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

Optimal. Leaf size=195 \[ -\frac {15 \sqrt {\frac {\pi }{2}} d^{5/2} \cos \left (a-\frac {b c}{d}\right ) C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{4 b^{7/2}}+\frac {15 \sqrt {\frac {\pi }{2}} d^{5/2} \sin \left (a-\frac {b c}{d}\right ) S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{4 b^{7/2}}+\frac {15 d^2 \sqrt {c+d x} \cos (a+b x)}{4 b^3}+\frac {5 d (c+d x)^{3/2} \sin (a+b x)}{2 b^2}-\frac {(c+d x)^{5/2} \cos (a+b x)}{b} \]

[Out]

-(d*x+c)^(5/2)*cos(b*x+a)/b+5/2*d*(d*x+c)^(3/2)*sin(b*x+a)/b^2-15/8*d^(5/2)*cos(a-b*c/d)*FresnelC(b^(1/2)*2^(1
/2)/Pi^(1/2)*(d*x+c)^(1/2)/d^(1/2))*2^(1/2)*Pi^(1/2)/b^(7/2)+15/8*d^(5/2)*FresnelS(b^(1/2)*2^(1/2)/Pi^(1/2)*(d
*x+c)^(1/2)/d^(1/2))*sin(a-b*c/d)*2^(1/2)*Pi^(1/2)/b^(7/2)+15/4*d^2*cos(b*x+a)*(d*x+c)^(1/2)/b^3

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Rubi [A]  time = 0.43, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3296, 3306, 3305, 3351, 3304, 3352} \[ -\frac {15 \sqrt {\frac {\pi }{2}} d^{5/2} \cos \left (a-\frac {b c}{d}\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{4 b^{7/2}}+\frac {15 \sqrt {\frac {\pi }{2}} d^{5/2} \sin \left (a-\frac {b c}{d}\right ) S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{4 b^{7/2}}+\frac {15 d^2 \sqrt {c+d x} \cos (a+b x)}{4 b^3}+\frac {5 d (c+d x)^{3/2} \sin (a+b x)}{2 b^2}-\frac {(c+d x)^{5/2} \cos (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(5/2)*Sin[a + b*x],x]

[Out]

(15*d^2*Sqrt[c + d*x]*Cos[a + b*x])/(4*b^3) - ((c + d*x)^(5/2)*Cos[a + b*x])/b - (15*d^(5/2)*Sqrt[Pi/2]*Cos[a
- (b*c)/d]*FresnelC[(Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(4*b^(7/2)) + (15*d^(5/2)*Sqrt[Pi/2]*FresnelS
[(Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[a - (b*c)/d])/(4*b^(7/2)) + (5*d*(c + d*x)^(3/2)*Sin[a + b*x]
)/(2*b^2)

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3304

Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[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 3305

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[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 3306

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d +
f*x]/Sqrt[c + d*x], x], x] + Dist[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 3351

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

Rule 3352

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

Rubi steps

\begin {align*} \int (c+d x)^{5/2} \sin (a+b x) \, dx &=-\frac {(c+d x)^{5/2} \cos (a+b x)}{b}+\frac {(5 d) \int (c+d x)^{3/2} \cos (a+b x) \, dx}{2 b}\\ &=-\frac {(c+d x)^{5/2} \cos (a+b x)}{b}+\frac {5 d (c+d x)^{3/2} \sin (a+b x)}{2 b^2}-\frac {\left (15 d^2\right ) \int \sqrt {c+d x} \sin (a+b x) \, dx}{4 b^2}\\ &=\frac {15 d^2 \sqrt {c+d x} \cos (a+b x)}{4 b^3}-\frac {(c+d x)^{5/2} \cos (a+b x)}{b}+\frac {5 d (c+d x)^{3/2} \sin (a+b x)}{2 b^2}-\frac {\left (15 d^3\right ) \int \frac {\cos (a+b x)}{\sqrt {c+d x}} \, dx}{8 b^3}\\ &=\frac {15 d^2 \sqrt {c+d x} \cos (a+b x)}{4 b^3}-\frac {(c+d x)^{5/2} \cos (a+b x)}{b}+\frac {5 d (c+d x)^{3/2} \sin (a+b x)}{2 b^2}-\frac {\left (15 d^3 \cos \left (a-\frac {b c}{d}\right )\right ) \int \frac {\cos \left (\frac {b c}{d}+b x\right )}{\sqrt {c+d x}} \, dx}{8 b^3}+\frac {\left (15 d^3 \sin \left (a-\frac {b c}{d}\right )\right ) \int \frac {\sin \left (\frac {b c}{d}+b x\right )}{\sqrt {c+d x}} \, dx}{8 b^3}\\ &=\frac {15 d^2 \sqrt {c+d x} \cos (a+b x)}{4 b^3}-\frac {(c+d x)^{5/2} \cos (a+b x)}{b}+\frac {5 d (c+d x)^{3/2} \sin (a+b x)}{2 b^2}-\frac {\left (15 d^2 \cos \left (a-\frac {b c}{d}\right )\right ) \operatorname {Subst}\left (\int \cos \left (\frac {b x^2}{d}\right ) \, dx,x,\sqrt {c+d x}\right )}{4 b^3}+\frac {\left (15 d^2 \sin \left (a-\frac {b c}{d}\right )\right ) \operatorname {Subst}\left (\int \sin \left (\frac {b x^2}{d}\right ) \, dx,x,\sqrt {c+d x}\right )}{4 b^3}\\ &=\frac {15 d^2 \sqrt {c+d x} \cos (a+b x)}{4 b^3}-\frac {(c+d x)^{5/2} \cos (a+b x)}{b}-\frac {15 d^{5/2} \sqrt {\frac {\pi }{2}} \cos \left (a-\frac {b c}{d}\right ) C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{4 b^{7/2}}+\frac {15 d^{5/2} \sqrt {\frac {\pi }{2}} S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right ) \sin \left (a-\frac {b c}{d}\right )}{4 b^{7/2}}+\frac {5 d (c+d x)^{3/2} \sin (a+b x)}{2 b^2}\\ \end {align*}

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Mathematica [C]  time = 0.12, size = 124, normalized size = 0.64 \[ \frac {d^2 \sqrt {c+d x} e^{-\frac {i (a d+b c)}{d}} \left (\frac {e^{2 i a} \Gamma \left (\frac {7}{2},-\frac {i b (c+d x)}{d}\right )}{\sqrt {-\frac {i b (c+d x)}{d}}}+\frac {e^{\frac {2 i b c}{d}} \Gamma \left (\frac {7}{2},\frac {i b (c+d x)}{d}\right )}{\sqrt {\frac {i b (c+d x)}{d}}}\right )}{2 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*Sqrt[c + d*x]*((E^((2*I)*a)*Gamma[7/2, ((-I)*b*(c + d*x))/d])/Sqrt[((-I)*b*(c + d*x))/d] + (E^(((2*I)*b*c
)/d)*Gamma[7/2, (I*b*(c + d*x))/d])/Sqrt[(I*b*(c + d*x))/d]))/(2*b^3*E^((I*(b*c + a*d))/d))

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fricas [A]  time = 0.76, size = 190, normalized size = 0.97 \[ -\frac {15 \, \sqrt {2} \pi d^{3} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {b c - a d}{d}\right ) \operatorname {C}\left (\sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) - 15 \, \sqrt {2} \pi d^{3} \sqrt {\frac {b}{\pi d}} \operatorname {S}\left (\sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {b c - a d}{d}\right ) + 2 \, \sqrt {d x + c} {\left ({\left (4 \, b^{3} d^{2} x^{2} + 8 \, b^{3} c d x + 4 \, b^{3} c^{2} - 15 \, b d^{2}\right )} \cos \left (b x + a\right ) - 10 \, {\left (b^{2} d^{2} x + b^{2} c d\right )} \sin \left (b x + a\right )\right )}}{8 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(15*sqrt(2)*pi*d^3*sqrt(b/(pi*d))*cos(-(b*c - a*d)/d)*fresnel_cos(sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d))) -
15*sqrt(2)*pi*d^3*sqrt(b/(pi*d))*fresnel_sin(sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-(b*c - a*d)/d) + 2*sq
rt(d*x + c)*((4*b^3*d^2*x^2 + 8*b^3*c*d*x + 4*b^3*c^2 - 15*b*d^2)*cos(b*x + a) - 10*(b^2*d^2*x + b^2*c*d)*sin(
b*x + a)))/b^4

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giac [C]  time = 1.95, size = 1246, normalized size = 6.39 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(8*(I*sqrt(2)*sqrt(pi)*d*erf(-1/2*sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((I*b*c
- I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)) - I*sqrt(2)*sqrt(pi)*d*erf(-1/2*sqrt(2)*sqrt(b*d)*sqrt(d*x
+ c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((-I*b*c + I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) + 1)))*c^3 + 6*c*d^
2*((I*sqrt(2)*sqrt(pi)*(4*b^2*c^2 + 4*I*b*c*d - 3*d^2)*d*erf(-1/2*sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(
b^2*d^2) + 1)/d)*e^((I*b*c - I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)*b^2) - 2*I*(2*I*(d*x + c)^(3/2)*b*
d - 4*I*sqrt(d*x + c)*b*c*d + 3*sqrt(d*x + c)*d^2)*e^((-I*(d*x + c)*b + I*b*c - I*a*d)/d)/b^2)/d^2 + (-I*sqrt(
2)*sqrt(pi)*(4*b^2*c^2 - 4*I*b*c*d - 3*d^2)*d*erf(-1/2*sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(-I*b*d/sqrt(b^2*d^2) +
1)/d)*e^((-I*b*c + I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) + 1)*b^2) - 2*I*(2*I*(d*x + c)^(3/2)*b*d - 4*I*
sqrt(d*x + c)*b*c*d - 3*sqrt(d*x + c)*d^2)*e^((I*(d*x + c)*b - I*b*c + I*a*d)/d)/b^2)/d^2) + d^3*((-I*sqrt(2)*
sqrt(pi)*(8*b^3*c^3 + 12*I*b^2*c^2*d - 18*b*c*d^2 - 15*I*d^3)*d*erf(-1/2*sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(I*b*
d/sqrt(b^2*d^2) + 1)/d)*e^((I*b*c - I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)*b^3) - 2*I*(4*I*(d*x + c)^(
5/2)*b^2*d - 12*I*(d*x + c)^(3/2)*b^2*c*d + 12*I*sqrt(d*x + c)*b^2*c^2*d + 10*(d*x + c)^(3/2)*b*d^2 - 18*sqrt(
d*x + c)*b*c*d^2 - 15*I*sqrt(d*x + c)*d^3)*e^((-I*(d*x + c)*b + I*b*c - I*a*d)/d)/b^3)/d^3 + (I*sqrt(2)*sqrt(p
i)*(8*b^3*c^3 - 12*I*b^2*c^2*d - 18*b*c*d^2 + 15*I*d^3)*d*erf(-1/2*sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(-I*b*d/sqr
t(b^2*d^2) + 1)/d)*e^((-I*b*c + I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) + 1)*b^3) - 2*I*(4*I*(d*x + c)^(5/2
)*b^2*d - 12*I*(d*x + c)^(3/2)*b^2*c*d + 12*I*sqrt(d*x + c)*b^2*c^2*d - 10*(d*x + c)^(3/2)*b*d^2 + 18*sqrt(d*x
+ c)*b*c*d^2 - 15*I*sqrt(d*x + c)*d^3)*e^((I*(d*x + c)*b - I*b*c + I*a*d)/d)/b^3)/d^3) + 12*(-I*sqrt(2)*sqrt(
pi)*(2*b*c + I*d)*d*erf(-1/2*sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((I*b*c - I*a*d)/d
)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)*b) + I*sqrt(2)*sqrt(pi)*(2*b*c - I*d)*d*erf(-1/2*sqrt(2)*sqrt(b*d)*sqrt
(d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((-I*b*c + I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) + 1)*b) + 2*sq
rt(d*x + c)*d*e^((I*(d*x + c)*b - I*b*c + I*a*d)/d)/b + 2*sqrt(d*x + c)*d*e^((-I*(d*x + c)*b + I*b*c - I*a*d)/
d)/b)*c^2)/d

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maple [A]  time = 0.02, size = 233, normalized size = 1.19 \[ \frac {-\frac {d \left (d x +c \right )^{\frac {5}{2}} \cos \left (\frac {\left (d x +c \right ) b}{d}+\frac {d a -c b}{d}\right )}{b}+\frac {5 d \left (\frac {d \left (d x +c \right )^{\frac {3}{2}} \sin \left (\frac {\left (d x +c \right ) b}{d}+\frac {d a -c b}{d}\right )}{2 b}-\frac {3 d \left (-\frac {d \sqrt {d x +c}\, \cos \left (\frac {\left (d x +c \right ) b}{d}+\frac {d a -c b}{d}\right )}{2 b}+\frac {d \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {d a -c b}{d}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {d a -c b}{d}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{4 b \sqrt {\frac {b}{d}}}\right )}{2 b}\right )}{b}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.76, size = 261, normalized size = 1.34 \[ \frac {\sqrt {2} {\left (40 \, \sqrt {2} {\left (d x + c\right )}^{\frac {3}{2}} b^{2} d \sin \left (\frac {{\left (d x + c\right )} b - b c + a d}{d}\right ) - 4 \, {\left (4 \, \sqrt {2} {\left (d x + c\right )}^{\frac {5}{2}} b^{3} - 15 \, \sqrt {2} \sqrt {d x + c} b d^{2}\right )} \cos \left (\frac {{\left (d x + c\right )} b - b c + a d}{d}\right ) + {\left (\left (15 i - 15\right ) \, \sqrt {\pi } d^{3} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {b c - a d}{d}\right ) + \left (15 i + 15\right ) \, \sqrt {\pi } d^{3} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {b c - a d}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {i \, b}{d}}\right ) + {\left (-\left (15 i + 15\right ) \, \sqrt {\pi } d^{3} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {b c - a d}{d}\right ) - \left (15 i - 15\right ) \, \sqrt {\pi } d^{3} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {b c - a d}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {i \, b}{d}}\right )\right )}}{32 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*sqrt(2)*(40*sqrt(2)*(d*x + c)^(3/2)*b^2*d*sin(((d*x + c)*b - b*c + a*d)/d) - 4*(4*sqrt(2)*(d*x + c)^(5/2)
*b^3 - 15*sqrt(2)*sqrt(d*x + c)*b*d^2)*cos(((d*x + c)*b - b*c + a*d)/d) + ((15*I - 15)*sqrt(pi)*d^3*(b^2/d^2)^
(1/4)*cos(-(b*c - a*d)/d) + (15*I + 15)*sqrt(pi)*d^3*(b^2/d^2)^(1/4)*sin(-(b*c - a*d)/d))*erf(sqrt(d*x + c)*sq
rt(I*b/d)) + (-(15*I + 15)*sqrt(pi)*d^3*(b^2/d^2)^(1/4)*cos(-(b*c - a*d)/d) - (15*I - 15)*sqrt(pi)*d^3*(b^2/d^
2)^(1/4)*sin(-(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(-I*b/d)))/b^4

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sin \left (a+b\,x\right )\,{\left (c+d\,x\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)*(c + d*x)^(5/2),x)

[Out]

int(sin(a + b*x)*(c + d*x)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c + d x\right )^{\frac {5}{2}} \sin {\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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