3.54 \(\int \frac{\cos ^2(a+b x)}{(c+d x)^{7/2}} \, dx\)

Optimal. Leaf size=216 \[ \frac{32 \sqrt{\pi } b^{5/2} \sin \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{\pi } \sqrt{d}}\right )}{15 d^{7/2}}+\frac{32 \sqrt{\pi } b^{5/2} \cos \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right )}{15 d^{7/2}}+\frac{32 b^2 \cos ^2(a+b x)}{15 d^3 \sqrt{c+d x}}+\frac{8 b \sin (a+b x) \cos (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac{2 \cos ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac{16 b^2}{15 d^3 \sqrt{c+d x}} \]

[Out]

(-16*b^2)/(15*d^3*Sqrt[c + d*x]) - (2*Cos[a + b*x]^2)/(5*d*(c + d*x)^(5/2)) + (32*b^2*Cos[a + b*x]^2)/(15*d^3*
Sqrt[c + d*x]) + (32*b^(5/2)*Sqrt[Pi]*Cos[2*a - (2*b*c)/d]*FresnelS[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi
])])/(15*d^(7/2)) + (32*b^(5/2)*Sqrt[Pi]*FresnelC[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])]*Sin[2*a - (2*b
*c)/d])/(15*d^(7/2)) + (8*b*Cos[a + b*x]*Sin[a + b*x])/(15*d^2*(c + d*x)^(3/2))

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Rubi [A]  time = 0.324499, antiderivative size = 216, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3314, 32, 3313, 12, 3306, 3305, 3351, 3304, 3352} \[ \frac{32 \sqrt{\pi } b^{5/2} \sin \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{\pi } \sqrt{d}}\right )}{15 d^{7/2}}+\frac{32 \sqrt{\pi } b^{5/2} \cos \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right )}{15 d^{7/2}}+\frac{32 b^2 \cos ^2(a+b x)}{15 d^3 \sqrt{c+d x}}+\frac{8 b \sin (a+b x) \cos (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac{2 \cos ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac{16 b^2}{15 d^3 \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^2/(c + d*x)^(7/2),x]

[Out]

(-16*b^2)/(15*d^3*Sqrt[c + d*x]) - (2*Cos[a + b*x]^2)/(5*d*(c + d*x)^(5/2)) + (32*b^2*Cos[a + b*x]^2)/(15*d^3*
Sqrt[c + d*x]) + (32*b^(5/2)*Sqrt[Pi]*Cos[2*a - (2*b*c)/d]*FresnelS[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi
])])/(15*d^(7/2)) + (32*b^(5/2)*Sqrt[Pi]*FresnelC[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])]*Sin[2*a - (2*b
*c)/d])/(15*d^(7/2)) + (8*b*Cos[a + b*x]*Sin[a + b*x])/(15*d^2*(c + d*x)^(3/2))

Rule 3314

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(b*Si
n[e + f*x])^n)/(d*(m + 1)), x] + (Dist[(b^2*f^2*n*(n - 1))/(d^2*(m + 1)*(m + 2)), Int[(c + d*x)^(m + 2)*(b*Sin
[e + f*x])^(n - 2), x], x] - Dist[(f^2*n^2)/(d^2*(m + 1)*(m + 2)), Int[(c + d*x)^(m + 2)*(b*Sin[e + f*x])^n, x
], x] - Simp[(b*f*n*(c + d*x)^(m + 2)*Cos[e + f*x]*(b*Sin[e + f*x])^(n - 1))/(d^2*(m + 1)*(m + 2)), x]) /; Fre
eQ[{b, c, d, e, f}, x] && GtQ[n, 1] && LtQ[m, -2]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 3313

