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

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

[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

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Rubi [A]  time = 0.422734, antiderivative size = 194, normalized size of antiderivative = 1., 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 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 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 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 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 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*}

Mathematica [C]  time = 0.056424, size = 124, normalized size = 0.64 \[ -\frac{d^3 e^{-\frac{i (a d+b c)}{d}} \left (e^{2 i a} \sqrt{-\frac{i b (c+d x)}{d}} \text{Gamma}\left (\frac{7}{2},-\frac{i b (c+d x)}{d}\right )+e^{\frac{2 i b c}{d}} \sqrt{\frac{i b (c+d x)}{d}} \text{Gamma}\left (\frac{7}{2},\frac{i b (c+d x)}{d}\right )\right )}{2 b^4 \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(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]))/(2*b^4*E^((I*(b*c + a*d))/d)*Sqrt[c + d*x])

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Maple [A]  time = 0.031, size = 232, normalized size = 1.2 \begin{align*} 2\,{\frac{1}{d} \left ( 1/2\,{\frac{d \left ( dx+c \right ) ^{5/2}}{b}\sin \left ({\frac{ \left ( dx+c \right ) b}{d}}+{\frac{da-cb}{d}} \right ) }-5/2\,{\frac{d}{b} \left ( -1/2\,{\frac{d \left ( dx+c \right ) ^{3/2}}{b}\cos \left ({\frac{ \left ( dx+c \right ) b}{d}}+{\frac{da-cb}{d}} \right ) }+3/2\,{\frac{d}{b} \left ( 1/2\,{\frac{d\sqrt{dx+c}}{b}\sin \left ({\frac{ \left ( dx+c \right ) b}{d}}+{\frac{da-cb}{d}} \right ) }-1/4\,{\frac{d\sqrt{2}\sqrt{\pi }}{b} \left ( \cos \left ({\frac{da-cb}{d}} \right ){\it FresnelS} \left ({\frac{\sqrt{2}\sqrt{dx+c}b}{\sqrt{\pi }d}{\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) +\sin \left ({\frac{da-cb}{d}} \right ){\it FresnelC} \left ({\frac{\sqrt{2}\sqrt{dx+c}b}{\sqrt{\pi }d}{\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) \right ){\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) } \right ) } \right ) } \end{align*}

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

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Maxima [C]  time = 2.24648, size = 886, normalized size = 4.57 \begin{align*} \text{result too large to display} \end{align*}

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*(80*(d*x + c)^(3/2)*b*d^2*sqrt(abs(b)/abs(d))*cos(((d*x + c)*b - b*c + a*d)/d) + ((15*I*sqrt(pi)*cos(1/4*
pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 15*I*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*ar
ctan2(0, d/sqrt(d^2))) + 15*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 15*sqrt(p
i)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d^3*cos(-(b*c - a*d)/d) + (15*sqrt(pi)*cos(
1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 15*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*
arctan2(0, d/sqrt(d^2))) - 15*I*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 15*I*
sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d^3*sin(-(b*c - a*d)/d))*erf(sqrt(d*x
+ c)*sqrt(I*b/d)) + ((-15*I*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 15*I*sqr
t(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 15*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0,
b) + 1/2*arctan2(0, d/sqrt(d^2))) - 15*sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))
))*d^3*cos(-(b*c - a*d)/d) + (15*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 15*s
qrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 15*I*sqrt(pi)*sin(1/4*pi + 1/2*arctan
2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 15*I*sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt
(d^2))))*d^3*sin(-(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(-I*b/d)) + 8*(4*(d*x + c)^(5/2)*b^2*d*sqrt(abs(b)/abs
(d)) - 15*sqrt(d*x + c)*d^3*sqrt(abs(b)/abs(d)))*sin(((d*x + c)*b - b*c + a*d)/d))/(b^3*d*sqrt(abs(b)/abs(d)))

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Fricas [A]  time = 1.48212, size = 466, normalized size = 2.4 \begin{align*} \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}} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [C]  time = 1.26431, size = 1374, normalized size = 7.08 \begin{align*} \text{result too large to display} \end{align*}

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*(4*(-I*sqrt(2)*sqrt(pi)*d^2*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)*d^2*erf(-1/2*sqrt(2)*sqrt(b*d)*sq
rt(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^2*((sqrt(2)*sqrt(pi)*(4*I*b^2*c^2*d - 12*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 + 8*I*(d*x + c)^(3/2)*b^2*c*d - 4*I*sqrt(d*x + c)*b^2*c^2*d - 10*(d*x + c)^(3/2)*
b*d^2 + 12*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^2 + (
sqrt(2)*sqrt(pi)*(-4*I*b^2*c^2*d - 12*b*c*d^2 + 15*I*d^3)*d*erf(-1/2*sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(-I*b*d/s
qrt(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 - 8*I*(d*x + c)^(3/2)*b^2*c*d + 4*I*sqrt(d*x + c)*b^2*c^2*d - 10*(d*x + c)^(3/2)*b*d^2 + 12*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^2) - 4*(sqrt(2)*sqrt(pi)*(-
2*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) + sqrt(2)*sqrt(pi)*(2*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 - 2*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 - 2*(-2*I*(d*x + c)^(3/2)*b*d + 2*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)*c)/d