3.200 \(\int \frac{\csc (c+d x)}{a-b \sin ^4(c+d x)} \, dx\)
Optimal. Leaf size=136 \[ -\frac{\sqrt [4]{b} \tan ^{-1}\left (\frac{\sqrt [4]{b} \cos (c+d x)}{\sqrt{\sqrt{a}-\sqrt{b}}}\right )}{2 a d \sqrt{\sqrt{a}-\sqrt{b}}}+\frac{\sqrt [4]{b} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \cos (c+d x)}{\sqrt{\sqrt{a}+\sqrt{b}}}\right )}{2 a d \sqrt{\sqrt{a}+\sqrt{b}}}-\frac{\tanh ^{-1}(\cos (c+d x))}{a d} \]
[Out]
-(b^(1/4)*ArcTan[(b^(1/4)*Cos[c + d*x])/Sqrt[Sqrt[a] - Sqrt[b]]])/(2*a*Sqrt[Sqrt[a] - Sqrt[b]]*d) - ArcTanh[Co
s[c + d*x]]/(a*d) + (b^(1/4)*ArcTanh[(b^(1/4)*Cos[c + d*x])/Sqrt[Sqrt[a] + Sqrt[b]]])/(2*a*Sqrt[Sqrt[a] + Sqrt
[b]]*d)
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Rubi [A] time = 0.178445, antiderivative size = 136, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used =
{3215, 1170, 207, 1166, 205, 208} \[ -\frac{\sqrt [4]{b} \tan ^{-1}\left (\frac{\sqrt [4]{b} \cos (c+d x)}{\sqrt{\sqrt{a}-\sqrt{b}}}\right )}{2 a d \sqrt{\sqrt{a}-\sqrt{b}}}+\frac{\sqrt [4]{b} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \cos (c+d x)}{\sqrt{\sqrt{a}+\sqrt{b}}}\right )}{2 a d \sqrt{\sqrt{a}+\sqrt{b}}}-\frac{\tanh ^{-1}(\cos (c+d x))}{a d} \]
Antiderivative was successfully verified.
[In]
Int[Csc[c + d*x]/(a - b*Sin[c + d*x]^4),x]
[Out]
-(b^(1/4)*ArcTan[(b^(1/4)*Cos[c + d*x])/Sqrt[Sqrt[a] - Sqrt[b]]])/(2*a*Sqrt[Sqrt[a] - Sqrt[b]]*d) - ArcTanh[Co
s[c + d*x]]/(a*d) + (b^(1/4)*ArcTanh[(b^(1/4)*Cos[c + d*x])/Sqrt[Sqrt[a] + Sqrt[b]]])/(2*a*Sqrt[Sqrt[a] + Sqrt
[b]]*d)
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]
Rule 1170
Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[ExpandIntegrand[(d + e*x
^2)^q/(a + b*x^2 + c*x^4), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a
*e^2, 0] && IntegerQ[q]
Rule 207
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
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 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]
Rubi steps
\begin{align*} \int \frac{\csc (c+d x)}{a-b \sin ^4(c+d x)} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right ) \left (a-b+2 b x^2-b x^4\right )} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int \left (-\frac{1}{a \left (-1+x^2\right )}+\frac{b-b x^2}{a \left (a-b+2 b x^2-b x^4\right )}\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\cos (c+d x)\right )}{a d}-\frac{\operatorname{Subst}\left (\int \frac{b-b x^2}{a-b+2 b x^2-b x^4} \, dx,x,\cos (c+d x)\right )}{a d}\\ &=-\frac{\tanh ^{-1}(\cos (c+d x))}{a d}+\frac{b \operatorname{Subst}\left (\int \frac{1}{-\sqrt{a} \sqrt{b}+b-b x^2} \, dx,x,\cos (c+d x)\right )}{2 a d}+\frac{b \operatorname{Subst}\left (\int \frac{1}{\sqrt{a} \sqrt{b}+b-b x^2} \, dx,x,\cos (c+d x)\right )}{2 a d}\\ &=-\frac{\sqrt [4]{b} \tan ^{-1}\left (\frac{\sqrt [4]{b} \cos (c+d x)}{\sqrt{\sqrt{a}-\sqrt{b}}}\right )}{2 a \sqrt{\sqrt{a}-\sqrt{b}} d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a d}+\frac{\sqrt [4]{b} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \cos (c+d x)}{\sqrt{\sqrt{a}+\sqrt{b}}}\right )}{2 a \sqrt{\sqrt{a}+\sqrt{b}} d}\\ \end{align*}
Mathematica [C] time = 0.250389, size = 318, normalized size = 2.34 \[ -\frac{i b \text{RootSum}\left [-16 \text{$\#$1}^4 a+\text{$\#$1}^8 b-4 \text{$\#$1}^6 b+6 \text{$\#$1}^4 b-4 \text{$\#$1}^2 b+b\& ,\frac{-i \text{$\#$1}^6 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )+3 i \text{$\#$1}^4 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )-3 i \text{$\#$1}^2 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )+i \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )+2 \text{$\#$1}^6 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )-6 \text{$\#$1}^4 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )+6 \text{$\#$1}^2 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )-2 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )}{-8 \text{$\#$1}^3 a+\text{$\#$1}^7 b-3 \text{$\#$1}^5 b+3 \text{$\#$1}^3 b-\text{$\#$1} b}\& \right ]-8 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+8 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{8 a d} \]
Antiderivative was successfully verified.
