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

Optimal. Leaf size=115 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} d \sqrt {\sqrt {a}-\sqrt {b}}}+\frac {\tan ^{-1}\left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} d \sqrt {\sqrt {a}+\sqrt {b}}} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3209, 1166, 205} \[ \frac {\tan ^{-1}\left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} d \sqrt {\sqrt {a}-\sqrt {b}}}+\frac {\tan ^{-1}\left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} d \sqrt {\sqrt {a}+\sqrt {b}}} \]

Antiderivative was successfully verified.

[In]

Int[(a - b*Sin[c + d*x]^4)^(-1),x]

[Out]

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

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 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rule 3209

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dis
t[ff/f, Subst[Int[(a + 2*a*ff^2*x^2 + (a + b)*ff^4*x^4)^p/(1 + ff^2*x^2)^(2*p + 1), x], x, Tan[e + f*x]/ff], x
]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[p]

Rubi steps

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

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Mathematica [A]  time = 0.26, size = 128, normalized size = 1.11 \[ \frac {\frac {\tan ^{-1}\left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tan (c+d x)}{\sqrt {\sqrt {a} \sqrt {b}+a}}\right )}{\sqrt {\sqrt {a} \sqrt {b}+a}}-\frac {\tanh ^{-1}\left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tan (c+d x)}{\sqrt {\sqrt {a} \sqrt {b}-a}}\right )}{\sqrt {\sqrt {a} \sqrt {b}-a}}}{2 \sqrt {a} d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - b*Sin[c + d*x]^4)^(-1),x]

[Out]

(ArcTan[((Sqrt[a] + Sqrt[b])*Tan[c + d*x])/Sqrt[a + Sqrt[a]*Sqrt[b]]]/Sqrt[a + Sqrt[a]*Sqrt[b]] - ArcTanh[((Sq
rt[a] - Sqrt[b])*Tan[c + d*x])/Sqrt[-a + Sqrt[a]*Sqrt[b]]]/Sqrt[-a + Sqrt[a]*Sqrt[b]])/(2*Sqrt[a]*d)

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fricas [B]  time = 0.60, size = 1079, normalized size = 9.38 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(-((a^2 - a*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) + 1)/((a^2 - a*b)*d^2))*log(1/4*b*cos(d*x +
c)^2 + 1/2*((a^4 - a^3*b)*d^3*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4))*cos(d*x + c)*sin(d*x + c) - a*b*d*cos(d
*x + c)*sin(d*x + c))*sqrt(-((a^2 - a*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) + 1)/((a^2 - a*b)*d^2)) -
1/4*(2*(a^3 - a^2*b)*d^2*cos(d*x + c)^2 - (a^3 - a^2*b)*d^2)*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) - 1/4*b)
- 1/8*sqrt(-((a^2 - a*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) + 1)/((a^2 - a*b)*d^2))*log(1/4*b*cos(d*
x + c)^2 - 1/2*((a^4 - a^3*b)*d^3*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4))*cos(d*x + c)*sin(d*x + c) - a*b*d*co
s(d*x + c)*sin(d*x + c))*sqrt(-((a^2 - a*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) + 1)/((a^2 - a*b)*d^2)
) - 1/4*(2*(a^3 - a^2*b)*d^2*cos(d*x + c)^2 - (a^3 - a^2*b)*d^2)*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) - 1/4
*b) + 1/8*sqrt(((a^2 - a*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) - 1)/((a^2 - a*b)*d^2))*log(-1/4*b*cos
(d*x + c)^2 + 1/2*((a^4 - a^3*b)*d^3*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4))*cos(d*x + c)*sin(d*x + c) + a*b*d
*cos(d*x + c)*sin(d*x + c))*sqrt(((a^2 - a*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) - 1)/((a^2 - a*b)*d^
2)) - 1/4*(2*(a^3 - a^2*b)*d^2*cos(d*x + c)^2 - (a^3 - a^2*b)*d^2)*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) + 1
/4*b) - 1/8*sqrt(((a^2 - a*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) - 1)/((a^2 - a*b)*d^2))*log(-1/4*b*c
os(d*x + c)^2 - 1/2*((a^4 - a^3*b)*d^3*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4))*cos(d*x + c)*sin(d*x + c) + a*b
*d*cos(d*x + c)*sin(d*x + c))*sqrt(((a^2 - a*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) - 1)/((a^2 - a*b)*
d^2)) - 1/4*(2*(a^3 - a^2*b)*d^2*cos(d*x + c)^2 - (a^3 - a^2*b)*d^2)*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) +
1/4*b)

