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

Optimal. Leaf size=194 \[ \frac {15 \sqrt {\frac {\pi }{2}} d^{5/2} \sin \left (a-\frac {b c}{d}\right ) C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{4 b^{7/2}}+\frac {15 \sqrt {\frac {\pi }{2}} d^{5/2} \cos \left (a-\frac {b c}{d}\right ) S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{4 b^{7/2}}-\frac {15 d^2 \sqrt {c+d x} \sin (a+b x)}{4 b^3}+\frac {5 d (c+d x)^{3/2} \cos (a+b x)}{2 b^2}+\frac {(c+d x)^{5/2} \sin (a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.42, antiderivative size = 194, 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} \[ \frac {15 \sqrt {\frac {\pi }{2}} d^{5/2} \sin \left (a-\frac {b c}{d}\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{4 b^{7/2}}+\frac {15 \sqrt {\frac {\pi }{2}} d^{5/2} \cos \left (a-\frac {b c}{d}\right ) S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{4 b^{7/2}}-\frac {15 d^2 \sqrt {c+d x} \sin (a+b x)}{4 b^3}+\frac {5 d (c+d x)^{3/2} \cos (a+b x)}{2 b^2}+\frac {(c+d x)^{5/2} \sin (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(5/2)*Cos[a + b*x],x]

[Out]

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

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

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Mathematica [C]  time = 0.06, size = 124, normalized size = 0.64 \[ -\frac {d^3 e^{-\frac {i (a d+b c)}{d}} \left (e^{2 i a} \sqrt {-\frac {i b (c+d x)}{d}} \Gamma \left (\frac {7}{2},-\frac {i b (c+d x)}{d}\right )+e^{\frac {2 i b c}{d}} \sqrt {\frac {i b (c+d x)}{d}} \Gamma \left (\frac {7}{2},\frac {i b (c+d x)}{d}\right )\right )}{2 b^4 \sqrt {c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(d^3*(E^((2*I)*a)*Sqrt[((-I)*b*(c + d*x))/d]*Gamma[7/2, ((-I)*b*(c + d*x))/d] + E^(((2*I)*b*c)/d)*Sqrt[(I
*b*(c + d*x))/d]*Gamma[7/2, (I*b*(c + d*x))/d]))/(b^4*E^((I*(b*c + a*d))/d)*Sqrt[c + d*x])

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fricas [A]  time = 1.07, size = 190, normalized size = 0.98 \[ \frac {15 \, \sqrt {2} \pi d^{3} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {b c - a d}{d}\right ) \operatorname {S}\left (\sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) + 15 \, \sqrt {2} \pi d^{3} \sqrt {\frac {b}{\pi d}} \operatorname {C}\left (\sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {b c - a d}{d}\right ) + 2 \, \sqrt {d x + c} {\left (10 \, {\left (b^{2} d^{2} x + b^{2} c d\right )} \cos \left (b x + a\right ) + {\left (4 \, b^{3} d^{2} x^{2} + 8 \, b^{3} c d x + 4 \, b^{3} c^{2} - 15 \, b d^{2}\right )} \sin \left (b x + a\right )\right )}}{8 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)*cos(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_sin(sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d))) +
15*sqrt(2)*pi*d^3*sqrt(b/(pi*d))*fresnel_cos(sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-(b*c - a*d)/d) + 2*sqr
t(d*x + c)*(10*(b^2*d^2*x + b^2*c*d)*cos(b*x + a) + (4*b^3*d^2*x^2 + 8*b^3*c*d*x + 4*b^3*c^2 - 15*b*d^2)*sin(b
*x + a)))/b^4

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giac [C]  time = 0.91, size = 1239, 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)*cos(b*x+a),x, algorithm="giac")

[Out]

-1/16*(8*(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)) + 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*((
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*(-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 + (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*(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*((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*(-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 + (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*(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*(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/sqr
t(b^2*d^2) + 1)*b) + 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*I*sqrt(d*x + c)*d*e^((I*(d*
x + c)*b - I*b*c + I*a*d)/d)/b + 2*I*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 = 0.04, size = 232, normalized size = 1.20 \[ \frac {\frac {d \left (d x +c \right )^{\frac {5}{2}} \sin \left (\frac {\left (d x +c \right ) b}{d}+\frac {d a -c b}{d}\right )}{b}-\frac {5 d \left (-\frac {d \left (d x +c \right )^{\frac {3}{2}} \cos \left (\frac {\left (d x +c \right ) b}{d}+\frac {d a -c b}{d}\right )}{2 b}+\frac {3 d \left (\frac {d \sqrt {d x +c}\, \sin \left (\frac {\left (d x +c \right ) b}{d}+\frac {d a -c b}{d}\right )}{2 b}-\frac {d \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {d a -c b}{d}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {d a -c b}{d}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{4 b \sqrt {\frac {b}{d}}}\right )}{2 b}\right )}{b}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.94, size = 261, normalized size = 1.35 \[ \frac {\sqrt {2} {\left (40 \, \sqrt {2} {\left (d x + c\right )}^{\frac {3}{2}} b^{2} d \cos \left (\frac {{\left (d x + c\right )} b - b c + a d}{d}\right ) + {\left (\left (15 i + 15\right ) \, \sqrt {\pi } d^{3} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {b c - a d}{d}\right ) - \left (15 i - 15\right ) \, \sqrt {\pi } d^{3} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {b c - a d}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {i \, b}{d}}\right ) + {\left (-\left (15 i - 15\right ) \, \sqrt {\pi } d^{3} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {b c - a d}{d}\right ) + \left (15 i + 15\right ) \, \sqrt {\pi } d^{3} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {b c - a d}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {i \, b}{d}}\right ) + 4 \, {\left (4 \, \sqrt {2} {\left (d x + c\right )}^{\frac {5}{2}} b^{3} - 15 \, \sqrt {2} \sqrt {d x + c} b d^{2}\right )} \sin \left (\frac {{\left (d x + c\right )} b - b c + a d}{d}\right )\right )}}{32 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*sqrt(2)*(40*sqrt(2)*(d*x + c)^(3/2)*b^2*d*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)*sqrt(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)) + 4*(4*sqrt(2)*(d*x + c)^(5/2)*b^3 - 15*
sqrt(2)*sqrt(d*x + c)*b*d^2)*sin(((d*x + c)*b - b*c + a*d)/d))/b^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)*(c + d*x)^(5/2),x)

[Out]

int(cos(a + b*x)*(c + d*x)^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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