[In]
Integrate[Csc[c + d*x]/(a - b*Sin[c + d*x]^4),x]
[Out]
-(8*Log[Cos[(c + d*x)/2]] - 8*Log[Sin[(c + d*x)/2]] + I*b*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)] + I*Log[1 - 2*Cos[c + d*x]*#1 + #1^2] + 6*ArcTan[
Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^2 - (3*I)*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^2 - 6*ArcTan[Sin[c + d*x]/
(Cos[c + d*x] - #1)]*#1^4 + (3*I)*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^4 + 2*ArcTan[Sin[c + d*x]/(Cos[c + d*x]
- #1)]*#1^6 - I*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^6)/(-(b*#1) - 8*a*#1^3 + 3*b*#1^3 - 3*b*#1^5 + b*#1^7) &
])/(8*a*d)
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Maple [A] time = 0.121, size = 120, normalized size = 0.9 \begin{align*}{\frac{\ln \left ( -1+\cos \left ( dx+c \right ) \right ) }{2\,da}}-{\frac{\ln \left ( 1+\cos \left ( dx+c \right ) \right ) }{2\,da}}-{\frac{b}{2\,da}\arctan \left ({b\cos \left ( dx+c \right ){\frac{1}{\sqrt{ \left ( \sqrt{ab}-b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( \sqrt{ab}-b \right ) b}}}}+{\frac{b}{2\,da}{\it Artanh} \left ({b\cos \left ( dx+c \right ){\frac{1}{\sqrt{ \left ( \sqrt{ab}+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( \sqrt{ab}+b \right ) b}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(csc(d*x+c)/(a-b*sin(d*x+c)^4),x)
[Out]
1/2/d/a*ln(-1+cos(d*x+c))-1/2/d/a*ln(1+cos(d*x+c))-1/2/d/a*b/(((a*b)^(1/2)-b)*b)^(1/2)*arctan(cos(d*x+c)*b/(((
a*b)^(1/2)-b)*b)^(1/2))+1/2/d/a*b/(((a*b)^(1/2)+b)*b)^(1/2)*arctanh(cos(d*x+c)*b/(((a*b)^(1/2)+b)*b)^(1/2))
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(d*x+c)/(a-b*sin(d*x+c)^4),x, algorithm="maxima")
[Out]
-1/2*(2*a*d*integrate(-2*(12*b^2*cos(3*d*x + 3*c)*sin(2*d*x + 2*c) - 4*b^2*cos(d*x + c)*sin(2*d*x + 2*c) + 4*b
^2*cos(2*d*x + 2*c)*sin(d*x + c) - b^2*sin(d*x + c) + (b^2*sin(7*d*x + 7*c) - 3*b^2*sin(5*d*x + 5*c) + 3*b^2*s
in(3*d*x + 3*c) - b^2*sin(d*x + c))*cos(8*d*x + 8*c) + 2*(2*b^2*sin(6*d*x + 6*c) + 2*b^2*sin(2*d*x + 2*c) + (8
*a*b - 3*b^2)*sin(4*d*x + 4*c))*cos(7*d*x + 7*c) + 4*(3*b^2*sin(5*d*x + 5*c) - 3*b^2*sin(3*d*x + 3*c) + b^2*si
n(d*x + c))*cos(6*d*x + 6*c) - 6*(2*b^2*sin(2*d*x + 2*c) + (8*a*b - 3*b^2)*sin(4*d*x + 4*c))*cos(5*d*x + 5*c)
- 2*(3*(8*a*b - 3*b^2)*sin(3*d*x + 3*c) - (8*a*b - 3*b^2)*sin(d*x + c))*cos(4*d*x + 4*c) - (b^2*cos(7*d*x + 7*
c) - 3*b^2*cos(5*d*x + 5*c) + 3*b^2*cos(3*d*x + 3*c) - b^2*cos(d*x + c))*sin(8*d*x + 8*c) - (4*b^2*cos(6*d*x +
6*c) + 4*b^2*cos(2*d*x + 2*c) - b^2 + 2*(8*a*b - 3*b^2)*cos(4*d*x + 4*c))*sin(7*d*x + 7*c) - 4*(3*b^2*cos(5*d
*x + 5*c) - 3*b^2*cos(3*d*x + 3*c) + b^2*cos(d*x + c))*sin(6*d*x + 6*c) + 3*(4*b^2*cos(2*d*x + 2*c) - b^2 + 2*
(8*a*b - 3*b^2)*cos(4*d*x + 4*c))*sin(5*d*x + 5*c) + 2*(3*(8*a*b - 3*b^2)*cos(3*d*x + 3*c) - (8*a*b - 3*b^2)*c
os(d*x + c))*sin(4*d*x + 4*c) - 3*(4*b^2*cos(2*d*x + 2*c) - b^2)*sin(3*d*x + 3*c))/(a*b^2*cos(8*d*x + 8*c)^2 +
