\(\int \frac {\sin ^7(c+d x)}{a-b \sin ^4(c+d x)} \, dx\) [145]

Optimal result

Integrand size = 24, antiderivative size = 148 \[ \int \frac {\sin ^7(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\frac {a \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}-\sqrt {b}}}\right )}{2 \sqrt {\sqrt {a}-\sqrt {b}} b^{7/4} d}+\frac {a \text {arctanh}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}+\sqrt {b}}}\right )}{2 \sqrt {\sqrt {a}+\sqrt {b}} b^{7/4} d}+\frac {\cos (c+d x)}{b d}-\frac {\cos ^3(c+d x)}{3 b d} \] Output:

-1/2*a*arctan(b^(1/4)*cos(d*x+c)/(a^(1/2)-b^(1/2))^(1/2))/(a^(1/2)-b^(1/2) 
)^(1/2)/b^(7/4)/d+1/2*a*arctanh(b^(1/4)*cos(d*x+c)/(a^(1/2)+b^(1/2))^(1/2) 
)/(a^(1/2)+b^(1/2))^(1/2)/b^(7/4)/d+cos(d*x+c)/b/d-1/3*cos(d*x+c)^3/b/d
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 2.41 (sec) , antiderivative size = 310, normalized size of antiderivative = 2.09 \[ \int \frac {\sin ^7(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {18 \cos (c+d x)-2 \cos (3 (c+d x))-3 i a \text {RootSum}\left [b-4 b \text {$\#$1}^2-16 a \text {$\#$1}^4+6 b \text {$\#$1}^4-4 b \text {$\#$1}^6+b \text {$\#$1}^8\&,\frac {-2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )+i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right )+6 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^2-3 i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^2-6 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^4+3 i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^4+2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^6-i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^6}{-b \text {$\#$1}-8 a \text {$\#$1}^3+3 b \text {$\#$1}^3-3 b \text {$\#$1}^5+b \text {$\#$1}^7}\&\right ]}{24 b d} \] Input:

Integrate[Sin[c + d*x]^7/(a - b*Sin[c + d*x]^4),x]
 

Output:

(18*Cos[c + d*x] - 2*Cos[3*(c + d*x)] - (3*I)*a*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) & ])/(24*b*d)
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 3694, 1484, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sin ^7(c+d x)}{a-b \sin ^4(c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (c+d x)^7}{a-b \sin (c+d x)^4}dx\)

\(\Big \downarrow \) 3694

\(\displaystyle -\frac {\int \frac {\left (1-\cos ^2(c+d x)\right )^3}{-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)+a-b}d\cos (c+d x)}{d}\)

\(\Big \downarrow \) 1484

\(\displaystyle -\frac {\int \left (\frac {\cos ^2(c+d x)}{b}-\frac {1}{b}+\frac {a-a \cos ^2(c+d x)}{b \left (-b \cos ^4(c+d x)+2 b \cos ^2(c+d x)+a-b\right )}\right )d\cos (c+d x)}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\frac {a \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}-\sqrt {b}}}\right )}{2 b^{7/4} \sqrt {\sqrt {a}-\sqrt {b}}}-\frac {a \text {arctanh}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}+\sqrt {b}}}\right )}{2 b^{7/4} \sqrt {\sqrt {a}+\sqrt {b}}}+\frac {\cos ^3(c+d x)}{3 b}-\frac {\cos (c+d x)}{b}}{d}\)

Input:

Int[Sin[c + d*x]^7/(a - b*Sin[c + d*x]^4),x]
 

Output:

-(((a*ArcTan[(b^(1/4)*Cos[c + d*x])/Sqrt[Sqrt[a] - Sqrt[b]]])/(2*Sqrt[Sqrt 
[a] - Sqrt[b]]*b^(7/4)) - (a*ArcTanh[(b^(1/4)*Cos[c + d*x])/Sqrt[Sqrt[a] + 
 Sqrt[b]]])/(2*Sqrt[Sqrt[a] + Sqrt[b]]*b^(7/4)) - Cos[c + d*x]/b + Cos[c + 
 d*x]^3/(3*b))/d)
 

Defintions of rubi rules used

rule 1484

Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symb 
ol] :> Int[ExpandIntegrand[(d + e*x^2)^q/(a + b*x^2 + c*x^4), x], x] /; Fre 
eQ[{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 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3694

