\(\int \frac {(a+b \sin (c+d x^2))^2}{x^4} \, dx\) [22]

Optimal result

Integrand size = 18, antiderivative size = 239 \[ \int \frac {\left (a+b \sin \left (c+d x^2\right )\right )^2}{x^4} \, dx=-\frac {2 a^2+b^2}{6 x^3}-\frac {4 a b d \cos \left (c+d x^2\right )}{3 x}+\frac {b^2 \cos \left (2 c+2 d x^2\right )}{6 x^3}+\frac {4}{3} b^2 d^{3/2} \sqrt {\pi } \cos (2 c) \operatorname {FresnelC}\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right )-\frac {4}{3} a b d^{3/2} \sqrt {2 \pi } \cos (c) \operatorname {FresnelS}\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )-\frac {4}{3} a b d^{3/2} \sqrt {2 \pi } \operatorname {FresnelC}\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right ) \sin (c)-\frac {4}{3} b^2 d^{3/2} \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right ) \sin (2 c)-\frac {2 a b \sin \left (c+d x^2\right )}{3 x^3}-\frac {2 b^2 d \sin \left (2 c+2 d x^2\right )}{3 x} \] Output:

-1/6*(2*a^2+b^2)/x^3-4/3*a*b*d*cos(d*x^2+c)/x+1/6*b^2*cos(2*d*x^2+2*c)/x^3 
+4/3*b^2*d^(3/2)*Pi^(1/2)*cos(2*c)*FresnelC(2*d^(1/2)*x/Pi^(1/2))-4/3*a*b* 
d^(3/2)*2^(1/2)*Pi^(1/2)*cos(c)*FresnelS(d^(1/2)*2^(1/2)/Pi^(1/2)*x)-4/3*a 
*b*d^(3/2)*2^(1/2)*Pi^(1/2)*FresnelC(d^(1/2)*2^(1/2)/Pi^(1/2)*x)*sin(c)-4/ 
3*b^2*d^(3/2)*Pi^(1/2)*FresnelS(2*d^(1/2)*x/Pi^(1/2))*sin(2*c)-2/3*a*b*sin 
(d*x^2+c)/x^3-2/3*b^2*d*sin(2*d*x^2+2*c)/x
 

Mathematica [A] (verified)

Time = 0.73 (sec) , antiderivative size = 226, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a+b \sin \left (c+d x^2\right )\right )^2}{x^4} \, dx=-\frac {2 a^2+b^2+8 a b d x^2 \cos \left (c+d x^2\right )-b^2 \cos \left (2 \left (c+d x^2\right )\right )-8 b^2 d^{3/2} \sqrt {\pi } x^3 \cos (2 c) \operatorname {FresnelC}\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right )+8 a b d^{3/2} \sqrt {2 \pi } x^3 \cos (c) \operatorname {FresnelS}\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )+8 a b d^{3/2} \sqrt {2 \pi } x^3 \operatorname {FresnelC}\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right ) \sin (c)+8 b^2 d^{3/2} \sqrt {\pi } x^3 \operatorname {FresnelS}\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right ) \sin (2 c)+4 a b \sin \left (c+d x^2\right )+4 b^2 d x^2 \sin \left (2 \left (c+d x^2\right )\right )}{6 x^3} \] Input:

Integrate[(a + b*Sin[c + d*x^2])^2/x^4,x]
 

Output:

-1/6*(2*a^2 + b^2 + 8*a*b*d*x^2*Cos[c + d*x^2] - b^2*Cos[2*(c + d*x^2)] - 
8*b^2*d^(3/2)*Sqrt[Pi]*x^3*Cos[2*c]*FresnelC[(2*Sqrt[d]*x)/Sqrt[Pi]] + 8*a 
*b*d^(3/2)*Sqrt[2*Pi]*x^3*Cos[c]*FresnelS[Sqrt[d]*Sqrt[2/Pi]*x] + 8*a*b*d^ 
(3/2)*Sqrt[2*Pi]*x^3*FresnelC[Sqrt[d]*Sqrt[2/Pi]*x]*Sin[c] + 8*b^2*d^(3/2) 
*Sqrt[Pi]*x^3*FresnelS[(2*Sqrt[d]*x)/Sqrt[Pi]]*Sin[2*c] + 4*a*b*Sin[c + d* 
x^2] + 4*b^2*d*x^2*Sin[2*(c + d*x^2)])/x^3
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3884, 6, 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.