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giac [B]  time = 0.42, size = 361, normalized size = 3.14 \[ \frac {\frac {{\left (3 \, \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} a^{2} - 6 \, \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} a b - \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \tan \left (d x + c\right )}{\sqrt {\frac {4 \, a + \sqrt {-16 \, {\left (a - b\right )} a + 16 \, a^{2}}}{a - b}}}\right )\right )} {\left | a - b \right |}}{3 \, a^{5} - 12 \, a^{4} b + 14 \, a^{3} b^{2} - 4 \, a^{2} b^{3} - a b^{4}} + \frac {{\left (3 \, \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} a^{2} - 6 \, \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} a b - \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \tan \left (d x + c\right )}{\sqrt {\frac {4 \, a - \sqrt {-16 \, {\left (a - b\right )} a + 16 \, a^{2}}}{a - b}}}\right )\right )} {\left | a - b \right |}}{3 \, a^{5} - 12 \, a^{4} b + 14 \, a^{3} b^{2} - 4 \, a^{2} b^{3} - a b^{4}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((3*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^2 - 6*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a*b - sqrt(a^2 - a*b +
sqrt(a*b)*(a - b))*b^2)*(pi*floor((d*x + c)/pi + 1/2) + arctan(2*tan(d*x + c)/sqrt((4*a + sqrt(-16*(a - b)*a
+ 16*a^2))/(a - b))))*abs(a - b)/(3*a^5 - 12*a^4*b + 14*a^3*b^2 - 4*a^2*b^3 - a*b^4) + (3*sqrt(a^2 - a*b - sqr
t(a*b)*(a - b))*a^2 - 6*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*a*b - sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*b^2)*(pi
*floor((d*x + c)/pi + 1/2) + arctan(2*tan(d*x + c)/sqrt((4*a - sqrt(-16*(a - b)*a + 16*a^2))/(a - b))))*abs(a
- b)/(3*a^5 - 12*a^4*b + 14*a^3*b^2 - 4*a^2*b^3 - a*b^4))/d

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maple [B]  time = 0.36, size = 492, normalized size = 4.28 \[ \frac {a \arctanh \left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 d \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}-\frac {a \arctanh \left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right ) b}{2 d \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}+\frac {a \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right ) b}{2 d \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}+\frac {a \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 d \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}-\frac {b \arctanh \left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 d \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}+\frac {\arctanh \left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right ) b^{2}}{2 d \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}-\frac {\arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right ) b^{2}}{2 d \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}-\frac {b \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 d \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {1}{b \sin \left (d x + c\right )^{4} - a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-integrate(1/(b*sin(d*x + c)^4 - a), x)