16*a*b^2*cos(6*d*x + 6*c)^2 + 16*a*b^2*cos(2*d*x + 2*c)^2 + a*b^2*sin(8*d*x + 8*c)^2 + 16*a*b^2*sin(6*d*x + 6
*c)^2 + 16*a*b^2*sin(2*d*x + 2*c)^2 - 8*a*b^2*cos(2*d*x + 2*c) + a*b^2 + 4*(64*a^3 - 48*a^2*b + 9*a*b^2)*cos(4
*d*x + 4*c)^2 + 4*(64*a^3 - 48*a^2*b + 9*a*b^2)*sin(4*d*x + 4*c)^2 + 16*(8*a^2*b - 3*a*b^2)*sin(4*d*x + 4*c)*s
in(2*d*x + 2*c) - 2*(4*a*b^2*cos(6*d*x + 6*c) + 4*a*b^2*cos(2*d*x + 2*c) - a*b^2 + 2*(8*a^2*b - 3*a*b^2)*cos(4
*d*x + 4*c))*cos(8*d*x + 8*c) + 8*(4*a*b^2*cos(2*d*x + 2*c) - a*b^2 + 2*(8*a^2*b - 3*a*b^2)*cos(4*d*x + 4*c))*
cos(6*d*x + 6*c) - 4*(8*a^2*b - 3*a*b^2 - 4*(8*a^2*b - 3*a*b^2)*cos(2*d*x + 2*c))*cos(4*d*x + 4*c) - 4*(2*a*b^
2*sin(6*d*x + 6*c) + 2*a*b^2*sin(2*d*x + 2*c) + (8*a^2*b - 3*a*b^2)*sin(4*d*x + 4*c))*sin(8*d*x + 8*c) + 16*(2
*a*b^2*sin(2*d*x + 2*c) + (8*a^2*b - 3*a*b^2)*sin(4*d*x + 4*c))*sin(6*d*x + 6*c)), x) + log(cos(d*x)^2 + 2*cos
(d*x)*cos(c) + cos(c)^2 + sin(d*x)^2 - 2*sin(d*x)*sin(c) + sin(c)^2) - log(cos(d*x)^2 - 2*cos(d*x)*cos(c) + co
s(c)^2 + sin(d*x)^2 + 2*sin(d*x)*sin(c) + sin(c)^2))/(a*d)
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Fricas [B] time = 3.18046, size = 1601, normalized size = 11.77 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(d*x+c)/(a-b*sin(d*x+c)^4),x, algorithm="fricas")
[Out]
1/4*(a*d*sqrt(-((a^3 - a^2*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) + b)/((a^3 - a^2*b)*d^2))*log(b*cos(
d*x + c) - ((a^4 - a^3*b)*d^3*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) - a*b*d)*sqrt(-((a^3 - a^2*b)*d^2*sqrt(b
/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) + b)/((a^3 - a^2*b)*d^2))) - a*d*sqrt(((a^3 - a^2*b)*d^2*sqrt(b/((a^5 - 2*a^
4*b + a^3*b^2)*d^4)) - b)/((a^3 - a^2*b)*d^2))*log(b*cos(d*x + c) - ((a^4 - a^3*b)*d^3*sqrt(b/((a^5 - 2*a^4*b
+ a^3*b^2)*d^4)) + a*b*d)*sqrt(((a^3 - a^2*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) - b)/((a^3 - a^2*b)*
d^2))) - a*d*sqrt(-((a^3 - a^2*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) + b)/((a^3 - a^2*b)*d^2))*log(-b
*cos(d*x + c) - ((a^4 - a^3*b)*d^3*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) - a*b*d)*sqrt(-((a^3 - a^2*b)*d^2*s
qrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) + b)/((a^3 - a^2*b)*d^2))) + a*d*sqrt(((a^3 - a^2*b)*d^2*sqrt(b/((a^5 -
2*a^4*b + a^3*b^2)*d^4)) - b)/((a^3 - a^2*b)*d^2))*log(-b*cos(d*x + c) - ((a^4 - a^3*b)*d^3*sqrt(b/((a^5 - 2*
a^4*b + a^3*b^2)*d^4)) + a*b*d)*sqrt(((a^3 - a^2*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) - b)/((a^3 - a
^2*b)*d^2))) - 2*log(1/2*cos(d*x + c) + 1/2) + 2*log(-1/2*cos(d*x + c) + 1/2))/(a*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(csc(d*x+c)/(a-b*sin(d*x+c)**4),x)
[Out]
Timed out
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(d*x+c)/(a-b*sin(d*x+c)^4),x, algorithm="giac")
[Out]
Exception raised: NotImplementedError