\(\displaystyle \int \frac {\left (a+b \sin \left (c+d x^2\right )\right )^2}{x^4} \, dx\)

\(\Big \downarrow \) 3884

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

\(\Big \downarrow \) 6

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 a^2+b^2}{6 x^3}-\frac {4}{3} \sqrt {2 \pi } a b d^{3/2} \sin (c) \operatorname {FresnelC}\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )-\frac {4}{3} \sqrt {2 \pi } a b d^{3/2} \cos (c) \operatorname {FresnelS}\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )-\frac {4 a b d \cos \left (c+d x^2\right )}{3 x}-\frac {2 a b \sin \left (c+d x^2\right )}{3 x^3}+\frac {4}{3} \sqrt {\pi } b^2 d^{3/2} \cos (2 c) \operatorname {FresnelC}\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right )-\frac {4}{3} \sqrt {\pi } b^2 d^{3/2} \sin (2 c) \operatorname {FresnelS}\left (\frac {2 \sqrt {d} x}{\sqrt {\pi }}\right )-\frac {2 b^2 d \sin \left (2 c+2 d x^2\right )}{3 x}+\frac {b^2 \cos \left (2 c+2 d x^2\right )}{6 x^3}\)

Input:

Int[(a + b*Sin[c + d*x^2])^2/x^4,x]
 

Output:

-1/6*(2*a^2 + b^2)/x^3 - (4*a*b*d*Cos[c + d*x^2])/(3*x) + (b^2*Cos[2*c + 2 
*d*x^2])/(6*x^3) + (4*b^2*d^(3/2)*Sqrt[Pi]*Cos[2*c]*FresnelC[(2*Sqrt[d]*x) 
/Sqrt[Pi]])/3 - (4*a*b*d^(3/2)*Sqrt[2*Pi]*Cos[c]*FresnelS[Sqrt[d]*Sqrt[2/P 
i]*x])/3 - (4*a*b*d^(3/2)*Sqrt[2*Pi]*FresnelC[Sqrt[d]*Sqrt[2/Pi]*x]*Sin[c] 
)/3 - (4*b^2*d^(3/2)*Sqrt[Pi]*FresnelS[(2*Sqrt[d]*x)/Sqrt[Pi]]*Sin[2*c])/3 
 - (2*a*b*Sin[c + d*x^2])/(3*x^3) - (2*b^2*d*Sin[2*c + 2*d*x^2])/(3*x)
 

Defintions of rubi rules used

rule 6

Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v 
+ (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] &&  !FreeQ[Fx, x]
 

rule 2009

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

rule 3884

Int[((e_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_), x 
_Symbol] :> Int[ExpandTrigReduce[(e*x)^m, (a + b*Sin[c + d*x^n])^p, x], x] 
/; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 1] && IGtQ[n, 0]
 

Maple [A] (verified)

Time = 1.90 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.72

method	result	size
parts	\(-\frac {a^{2}}{3 x^{3}}+b^{2} \left (-\frac {1}{6 x^{3}}+\frac {\cos \left (2 d \,x^{2}+2 c \right )}{6 x^{3}}+\frac {2 d \left (-\frac {\sin \left (2 d \,x^{2}+2 c \right )}{x}+2 \sqrt {d}\, \sqrt {\pi }\, \left (\cos \left (2 c \right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {d}\, x}{\sqrt {\pi }}\right )-\sin \left (2 c \right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {d}\, x}{\sqrt {\pi }}\right )\right )\right )}{3}\right )+2 a b \left (-\frac {\sin \left (d \,x^{2}+c \right )}{3 x^{3}}+\frac {2 d \left (-\frac {\cos \left (d \,x^{2}+c \right )}{x}-\sqrt {d}\, \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (c \right ) \operatorname {FresnelS}\left (\frac {\sqrt {d}\, \sqrt {2}\, x}{\sqrt {\pi }}\right )+\sin \left (c \right ) \operatorname {FresnelC}\left (\frac {\sqrt {d}\, \sqrt {2}\, x}{\sqrt {\pi }}\right )\right )\right )}{3}\right )\)	\(173\)
default	\(-\frac {a^{2}+\frac {b^{2}}{2}}{3 x^{3}}-\frac {b^{2} \left (-\frac {\cos \left (2 d \,x^{2}+2 c \right )}{3 x^{3}}-\frac {4 d \left (-\frac {\sin \left (2 d \,x^{2}+2 c \right )}{x}+2 \sqrt {d}\, \sqrt {\pi }\, \left (\cos \left (2 c \right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {d}\, x}{\sqrt {\pi }}\right )-\sin \left (2 c \right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {d}\, x}{\sqrt {\pi }}\right )\right )\right )}{3}\right )}{2}+2 a b \left (-\frac {\sin \left (d \,x^{2}+c \right )}{3 x^{3}}+\frac {2 d \left (-\frac {\cos \left (d \,x^{2}+c \right )}{x}-\sqrt {d}\, \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (c \right ) \operatorname {FresnelS}\left (\frac {\sqrt {d}\, \sqrt {2}\, x}{\sqrt {\pi }}\right )+\sin \left (c \right ) \operatorname {FresnelC}\left (\frac {\sqrt {d}\, \sqrt {2}\, x}{\sqrt {\pi }}\right )\right )\right )}{3}\right )\)	\(175\)
risch	\(-\frac {2 i a b \,d^{2} \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {i d}\, x \right ) {\mathrm e}^{-i c}}{3 \sqrt {i d}}-\frac {a^{2}}{3 x^{3}}-\frac {b^{2}}{6 x^{3}}+\frac {b^{2} d^{2} \sqrt {\pi }\, \sqrt {2}\, \operatorname {erf}\left (\sqrt {2}\, \sqrt {i d}\, x \right ) {\mathrm e}^{-2 i c}}{3 \sqrt {i d}}+\frac {2 b^{2} d^{2} \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-2 i d}\, x \right ) {\mathrm e}^{2 i c}}{3 \sqrt {-2 i d}}+\frac {2 i a b \,d^{2} \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-i d}\, x \right ) {\mathrm e}^{i c}}{3 \sqrt {-i d}}-\frac {4 a b d \cos \left (d \,x^{2}+c \right )}{3 x}-\frac {2 a b \sin \left (d \,x^{2}+c \right )}{3 x^{3}}+\frac {b^{2} \cos \left (2 d \,x^{2}+2 c \right )}{6 x^{3}}-\frac {2 b^{2} d \sin \left (2 d \,x^{2}+2 c \right )}{3 x}\)	\(218\)

