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

Optimal. Leaf size=125 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}-\sqrt {b}}}\right )}{2 \sqrt {a} \sqrt [4]{b} d \sqrt {\sqrt {a}-\sqrt {b}}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}+\sqrt {b}}}\right )}{2 \sqrt {a} \sqrt [4]{b} d \sqrt {\sqrt {a}+\sqrt {b}}} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3215, 1093, 205, 208} \[ -\frac {\tan ^{-1}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}-\sqrt {b}}}\right )}{2 \sqrt {a} \sqrt [4]{b} d \sqrt {\sqrt {a}-\sqrt {b}}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}+\sqrt {b}}}\right )}{2 \sqrt {a} \sqrt [4]{b} d \sqrt {\sqrt {a}+\sqrt {b}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1093

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

Rule 3215

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(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 - 2*b*ff^2*x^2 + b*ff^4*x^4
)^p, x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

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

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Mathematica [C]  time = 0.16, size = 183, normalized size = 1.46 \[ \frac {i \text {RootSum}\left [\text {$\#$1}^8 b-4 \text {$\#$1}^6 b-16 \text {$\#$1}^4 a+6 \text {$\#$1}^4 b-4 \text {$\#$1}^2 b+b\& ,\frac {2 \text {$\#$1}^3 \tan ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )+i \text {$\#$1} \log \left (\text {$\#$1}^2-2 \text {$\#$1} \cos (c+d x)+1\right )-i \text {$\#$1}^3 \log \left (\text {$\#$1}^2-2 \text {$\#$1} \cos (c+d x)+1\right )-2 \text {$\#$1} \tan ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )}{\text {$\#$1}^6 b-3 \text {$\#$1}^4 b-8 \text {$\#$1}^2 a+3 \text {$\#$1}^2 b-b}\& \right ]}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*RootSum[b - 4*b*#1^2 - 16*a*#1^4 + 6*b*#1^4 - 4*b*#1^6 + b*#1^8 & , (-2*ArcTan[Sin[c + d*x]/(Cos[c + d*
x] - #1)]*#1 + I*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1 + 2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^3 - I*Lo
g[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^3)/(-b - 8*a*#1^2 + 3*b*#1^2 - 3*b*#1^4 + b*#1^6) & ])/d

