^2 - (a^11 + 7*a^10*b + 21*a^9
*b^2 + 35*a^8*b^3 + 35*a^7*b^4 + 21*a^6*b^5 + 7*a^5*b^6 + a^4*b^7)*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(1/(a+b*sin(d*x+c)**2)**4,x)

[Out]

Timed out

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Giac [A]  time = 1.12537, size = 464, normalized size = 2.25 \begin{align*} \frac{\frac{3 \,{\left (16 \, a^{3} + 24 \, a^{2} b + 18 \, a b^{2} + 5 \, b^{3}\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^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}\right )} \sqrt{a^{2} + a b}} + \frac{72 \, a^{4} b \tan \left (d x + c\right )^{5} + 198 \, a^{3} b^{2} \tan \left (d x + c\right )^{5} + 195 \, a^{2} b^{3} \tan \left (d x + c\right )^{5} + 84 \, a b^{4} \tan \left (d x + c\right )^{5} + 15 \, b^{5} \tan \left (d x + c\right )^{5} + 144 \, a^{4} b \tan \left (d x + c\right )^{3} + 288 \, a^{3} b^{2} \tan \left (d x + c\right )^{3} + 184 \, a^{2} b^{3} \tan \left (d x + c\right )^{3} + 40 \, a b^{4} \tan \left (d x + c\right )^{3} + 72 \, a^{4} b \tan \left (d x + c\right ) + 90 \, a^{3} b^{2} \tan \left (d x + c\right ) + 33 \, a^{2} b^{3} \tan \left (d x + c\right )}{{\left (a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}\right )}{\left (a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right )^{2} + a\right )}^{3}}}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(d*x+c)^2)^4,x, algorithm="giac")

[Out]

1/48*(3*(16*a^3 + 24*a^2*b + 18*a*b^2 + 5*b^3)*(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^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*sqrt(a^2 + a*b)) + (72*a^4*b
*tan(d*x + c)^5 + 198*a^3*b^2*tan(d*x + c)^5 + 195*a^2*b^3*tan(d*x + c)^5 + 84*a*b^4*tan(d*x + c)^5 + 15*b^5*t
an(d*x + c)^5 + 144*a^4*b*tan(d*x + c)^3 + 288*a^3*b^2*tan(d*x + c)^3 + 184*a^2*b^3*tan(d*x + c)^3 + 40*a*b^4*
tan(d*x + c)^3 + 72*a^4*b*tan(d*x + c) + 90*a^3*b^2*tan(d*x + c) + 33*a^2*b^3*tan(d*x + c))/((a^6 + 3*a^5*b +
3*a^4*b^2 + a^3*b^3)*(a*tan(d*x + c)^2 + b*tan(d*x + c)^2 + a)^3))/d