Input:

int((a+b*sin(d*x^2+c))^2/x^4,x,method=_RETURNVERBOSE)
 

Output:

-1/3/x^3*a^2+b^2*(-1/6/x^3+1/6/x^3*cos(2*d*x^2+2*c)+2/3*d*(-1/x*sin(2*d*x^ 
2+2*c)+2*d^(1/2)*Pi^(1/2)*(cos(2*c)*FresnelC(2*d^(1/2)*x/Pi^(1/2))-sin(2*c 
)*FresnelS(2*d^(1/2)*x/Pi^(1/2)))))+2*a*b*(-1/3*sin(d*x^2+c)/x^3+2/3*d*(-1 
/x*cos(d*x^2+c)-d^(1/2)*2^(1/2)*Pi^(1/2)*(cos(c)*FresnelS(d^(1/2)*2^(1/2)/ 
Pi^(1/2)*x)+sin(c)*FresnelC(d^(1/2)*2^(1/2)/Pi^(1/2)*x))))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a+b \sin \left (c+d x^2\right )\right )^2}{x^4} \, dx=-\frac {4 \, \sqrt {2} \pi a b d x^{3} \sqrt {\frac {d}{\pi }} \cos \left (c\right ) \operatorname {S}\left (\sqrt {2} x \sqrt {\frac {d}{\pi }}\right ) + 4 \, \sqrt {2} \pi a b d x^{3} \sqrt {\frac {d}{\pi }} \operatorname {C}\left (\sqrt {2} x \sqrt {\frac {d}{\pi }}\right ) \sin \left (c\right ) - 4 \, \pi b^{2} d x^{3} \sqrt {\frac {d}{\pi }} \cos \left (2 \, c\right ) \operatorname {C}\left (2 \, x \sqrt {\frac {d}{\pi }}\right ) + 4 \, \pi b^{2} d x^{3} \sqrt {\frac {d}{\pi }} \operatorname {S}\left (2 \, x \sqrt {\frac {d}{\pi }}\right ) \sin \left (2 \, c\right ) + 4 \, a b d x^{2} \cos \left (d x^{2} + c\right ) - b^{2} \cos \left (d x^{2} + c\right )^{2} + a^{2} + b^{2} + 2 \, {\left (2 \, b^{2} d x^{2} \cos \left (d x^{2} + c\right ) + a b\right )} \sin \left (d x^{2} + c\right )}{3 \, x^{3}} \] Input:

integrate((a+b*sin(d*x^2+c))^2/x^4,x, algorithm="fricas")
 

Output:

-1/3*(4*sqrt(2)*pi*a*b*d*x^3*sqrt(d/pi)*cos(c)*fresnel_sin(sqrt(2)*x*sqrt( 
d/pi)) + 4*sqrt(2)*pi*a*b*d*x^3*sqrt(d/pi)*fresnel_cos(sqrt(2)*x*sqrt(d/pi 
))*sin(c) - 4*pi*b^2*d*x^3*sqrt(d/pi)*cos(2*c)*fresnel_cos(2*x*sqrt(d/pi)) 
 + 4*pi*b^2*d*x^3*sqrt(d/pi)*fresnel_sin(2*x*sqrt(d/pi))*sin(2*c) + 4*a*b* 
d*x^2*cos(d*x^2 + c) - b^2*cos(d*x^2 + c)^2 + a^2 + b^2 + 2*(2*b^2*d*x^2*c 
os(d*x^2 + c) + a*b)*sin(d*x^2 + c))/x^3
 