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fricas [B]  time = 0.56, size = 703, normalized size = 5.62 \[ -\frac {1}{4} \, \sqrt {-\frac {{\left (a^{2} - a b\right )} d^{2} \sqrt {\frac {1}{{\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3}\right )} d^{4}}} + 1}{{\left (a^{2} - a b\right )} d^{2}}} \log \left (-{\left ({\left (a^{2} b - a b^{2}\right )} d^{3} \sqrt {\frac {1}{{\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3}\right )} d^{4}}} - a d\right )} \sqrt {-\frac {{\left (a^{2} - a b\right )} d^{2} \sqrt {\frac {1}{{\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3}\right )} d^{4}}} + 1}{{\left (a^{2} - a b\right )} d^{2}}} + \cos \left (d x + c\right )\right ) + \frac {1}{4} \, \sqrt {-\frac {{\left (a^{2} - a b\right )} d^{2} \sqrt {\frac {1}{{\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3}\right )} d^{4}}} + 1}{{\left (a^{2} - a b\right )} d^{2}}} \log \left (-{\left ({\left (a^{2} b - a b^{2}\right )} d^{3} \sqrt {\frac {1}{{\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3}\right )} d^{4}}} - a d\right )} \sqrt {-\frac {{\left (a^{2} - a b\right )} d^{2} \sqrt {\frac {1}{{\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3}\right )} d^{4}}} + 1}{{\left (a^{2} - a b\right )} d^{2}}} - \cos \left (d x + c\right )\right ) + \frac {1}{4} \, \sqrt {\frac {{\left (a^{2} - a b\right )} d^{2} \sqrt {\frac {1}{{\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3}\right )} d^{4}}} - 1}{{\left (a^{2} - a b\right )} d^{2}}} \log \left (-{\left ({\left (a^{2} b - a b^{2}\right )} d^{3} \sqrt {\frac {1}{{\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3}\right )} d^{4}}} + a d\right )} \sqrt {\frac {{\left (a^{2} - a b\right )} d^{2} \sqrt {\frac {1}{{\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3}\right )} d^{4}}} - 1}{{\left (a^{2} - a b\right )} d^{2}}} + \cos \left (d x + c\right )\right ) - \frac {1}{4} \, \sqrt {\frac {{\left (a^{2} - a b\right )} d^{2} \sqrt {\frac {1}{{\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3}\right )} d^{4}}} - 1}{{\left (a^{2} - a b\right )} d^{2}}} \log \left (-{\left ({\left (a^{2} b - a b^{2}\right )} d^{3} \sqrt {\frac {1}{{\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3}\right )} d^{4}}} + a d\right )} \sqrt {\frac {{\left (a^{2} - a b\right )} d^{2} \sqrt {\frac {1}{{\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3}\right )} d^{4}}} - 1}{{\left (a^{2} - a b\right )} d^{2}}} - \cos \left (d x + c\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*sqrt(-((a^2 - a*b)*d^2*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) + 1)/((a^2 - a*b)*d^2))*log(-((a^2*b - a
*b^2)*d^3*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) - a*d)*sqrt(-((a^2 - a*b)*d^2*sqrt(1/((a^3*b - 2*a^2*b^2 +
a*b^3)*d^4)) + 1)/((a^2 - a*b)*d^2)) + cos(d*x + c)) + 1/4*sqrt(-((a^2 - a*b)*d^2*sqrt(1/((a^3*b - 2*a^2*b^2
+ a*b^3)*d^4)) + 1)/((a^2 - a*b)*d^2))*log(-((a^2*b - a*b^2)*d^3*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) - a
*d)*sqrt(-((a^2 - a*b)*d^2*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) + 1)/((a^2 - a*b)*d^2)) - cos(d*x + c)) +
1/4*sqrt(((a^2 - a*b)*d^2*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) - 1)/((a^2 - a*b)*d^2))*log(-((a^2*b - a*
b^2)*d^3*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) + a*d)*sqrt(((a^2 - a*b)*d^2*sqrt(1/((a^3*b - 2*a^2*b^2 + a
*b^3)*d^4)) - 1)/((a^2 - a*b)*d^2)) + cos(d*x + c)) - 1/4*sqrt(((a^2 - a*b)*d^2*sqrt(1/((a^3*b - 2*a^2*b^2 + a
*b^3)*d^4)) - 1)/((a^2 - a*b)*d^2))*log(-((a^2*b - a*b^2)*d^3*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) + a*d)
*sqrt(((a^2 - a*b)*d^2*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) - 1)/((a^2 - a*b)*d^2)) - cos(d*x + c))

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giac [B]  time = 0.76, size = 183, normalized size = 1.46 \[ -\frac {\sqrt {a b} \sqrt {-b^{2} - \sqrt {a b} b} \arctan \left (\frac {\cos \left (d x + c\right )}{d \sqrt {-\frac {b d^{2} + \sqrt {{\left (a - b\right )} b d^{4} + b^{2} d^{4}}}{b d^{4}}}}\right )}{2 \, {\left (a b + \sqrt {a b} a\right )} d {\left | b \right |}} + \frac {\sqrt {a b} \sqrt {-b^{2} + \sqrt {a b} b} \arctan \left (\frac {\cos \left (d x + c\right )}{d \sqrt {-\frac {b d^{2} - \sqrt {{\left (a - b\right )} b d^{4} + b^{2} d^{4}}}{b d^{4}}}}\right )}{2 \, {\left (a b - \sqrt {a b} a\right )} d {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(a*b)*sqrt(-b^2 - sqrt(a*b)*b)*arctan(cos(d*x + c)/(d*sqrt(-(b*d^2 + sqrt((a - b)*b*d^4 + b^2*d^4))/(
b*d^4))))/((a*b + sqrt(a*b)*a)*d*abs(b)) + 1/2*sqrt(a*b)*sqrt(-b^2 + sqrt(a*b)*b)*arctan(cos(d*x + c)/(d*sqrt(
-(b*d^2 - sqrt((a - b)*b*d^4 + b^2*d^4))/(b*d^4))))/((a*b - sqrt(a*b)*a)*d*abs(b))

