3.198 \(\int \frac{\sin ^3(c+d x)}{a-b \sin ^4(c+d x)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.117166, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3215, 1166, 205, 208} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt [4]{b} \cos (c+d x)}{\sqrt{\sqrt{a}+\sqrt{b}}}\right )}{2 b^{3/4} d \sqrt{\sqrt{a}+\sqrt{b}}}-\frac{\tan ^{-1}\left (\frac{\sqrt [4]{b} \cos (c+d x)}{\sqrt{\sqrt{a}-\sqrt{b}}}\right )}{2 b^{3/4} d \sqrt{\sqrt{a}-\sqrt{b}}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^3/(a - b*Sin[c + d*x]^4),x]

[Out]

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

Mathematica [C]  time = 0.170779, size = 285, normalized size = 2.48 \[ -\frac{i \text{RootSum}\left [-16 \text{$\#$1}^4 a+\text{$\#$1}^8 b-4 \text{$\#$1}^6 b+6 \text{$\#$1}^4 b-4 \text{$\#$1}^2 b+b\& ,\frac{-i \text{$\#$1}^6 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )+3 i \text{$\#$1}^4 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )-3 i \text{$\#$1}^2 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )+i \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )+2 \text{$\#$1}^6 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )-6 \text{$\#$1}^4 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )+6 \text{$\#$1}^2 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )-2 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )}{-8 \text{$\#$1}^3 a+\text{$\#$1}^7 b-3 \text{$\#$1}^5 b+3 \text{$\#$1}^3 b-\text{$\#$1} b}\& \right ]}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^3/(a - b*Sin[c + d*x]^4),x]

[Out]

((-I/8)*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) & ])/d

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Maple [A]  time = 0.085, size = 78, normalized size = 0.7 \begin{align*} -{\frac{1}{2\,d}\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{1}{2\,d}{\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(sin(d*x+c)^3/(a-b*sin(d*x+c)^4),x)

[Out]

-1/2/d/(((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)^(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*} -\int \frac{\sin \left (d x + c\right )^{3}}{b \sin \left (d x + c\right )^{4} - a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^3/(a-b*sin(d*x+c)^4),x, algorithm="maxima")

[Out]

-integrate(sin(d*x + c)^3/(b*sin(d*x + c)^4 - a), x)

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Fricas [B]  time = 2.41604, size = 1435, normalized size = 12.48 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^3/(a-b*sin(d*x+c)^4),x, algorithm="fricas")

[Out]

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

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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(sin(d*x+c)**3/(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(sin(d*x+c)^3/(a-b*sin(d*x+c)^4),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError