\(\int \frac {\sin (a+b x)}{(c+d x)^{5/2}} \, dx\) [43]

Optimal result

Integrand size = 16, antiderivative size = 168 \[ \int \frac {\sin (a+b x)}{(c+d x)^{5/2}} \, dx=-\frac {4 b \cos (a+b x)}{3 d^2 \sqrt {c+d x}}-\frac {4 b^{3/2} \sqrt {2 \pi } \cos \left (a-\frac {b c}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{3 d^{5/2}}-\frac {4 b^{3/2} \sqrt {2 \pi } \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right ) \sin \left (a-\frac {b c}{d}\right )}{3 d^{5/2}}-\frac {2 \sin (a+b x)}{3 d (c+d x)^{3/2}} \] Output:

-4/3*b*cos(b*x+a)/d^2/(d*x+c)^(1/2)-4/3*b^(3/2)*2^(1/2)*Pi^(1/2)*cos(a-b*c 
/d)*FresnelS(b^(1/2)*2^(1/2)/Pi^(1/2)*(d*x+c)^(1/2)/d^(1/2))/d^(5/2)-4/3*b 
^(3/2)*2^(1/2)*Pi^(1/2)*FresnelC(b^(1/2)*2^(1/2)/Pi^(1/2)*(d*x+c)^(1/2)/d^ 
(1/2))*sin(a-b*c/d)/d^(5/2)-2/3*sin(b*x+a)/d/(d*x+c)^(3/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.63 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.96 \[ \int \frac {\sin (a+b x)}{(c+d x)^{5/2}} \, dx=\frac {2 \left (-b (c+d x) \left (-e^{i \left (a-\frac {b c}{d}\right )} \sqrt {-\frac {i b (c+d x)}{d}} \Gamma \left (\frac {1}{2},-\frac {i b (c+d x)}{d}\right )+e^{-i (a+b x)} \left (1+e^{2 i (a+b x)}-e^{\frac {i b (c+d x)}{d}} \sqrt {\frac {i b (c+d x)}{d}} \Gamma \left (\frac {1}{2},\frac {i b (c+d x)}{d}\right )\right )\right )-d \sin (a+b x)\right )}{3 d^2 (c+d x)^{3/2}} \] Input:

Integrate[Sin[a + b*x]/(c + d*x)^(5/2),x]
 

Output:

(2*(-(b*(c + d*x)*(-(E^(I*(a - (b*c)/d))*Sqrt[((-I)*b*(c + d*x))/d]*Gamma[ 
1/2, ((-I)*b*(c + d*x))/d]) + (1 + E^((2*I)*(a + b*x)) - E^((I*b*(c + d*x) 
)/d)*Sqrt[(I*b*(c + d*x))/d]*Gamma[1/2, (I*b*(c + d*x))/d])/E^(I*(a + b*x) 
))) - d*Sin[a + b*x]))/(3*d^2*(c + d*x)^(3/2))
 

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.04, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3042, 3778, 3042, 3778, 25, 3042, 3787, 3042, 3785, 3786, 3832, 3833}

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.

\(\displaystyle \int \frac {\sin (a+b x)}{(c+d x)^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (a+b x)}{(c+d x)^{5/2}}dx\)

\(\Big \downarrow \) 3778

\(\displaystyle \frac {2 b \int \frac {\cos (a+b x)}{(c+d x)^{3/2}}dx}{3 d}-\frac {2 \sin (a+b x)}{3 d (c+d x)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b \int \frac {\sin \left (a+b x+\frac {\pi }{2}\right )}{(c+d x)^{3/2}}dx}{3 d}-\frac {2 \sin (a+b x)}{3 d (c+d x)^{3/2}}\)

\(\Big \downarrow \) 3778

\(\displaystyle \frac {2 b \left (\frac {2 b \int -\frac {\sin (a+b x)}{\sqrt {c+d x}}dx}{d}-\frac {2 \cos (a+b x)}{d \sqrt {c+d x}}\right )}{3 d}-\frac {2 \sin (a+b x)}{3 d (c+d x)^{3/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 b \left (-\frac {2 b \int \frac {\sin (a+b x)}{\sqrt {c+d x}}dx}{d}-\frac {2 \cos (a+b x)}{d \sqrt {c+d x}}\right )}{3 d}-\frac {2 \sin (a+b x)}{3 d (c+d x)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b \left (-\frac {2 b \int \frac {\sin (a+b x)}{\sqrt {c+d x}}dx}{d}-\frac {2 \cos (a+b x)}{d \sqrt {c+d x}}\right )}{3 d}-\frac {2 \sin (a+b x)}{3 d (c+d x)^{3/2}}\)

\(\Big \downarrow \) 3787

\(\displaystyle \frac {2 b \left (-\frac {2 b \left (\sin \left (a-\frac {b c}{d}\right ) \int \frac {\cos \left (\frac {b c}{d}+b x\right )}{\sqrt {c+d x}}dx+\cos \left (a-\frac {b c}{d}\right ) \int \frac {\sin \left (\frac {b c}{d}+b x\right )}{\sqrt {c+d x}}dx\right )}{d}-\frac {2 \cos (a+b x)}{d \sqrt {c+d x}}\right )}{3 d}-\frac {2 \sin (a+b x)}{3 d (c+d x)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b \left (-\frac {2 b \left (\sin \left (a-\frac {b c}{d}\right ) \int \frac {\sin \left (\frac {b c}{d}+b x+\frac {\pi }{2}\right )}{\sqrt {c+d x}}dx+\cos \left (a-\frac {b c}{d}\right ) \int \frac {\sin \left (\frac {b c}{d}+b x\right )}{\sqrt {c+d x}}dx\right )}{d}-\frac {2 \cos (a+b x)}{d \sqrt {c+d x}}\right )}{3 d}-\frac {2 \sin (a+b x)}{3 d (c+d x)^{3/2}}\)

\(\Big \downarrow \) 3785

\(\displaystyle \frac {2 b \left (-\frac {2 b \left (\frac {2 \sin \left (a-\frac {b c}{d}\right ) \int \cos \left (\frac {b (c+d x)}{d}\right )d\sqrt {c+d x}}{d}+\cos \left (a-\frac {b c}{d}\right ) \int \frac {\sin \left (\frac {b c}{d}+b x\right )}{\sqrt {c+d x}}dx\right )}{d}-\frac {2 \cos (a+b x)}{d \sqrt {c+d x}}\right )}{3 d}-\frac {2 \sin (a+b x)}{3 d (c+d x)^{3/2}}\)

\(\Big \downarrow \) 3786

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

\(\Big \downarrow \) 3832

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

\(\Big \downarrow \) 3833

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

Input:

Int[Sin[a + b*x]/(c + d*x)^(5/2),x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042

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

rule 3778

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(c 
 + d*x)^(m + 1)*(Sin[e + f*x]/(d*(m + 1))), x] - Simp[f/(d*(m + 1))   Int[( 
c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[m, - 
1]
 

rule 3785

Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> S 
imp[2/d   Subst[Int[Cos[f*(x^2/d)], x], x, Sqrt[c + d*x]], x] /; FreeQ[{c, 
d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]
 

rule 3786

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[2/d 
   Subst[Int[Sin[f*(x^2/d)], x], x, Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f 
}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]
 

rule 3787

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Cos 
[(d*e - c*f)/d]   Int[Sin[c*(f/d) + f*x]/Sqrt[c + d*x], x], x] + Simp[Sin[( 
d*e - c*f)/d]   Int[Cos[c*(f/d) + f*x]/Sqrt[c + d*x], x], x] /; FreeQ[{c, d 
, e, f}, x] && ComplexFreeQ[f] && NeQ[d*e - c*f, 0]
 

rule 3832

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ 
d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]
 

rule 3833

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ 
d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]
 

Maple [A] (verified)

Time = 0.75 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.07

method	result	size
derivativedivides	\(\frac {-\frac {2 \sin \left (\frac {b \left (d x +c \right )}{d}+\frac {a d -b c}{d}\right )}{3 \left (d x +c \right )^{\frac {3}{2}}}+\frac {4 b \left (-\frac {\cos \left (\frac {b \left (d x +c \right )}{d}+\frac {a d -b c}{d}\right )}{\sqrt {d x +c}}-\frac {b \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {a d -b c}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {a d -b c}{d}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{d \sqrt {\frac {b}{d}}}\right )}{3 d}}{d}\)	\(180\)
default	\(\frac {-\frac {2 \sin \left (\frac {b \left (d x +c \right )}{d}+\frac {a d -b c}{d}\right )}{3 \left (d x +c \right )^{\frac {3}{2}}}+\frac {4 b \left (-\frac {\cos \left (\frac {b \left (d x +c \right )}{d}+\frac {a d -b c}{d}\right )}{\sqrt {d x +c}}-\frac {b \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {a d -b c}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {a d -b c}{d}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{d \sqrt {\frac {b}{d}}}\right )}{3 d}}{d}\)	\(180\)

Input:

int(sin(b*x+a)/(d*x+c)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.24 \[ \int \frac {\sin (a+b x)}{(c+d x)^{5/2}} \, dx=-\frac {2 \, {\left (2 \, \sqrt {2} {\left (\pi b d^{2} x^{2} + 2 \, \pi b c d x + \pi b c^{2}\right )} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {b c - a d}{d}\right ) \operatorname {S}\left (\sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) + 2 \, \sqrt {2} {\left (\pi b d^{2} x^{2} + 2 \, \pi b c d x + \pi b c^{2}\right )} \sqrt {\frac {b}{\pi d}} \operatorname {C}\left (\sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {b c - a d}{d}\right ) + \sqrt {d x + c} {\left (2 \, {\left (b d x + b c\right )} \cos \left (b x + a\right ) + d \sin \left (b x + a\right )\right )}\right )}}{3 \, {\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )}} \] Input:

integrate(sin(b*x+a)/(d*x+c)^(5/2),x, algorithm="fricas")
 

Output:

-2/3*(2*sqrt(2)*(pi*b*d^2*x^2 + 2*pi*b*c*d*x + pi*b*c^2)*sqrt(b/(pi*d))*co 
s(-(b*c - a*d)/d)*fresnel_sin(sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d))) + 2*sq 
rt(2)*(pi*b*d^2*x^2 + 2*pi*b*c*d*x + pi*b*c^2)*sqrt(b/(pi*d))*fresnel_cos( 
sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-(b*c - a*d)/d) + sqrt(d*x + c)* 
(2*(b*d*x + b*c)*cos(b*x + a) + d*sin(b*x + a)))/(d^4*x^2 + 2*c*d^3*x + c^ 
2*d^2)
 

Sympy [F]

\[ \int \frac {\sin (a+b x)}{(c+d x)^{5/2}} \, dx=\int \frac {\sin {\left (a + b x \right )}}{\left (c + d x\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(sin(b*x+a)/(d*x+c)**(5/2),x)
 

Output:

Integral(sin(a + b*x)/(c + d*x)**(5/2), x)
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.24 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.77 \[ \int \frac {\sin (a+b x)}{(c+d x)^{5/2}} \, dx=-\frac {{\left ({\left (-\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, \frac {i \, {\left (d x + c\right )} b}{d}\right ) + \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, -\frac {i \, {\left (d x + c\right )} b}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) + {\left (\left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, \frac {i \, {\left (d x + c\right )} b}{d}\right ) - \left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, -\frac {i \, {\left (d x + c\right )} b}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right )\right )} \left (\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {3}{2}}}{4 \, {\left (d x + c\right )}^{\frac {3}{2}} d} \] Input:

integrate(sin(b*x+a)/(d*x+c)^(5/2),x, algorithm="maxima")
 

Output:

-1/4*((-(I + 1)*sqrt(2)*gamma(-3/2, I*(d*x + c)*b/d) + (I - 1)*sqrt(2)*gam 
ma(-3/2, -I*(d*x + c)*b/d))*cos(-(b*c - a*d)/d) + ((I - 1)*sqrt(2)*gamma(- 
3/2, I*(d*x + c)*b/d) - (I + 1)*sqrt(2)*gamma(-3/2, -I*(d*x + c)*b/d))*sin 
(-(b*c - a*d)/d))*((d*x + c)*b/d)^(3/2)/((d*x + c)^(3/2)*d)
 

Giac [F]

\[ \int \frac {\sin (a+b x)}{(c+d x)^{5/2}} \, dx=\int { \frac {\sin \left (b x + a\right )}{{\left (d x + c\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(sin(b*x+a)/(d*x+c)^(5/2),x, algorithm="giac")
 

Output:

integrate(sin(b*x + a)/(d*x + c)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sin (a+b x)}{(c+d x)^{5/2}} \, dx=\int \frac {\sin \left (a+b\,x\right )}{{\left (c+d\,x\right )}^{5/2}} \,d x \] Input:

int(sin(a + b*x)/(c + d*x)^(5/2),x)
 

Output:

int(sin(a + b*x)/(c + d*x)^(5/2), x)
 

Reduce [F]

\[ \int \frac {\sin (a+b x)}{(c+d x)^{5/2}} \, dx=\int \frac {\sin \left (b x +a \right )}{\sqrt {d x +c}\, c^{2}+2 \sqrt {d x +c}\, c d x +\sqrt {d x +c}\, d^{2} x^{2}}d x \] Input:

int(sin(b*x+a)/(d*x+c)^(5/2),x)
 

Output:

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