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maple [A]  time = 0.36, size = 90, normalized size = 0.72 \[ -\frac {b \arctanh \left (\frac {\cos \left (d x +c \right ) b}{\sqrt {\left (\sqrt {a b}+b \right ) b}}\right )}{2 d \sqrt {a b}\, \sqrt {\left (\sqrt {a b}+b \right ) b}}-\frac {b \arctan \left (\frac {\cos \left (d x +c \right ) b}{\sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{2 d \sqrt {a b}\, \sqrt {\left (\sqrt {a b}-b \right ) b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 15.11, size = 361, normalized size = 2.89 \[ \frac {\ln \left (4\,a\,b^3\,\sqrt {\frac {1}{a\,b+\sqrt {a^3\,b}}}-4\,b^3\,\cos \left (c+d\,x\right )+\frac {4\,a\,b^4\,\cos \left (c+d\,x\right )}{a\,b+\sqrt {a^3\,b}}\right )\,\sqrt {-\frac {a\,b-\sqrt {a^3\,b}}{16\,\left (a^3\,b-a^2\,b^2\right )}}}{d}+\frac {\ln \left (4\,b^3\,\cos \left (c+d\,x\right )-4\,a\,b^3\,\sqrt {\frac {1}{a\,b-\sqrt {a^3\,b}}}-\frac {4\,a\,b^4\,\cos \left (c+d\,x\right )}{a\,b-\sqrt {a^3\,b}}\right )\,\sqrt {-\frac {a\,b+\sqrt {a^3\,b}}{16\,\left (a^3\,b-a^2\,b^2\right )}}}{d}-\frac {\ln \left (4\,b^3\,\cos \left (c+d\,x\right )+4\,a\,b^3\,\sqrt {\frac {1}{a\,b+\sqrt {a^3\,b}}}-\frac {4\,a\,b^4\,\cos \left (c+d\,x\right )}{a\,b+\sqrt {a^3\,b}}\right )\,\sqrt {\frac {1}{a\,b+\sqrt {a^3\,b}}}}{4\,d}-\frac {\ln \left (4\,b^3\,\cos \left (c+d\,x\right )+4\,a\,b^3\,\sqrt {\frac {1}{a\,b-\sqrt {a^3\,b}}}-\frac {4\,a\,b^4\,\cos \left (c+d\,x\right )}{a\,b-\sqrt {a^3\,b}}\right )\,\sqrt {\frac {1}{a\,b-\sqrt {a^3\,b}}}}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(4*a*b^3*(1/(a*b + (a^3*b)^(1/2)))^(1/2) - 4*b^3*cos(c + d*x) + (4*a*b^4*cos(c + d*x))/(a*b + (a^3*b)^(1/2
)))*(-(a*b - (a^3*b)^(1/2))/(16*(a^3*b - a^2*b^2)))^(1/2))/d + (log(4*b^3*cos(c + d*x) - 4*a*b^3*(1/(a*b - (a^
3*b)^(1/2)))^(1/2) - (4*a*b^4*cos(c + d*x))/(a*b - (a^3*b)^(1/2)))*(-(a*b + (a^3*b)^(1/2))/(16*(a^3*b - a^2*b^
2)))^(1/2))/d - (log(4*b^3*cos(c + d*x) + 4*a*b^3*(1/(a*b + (a^3*b)^(1/2)))^(1/2) - (4*a*b^4*cos(c + d*x))/(a*
b + (a^3*b)^(1/2)))*(1/(a*b + (a^3*b)^(1/2)))^(1/2))/(4*d) - (log(4*b^3*cos(c + d*x) + 4*a*b^3*(1/(a*b - (a^3*
b)^(1/2)))^(1/2) - (4*a*b^4*cos(c + d*x))/(a*b - (a^3*b)^(1/2)))*(1/(a*b - (a^3*b)^(1/2)))^(1/2))/(4*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(sin(d*x+c)/(a-b*sin(d*x+c)**4),x)

[Out]

Timed out

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