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^ 
(p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-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]
 

Maple [A] (verified)

Time = 2.48 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(\frac {-\frac {\frac {\cos \left (d x +c \right )^{3}}{3}-\cos \left (d x +c \right )}{b}-a \left (\frac {\arctan \left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{2 b \sqrt {\left (\sqrt {a b}-b \right ) b}}-\frac {\operatorname {arctanh}\left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+b \right ) b}}\right )}{2 b \sqrt {\left (\sqrt {a b}+b \right ) b}}\right )}{d}\)	\(110\)
default	\(\frac {-\frac {\frac {\cos \left (d x +c \right )^{3}}{3}-\cos \left (d x +c \right )}{b}-a \left (\frac {\arctan \left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{2 b \sqrt {\left (\sqrt {a b}-b \right ) b}}-\frac {\operatorname {arctanh}\left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+b \right ) b}}\right )}{2 b \sqrt {\left (\sqrt {a b}+b \right ) b}}\right )}{d}\)	\(110\)
risch	\(\frac {3 \,{\mathrm e}^{i \left (d x +c \right )}}{8 b d}+\frac {3 \,{\mathrm e}^{-i \left (d x +c \right )}}{8 b d}+\frac {i \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (a \,b^{7} d^{4}-b^{8} d^{4}\right ) \textit {\_Z}^{4}-2048 a^{2} b^{4} d^{2} \textit {\_Z}^{2}-1048576 a^{4}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\left (\left (-\frac {i b^{5} d^{3}}{16384 a^{2}}+\frac {i b^{6} d^{3}}{16384 a^{3}}\right ) \textit {\_R}^{3}+\frac {i d \,b^{2} \textit {\_R}}{8 a}\right ) {\mathrm e}^{i \left (d x +c \right )}+1\right )\right )}{128}-\frac {\cos \left (3 d x +3 c \right )}{12 b d}\)	\(166\)

Input:

int(sin(d*x+c)^7/(a-b*sin(d*x+c)^4),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 849 vs. \(2 (110) = 220\).

Time = 0.14 (sec) , antiderivative size = 849, normalized size of antiderivative = 5.74 \[ \int \frac {\sin ^7(c+d x)}{a-b \sin ^4(c+d x)} \, dx =\text {Too large to display} \] Input:

integrate(sin(d*x+c)^7/(a-b*sin(d*x+c)^4),x, algorithm="fricas")
 

Output:

1/12*(3*b*d*sqrt(-((a*b^3 - b^4)*d^2*sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b^9)*d 
^4)) + a^2)/((a*b^3 - b^4)*d^2))*log(a^3*cos(d*x + c) + (a^2*b^2*d - (a*b^ 
5 - b^6)*d^3*sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b^9)*d^4)))*sqrt(-((a*b^3 - b^ 
4)*d^2*sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b^9)*d^4)) + a^2)/((a*b^3 - b^4)*d^2 
))) - 3*b*d*sqrt(((a*b^3 - b^4)*d^2*sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b^9)*d^ 
4)) - a^2)/((a*b^3 - b^4)*d^2))*log(a^3*cos(d*x + c) - (a^2*b^2*d + (a*b^5 
 - b^6)*d^3*sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b^9)*d^4)))*sqrt(((a*b^3 - b^4) 
*d^2*sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b^9)*d^4)) - a^2)/((a*b^3 - b^4)*d^2)) 
) - 3*b*d*sqrt(-((a*b^3 - b^4)*d^2*sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b^9)*d^4 
)) + a^2)/((a*b^3 - b^4)*d^2))*log(-a^3*cos(d*x + c) + (a^2*b^2*d - (a*b^5 
 - b^6)*d^3*sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b^9)*d^4)))*sqrt(-((a*b^3 - b^4 
)*d^2*sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b^9)*d^4)) + a^2)/((a*b^3 - b^4)*d^2) 
)) + 3*b*d*sqrt(((a*b^3 - b^4)*d^2*sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b^9)*d^4 
)) - a^2)/((a*b^3 - b^4)*d^2))*log(-a^3*cos(d*x + c) - (a^2*b^2*d + (a*b^5 
 - b^6)*d^3*sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b^9)*d^4)))*sqrt(((a*b^3 - b^4) 
*d^2*sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b^9)*d^4)) - a^2)/((a*b^3 - b^4)*d^2)) 
) - 4*cos(d*x + c)^3 + 12*cos(d*x + c))/(b*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^7(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Timed out} \] Input:

integrate(sin(d*x+c)**7/(a-b*sin(d*x+c)**4),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\sin ^7(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\int { -\frac {\sin \left (d x + c\right )^{7}}{b \sin \left (d x + c\right )^{4} - a} \,d x } \] Input:

integrate(sin(d*x+c)^7/(a-b*sin(d*x+c)^4),x, algorithm="maxima")
 