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mupad [B]  time = 14.99, size = 671, normalized size = 5.83 \[ \frac {\mathrm {atan}\left (\frac {a^3\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}}\,4{}\mathrm {i}+a^5\,\mathrm {tan}\left (c+d\,x\right )\,{\left (-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}\right )}^{3/2}\,64{}\mathrm {i}+a^3\,\mathrm {tan}\left (c+d\,x\right )\,{\left (-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}\right )}^{3/2}\,\sqrt {a^3\,b}\,64{}\mathrm {i}+a^2\,b\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}}\,4{}\mathrm {i}-a^4\,b\,\mathrm {tan}\left (c+d\,x\right )\,{\left (-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}\right )}^{3/2}\,64{}\mathrm {i}+a\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}}\,\sqrt {a^3\,b}\,4{}\mathrm {i}+b\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}}\,\sqrt {a^3\,b}\,4{}\mathrm {i}-a^2\,b\,\mathrm {tan}\left (c+d\,x\right )\,{\left (-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}\right )}^{3/2}\,\sqrt {a^3\,b}\,64{}\mathrm {i}}{a\,b+\sqrt {a^3\,b}}\right )\,\sqrt {-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}}\,2{}\mathrm {i}}{d}+\frac {\mathrm {atan}\left (\frac {a^3\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}}\,4{}\mathrm {i}+a^5\,\mathrm {tan}\left (c+d\,x\right )\,{\left (-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}\right )}^{3/2}\,64{}\mathrm {i}-a^3\,\mathrm {tan}\left (c+d\,x\right )\,{\left (-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}\right )}^{3/2}\,\sqrt {a^3\,b}\,64{}\mathrm {i}+a^2\,b\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}}\,4{}\mathrm {i}-a^4\,b\,\mathrm {tan}\left (c+d\,x\right )\,{\left (-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}\right )}^{3/2}\,64{}\mathrm {i}-a\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}}\,\sqrt {a^3\,b}\,4{}\mathrm {i}-b\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}}\,\sqrt {a^3\,b}\,4{}\mathrm {i}+a^2\,b\,\mathrm {tan}\left (c+d\,x\right )\,{\left (-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}\right )}^{3/2}\,\sqrt {a^3\,b}\,64{}\mathrm {i}}{a\,b-\sqrt {a^3\,b}}\right )\,\sqrt {-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}}\,2{}\mathrm {i}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a - b*sin(c + d*x)^4),x)

[Out]

(atan((a^3*tan(c + d*x)*(-1/(16*a^2 + 16*(a^3*b)^(1/2)))^(1/2)*4i + a^5*tan(c + d*x)*(-1/(16*a^2 + 16*(a^3*b)^
(1/2)))^(3/2)*64i + a^3*tan(c + d*x)*(-1/(16*a^2 + 16*(a^3*b)^(1/2)))^(3/2)*(a^3*b)^(1/2)*64i + a^2*b*tan(c +
d*x)*(-1/(16*a^2 + 16*(a^3*b)^(1/2)))^(1/2)*4i - a^4*b*tan(c + d*x)*(-1/(16*a^2 + 16*(a^3*b)^(1/2)))^(3/2)*64i
+ a*tan(c + d*x)*(-1/(16*a^2 + 16*(a^3*b)^(1/2)))^(1/2)*(a^3*b)^(1/2)*4i + b*tan(c + d*x)*(-1/(16*a^2 + 16*(a
^3*b)^(1/2)))^(1/2)*(a^3*b)^(1/2)*4i - a^2*b*tan(c + d*x)*(-1/(16*a^2 + 16*(a^3*b)^(1/2)))^(3/2)*(a^3*b)^(1/2)
*64i)/(a*b + (a^3*b)^(1/2)))*(-1/(16*a^2 + 16*(a^3*b)^(1/2)))^(1/2)*2i)/d + (atan((a^3*tan(c + d*x)*(-1/(16*a^
2 - 16*(a^3*b)^(1/2)))^(1/2)*4i + a^5*tan(c + d*x)*(-1/(16*a^2 - 16*(a^3*b)^(1/2)))^(3/2)*64i - a^3*tan(c + d*
x)*(-1/(16*a^2 - 16*(a^3*b)^(1/2)))^(3/2)*(a^3*b)^(1/2)*64i + a^2*b*tan(c + d*x)*(-1/(16*a^2 - 16*(a^3*b)^(1/2
)))^(1/2)*4i - a^4*b*tan(c + d*x)*(-1/(16*a^2 - 16*(a^3*b)^(1/2)))^(3/2)*64i - a*tan(c + d*x)*(-1/(16*a^2 - 16
*(a^3*b)^(1/2)))^(1/2)*(a^3*b)^(1/2)*4i - b*tan(c + d*x)*(-1/(16*a^2 - 16*(a^3*b)^(1/2)))^(1/2)*(a^3*b)^(1/2)*
4i + a^2*b*tan(c + d*x)*(-1/(16*a^2 - 16*(a^3*b)^(1/2)))^(3/2)*(a^3*b)^(1/2)*64i)/(a*b - (a^3*b)^(1/2)))*(-1/(
16*a^2 - 16*(a^3*b)^(1/2)))^(1/2)*2i)/d

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

[Out]

Timed out

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