Sympy [F]

\[ \int \frac {\left (a+b \sin \left (c+d x^2\right )\right )^2}{x^4} \, dx=\int \frac {\left (a + b \sin {\left (c + d x^{2} \right )}\right )^{2}}{x^{4}}\, dx \] Input:

integrate((a+b*sin(d*x**2+c))**2/x**4,x)
 

Output:

Integral((a + b*sin(c + d*x**2))**2/x**4, x)
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.24 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.74 \[ \int \frac {\left (a+b \sin \left (c+d x^2\right )\right )^2}{x^4} \, dx=-\frac {\sqrt {d x^{2}} {\left ({\left (-\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, i \, d x^{2}\right ) + \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, -i \, d x^{2}\right )\right )} \cos \left (c\right ) + {\left (\left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, i \, d x^{2}\right ) - \left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, -i \, d x^{2}\right )\right )} \sin \left (c\right )\right )} a b d}{4 \, x} - \frac {{\left (3 \, \sqrt {2} \sqrt {d x^{2}} {\left ({\left (-\left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, 2 i \, d x^{2}\right ) + \left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, -2 i \, d x^{2}\right )\right )} \cos \left (2 \, c\right ) + {\left (-\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, 2 i \, d x^{2}\right ) + \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, -2 i \, d x^{2}\right )\right )} \sin \left (2 \, c\right )\right )} d x^{2} + 4\right )} b^{2}}{24 \, x^{3}} - \frac {a^{2}}{3 \, x^{3}} \] Input:

integrate((a+b*sin(d*x^2+c))^2/x^4,x, algorithm="maxima")
 

Output:

-1/4*sqrt(d*x^2)*((-(I + 1)*sqrt(2)*gamma(-3/2, I*d*x^2) + (I - 1)*sqrt(2) 
*gamma(-3/2, -I*d*x^2))*cos(c) + ((I - 1)*sqrt(2)*gamma(-3/2, I*d*x^2) - ( 
I + 1)*sqrt(2)*gamma(-3/2, -I*d*x^2))*sin(c))*a*b*d/x - 1/24*(3*sqrt(2)*sq 
rt(d*x^2)*((-(I - 1)*sqrt(2)*gamma(-3/2, 2*I*d*x^2) + (I + 1)*sqrt(2)*gamm 
a(-3/2, -2*I*d*x^2))*cos(2*c) + (-(I + 1)*sqrt(2)*gamma(-3/2, 2*I*d*x^2) + 
 (I - 1)*sqrt(2)*gamma(-3/2, -2*I*d*x^2))*sin(2*c))*d*x^2 + 4)*b^2/x^3 - 1 
/3*a^2/x^3
 

Giac [F]

\[ \int \frac {\left (a+b \sin \left (c+d x^2\right )\right )^2}{x^4} \, dx=\int { \frac {{\left (b \sin \left (d x^{2} + c\right ) + a\right )}^{2}}{x^{4}} \,d x } \] Input:

integrate((a+b*sin(d*x^2+c))^2/x^4,x, algorithm="giac")
 

Output:

integrate((b*sin(d*x^2 + c) + a)^2/x^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b \sin \left (c+d x^2\right )\right )^2}{x^4} \, dx=\int \frac {{\left (a+b\,\sin \left (d\,x^2+c\right )\right )}^2}{x^4} \,d x \] Input:

int((a + b*sin(c + d*x^2))^2/x^4,x)
 

Output:

int((a + b*sin(c + d*x^2))^2/x^4, x)
 

Reduce [F]

\[ \int \frac {\left (a+b \sin \left (c+d x^2\right )\right )^2}{x^4} \, dx=\frac {3 \left (\int \frac {\sin \left (d \,x^{2}+c \right )^{2}}{x^{4}}d x \right ) b^{2} x^{3}+6 \left (\int \frac {\sin \left (d \,x^{2}+c \right )}{x^{4}}d x \right ) a b \,x^{3}-a^{2}}{3 x^{3}} \] Input:

int((a+b*sin(d*x^2+c))^2/x^4,x)
 

Output:

(3*int(sin(c + d*x**2)**2/x**4,x)*b**2*x**3 + 6*int(sin(c + d*x**2)/x**4,x 
)*a*b*x**3 - a**2)/(3*x**3)