Output:

-1/12*(12*b*d*integrate(-2*(12*a*b*cos(3*d*x + 3*c)*sin(2*d*x + 2*c) - 4*a 
*b*cos(d*x + c)*sin(2*d*x + 2*c) + 4*a*b*cos(2*d*x + 2*c)*sin(d*x + c) - a 
*b*sin(d*x + c) + (a*b*sin(7*d*x + 7*c) - 3*a*b*sin(5*d*x + 5*c) + 3*a*b*s 
in(3*d*x + 3*c) - a*b*sin(d*x + c))*cos(8*d*x + 8*c) + 2*(2*a*b*sin(6*d*x 
+ 6*c) + 2*a*b*sin(2*d*x + 2*c) + (8*a^2 - 3*a*b)*sin(4*d*x + 4*c))*cos(7* 
d*x + 7*c) + 4*(3*a*b*sin(5*d*x + 5*c) - 3*a*b*sin(3*d*x + 3*c) + a*b*sin( 
d*x + c))*cos(6*d*x + 6*c) - 6*(2*a*b*sin(2*d*x + 2*c) + (8*a^2 - 3*a*b)*s 
in(4*d*x + 4*c))*cos(5*d*x + 5*c) - 2*(3*(8*a^2 - 3*a*b)*sin(3*d*x + 3*c) 
- (8*a^2 - 3*a*b)*sin(d*x + c))*cos(4*d*x + 4*c) - (a*b*cos(7*d*x + 7*c) - 
 3*a*b*cos(5*d*x + 5*c) + 3*a*b*cos(3*d*x + 3*c) - a*b*cos(d*x + c))*sin(8 
*d*x + 8*c) - (4*a*b*cos(6*d*x + 6*c) + 4*a*b*cos(2*d*x + 2*c) - a*b + 2*( 
8*a^2 - 3*a*b)*cos(4*d*x + 4*c))*sin(7*d*x + 7*c) - 4*(3*a*b*cos(5*d*x + 5 
*c) - 3*a*b*cos(3*d*x + 3*c) + a*b*cos(d*x + c))*sin(6*d*x + 6*c) + 3*(4*a 
*b*cos(2*d*x + 2*c) - a*b + 2*(8*a^2 - 3*a*b)*cos(4*d*x + 4*c))*sin(5*d*x 
+ 5*c) + 2*(3*(8*a^2 - 3*a*b)*cos(3*d*x + 3*c) - (8*a^2 - 3*a*b)*cos(d*x + 
 c))*sin(4*d*x + 4*c) - 3*(4*a*b*cos(2*d*x + 2*c) - a*b)*sin(3*d*x + 3*c)) 
/(b^3*cos(8*d*x + 8*c)^2 + 16*b^3*cos(6*d*x + 6*c)^2 + 16*b^3*cos(2*d*x + 
2*c)^2 + b^3*sin(8*d*x + 8*c)^2 + 16*b^3*sin(6*d*x + 6*c)^2 + 16*b^3*sin(2 
*d*x + 2*c)^2 - 8*b^3*cos(2*d*x + 2*c) + b^3 + 4*(64*a^2*b - 48*a*b^2 + 9* 
b^3)*cos(4*d*x + 4*c)^2 + 4*(64*a^2*b - 48*a*b^2 + 9*b^3)*sin(4*d*x + 4...
 

Giac [F]

\[ \int \frac {\sin ^7(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\int { -\frac {\sin \left (d x + c\right )^{7}}{b \sin \left (d x + c\right )^{4} - a} \,d x } \] Input:

integrate(sin(d*x+c)^7/(a-b*sin(d*x+c)^4),x, algorithm="giac")
 

Output:

sage0*x
 

Mupad [B] (verification not implemented)

Time = 38.26 (sec) , antiderivative size = 1119, normalized size of antiderivative = 7.56 \[ \int \frac {\sin ^7(c+d x)}{a-b \sin ^4(c+d x)} \, dx =\text {Too large to display} \] Input:

int(sin(c + d*x)^7/(a - b*sin(c + d*x)^4),x)
 

Output:

cos(c + d*x)/(b*d) - cos(c + d*x)^3/(3*b*d) + (atan((a^3*b^8*cos(c + d*x)* 
(- (a^5*b^7)^(1/2)/(16*(a*b^7 - b^8)) - (a^2*b^4)/(16*(a*b^7 - b^8)))^(1/2 
)*8i)/(2*a^4*b^6 - 2*a^5*b^5 + (2*a^4*b^14)/(a*b^7 - b^8) - (2*a^5*b^13)/( 
a*b^7 - b^8) + (2*a^2*b^10*(a^5*b^7)^(1/2))/(a*b^7 - b^8) - (2*a^3*b^9*(a^ 
5*b^7)^(1/2))/(a*b^7 - b^8)) - (a^3*cos(c + d*x)*(- (a^5*b^7)^(1/2)/(16*(a 
*b^7 - b^8)) - (a^2*b^4)/(16*(a*b^7 - b^8)))^(1/2)*8i)/((2*a^4)/b^2 + (2*a 
^4*b^6)/(a*b^7 - b^8) + (2*a^2*b^2*(a^5*b^7)^(1/2))/(a*b^7 - b^8)) + (a*b^ 
4*cos(c + d*x)*(- (a^5*b^7)^(1/2)/(16*(a*b^7 - b^8)) - (a^2*b^4)/(16*(a*b^ 
7 - b^8)))^(1/2)*(a^5*b^7)^(1/2)*8i)/(2*a^4*b^6 - 2*a^5*b^5 + (2*a^4*b^14) 
/(a*b^7 - b^8) - (2*a^5*b^13)/(a*b^7 - b^8) + (2*a^2*b^10*(a^5*b^7)^(1/2)) 
/(a*b^7 - b^8) - (2*a^3*b^9*(a^5*b^7)^(1/2))/(a*b^7 - b^8)))*(-((a^5*b^7)^ 
(1/2) + a^2*b^4)/(16*(a*b^7 - b^8)))^(1/2)*2i)/d - (atan((a^3*cos(c + d*x) 
*((a^5*b^7)^(1/2)/(16*(a*b^7 - b^8)) - (a^2*b^4)/(16*(a*b^7 - b^8)))^(1/2) 
*8i)/((2*a^4)/b^2 + (2*a^4*b^6)/(a*b^7 - b^8) - (2*a^2*b^2*(a^5*b^7)^(1/2) 
)/(a*b^7 - b^8)) - (a^3*b^8*cos(c + d*x)*((a^5*b^7)^(1/2)/(16*(a*b^7 - b^8 
)) - (a^2*b^4)/(16*(a*b^7 - b^8)))^(1/2)*8i)/(2*a^4*b^6 - 2*a^5*b^5 + (2*a 
^4*b^14)/(a*b^7 - b^8) - (2*a^5*b^13)/(a*b^7 - b^8) - (2*a^2*b^10*(a^5*b^7 
)^(1/2))/(a*b^7 - b^8) + (2*a^3*b^9*(a^5*b^7)^(1/2))/(a*b^7 - b^8)) + (a*b 
^4*cos(c + d*x)*((a^5*b^7)^(1/2)/(16*(a*b^7 - b^8)) - (a^2*b^4)/(16*(a*b^7 
 - b^8)))^(1/2)*(a^5*b^7)^(1/2)*8i)/(2*a^4*b^6 - 2*a^5*b^5 + (2*a^4*b^1...
 

Reduce [F]

\[ \int \frac {\sin ^7(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\left (\int \frac {\sin \left (d x +c \right )^{7}}{\sin \left (d x +c \right )^{4} b -a}d x \right ) \] Input:

int(sin(d*x+c)^7/(a-b*sin(d*x+c)^4),x)
 

Output:

 - int(sin(c + d*x)**7/(sin(c + d*x)**4*b - a),x)