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Simp[((c + d*x)^(m + 1)*Sin[e + f*x]^
n)/(d*(m + 1)), x] - Dist[(f*n)/(d*(m + 1)), Int[ExpandTrigReduce[(c + d*x)^(m + 1), Cos[e + f*x]*Sin[e + f*x]
^(n - 1), x], x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && GeQ[m, -2] && LtQ[m, -1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 \frac{\cos ^2(a+b x)}{(c+d x)^{7/2}} \, dx &=-\frac{2 \cos ^2(a+b x)}{5 d (c+d x)^{5/2}}+\frac{8 b \cos (a+b x) \sin (a+b x)}{15 d^2 (c+d x)^{3/2}}+\frac{\left (8 b^2\right ) \int \frac{1}{(c+d x)^{3/2}} \, dx}{15 d^2}-\frac{\left (16 b^2\right ) \int \frac{\cos ^2(a+b x)}{(c+d x)^{3/2}} \, dx}{15 d^2}\\ &=-\frac{16 b^2}{15 d^3 \sqrt{c+d x}}-\frac{2 \cos ^2(a+b x)}{5 d (c+d x)^{5/2}}+\frac{32 b^2 \cos ^2(a+b x)}{15 d^3 \sqrt{c+d x}}+\frac{8 b \cos (a+b x) \sin (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac{\left (64 b^3\right ) \int -\frac{\sin (2 a+2 b x)}{2 \sqrt{c+d x}} \, dx}{15 d^3}\\ &=-\frac{16 b^2}{15 d^3 \sqrt{c+d x}}-\frac{2 \cos ^2(a+b x)}{5 d (c+d x)^{5/2}}+\frac{32 b^2 \cos ^2(a+b x)}{15 d^3 \sqrt{c+d x}}+\frac{8 b \cos (a+b x) \sin (a+b x)}{15 d^2 (c+d x)^{3/2}}+\frac{\left (32 b^3\right ) \int \frac{\sin (2 a+2 b x)}{\sqrt{c+d x}} \, dx}{15 d^3}\\ &=-\frac{16 b^2}{15 d^3 \sqrt{c+d x}}-\frac{2 \cos ^2(a+b x)}{5 d (c+d x)^{5/2}}+\frac{32 b^2 \cos ^2(a+b x)}{15 d^3 \sqrt{c+d x}}+\frac{8 b \cos (a+b x) \sin (a+b x)}{15 d^2 (c+d x)^{3/2}}+\frac{\left (32 b^3 \cos \left (2 a-\frac{2 b c}{d}\right )\right ) \int \frac{\sin \left (\frac{2 b c}{d}+2 b x\right )}{\sqrt{c+d x}} \, dx}{15 d^3}+\frac{\left (32 b^3 \sin \left (2 a-\frac{2 b c}{d}\right )\right ) \int \frac{\cos \left (\frac{2 b c}{d}+2 b x\right )}{\sqrt{c+d x}} \, dx}{15 d^3}\\ &=-\frac{16 b^2}{15 d^3 \sqrt{c+d x}}-\frac{2 \cos ^2(a+b x)}{5 d (c+d x)^{5/2}}+\frac{32 b^2 \cos ^2(a+b x)}{15 d^3 \sqrt{c+d x}}+\frac{8 b \cos (a+b x) \sin (a+b x)}{15 d^2 (c+d x)^{3/2}}+\frac{\left (64 b^3 \cos \left (2 a-\frac{2 b c}{d}\right )\right ) \operatorname{Subst}\left (\int \sin \left (\frac{2 b x^2}{d}\right ) \, dx,x,\sqrt{c+d x}\right )}{15 d^4}+\frac{\left (64 b^3 \sin \left (2 a-\frac{2 b c}{d}\right )\right ) \operatorname{Subst}\left (\int \cos \left (\frac{2 b x^2}{d}\right ) \, dx,x,\sqrt{c+d x}\right )}{15 d^4}\\ &=-\frac{16 b^2}{15 d^3 \sqrt{c+d x}}-\frac{2 \cos ^2(a+b x)}{5 d (c+d x)^{5/2}}+\frac{32 b^2 \cos ^2(a+b x)}{15 d^3 \sqrt{c+d x}}+\frac{32 b^{5/2} \sqrt{\pi } \cos \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right )}{15 d^{7/2}}+\frac{32 b^{5/2} \sqrt{\pi } C\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right ) \sin \left (2 a-\frac{2 b c}{d}\right )}{15 d^{7/2}}+\frac{8 b \cos (a+b x) \sin (a+b x)}{15 d^2 (c+d x)^{3/2}}\\ \end{align*}

Mathematica [A]  time = 1.28978, size = 244, normalized size = 1.13 \[ \frac{16 b^2 c^2 \cos (2 (a+b x))+32 b^2 c d x \cos (2 (a+b x))+16 b^2 d^2 x^2 \cos (2 (a+b x))+32 \sqrt{\pi } b d \left (\frac{b}{d}\right )^{3/2} (c+d x)^{5/2} \sin \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{\frac{b}{d}} \sqrt{c+d x}}{\sqrt{\pi }}\right )+32 \sqrt{\pi } b d \left (\frac{b}{d}\right )^{3/2} (c+d x)^{5/2} \cos \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{\frac{b}{d}} \sqrt{c+d x}}{\sqrt{\pi }}\right )+4 b c d \sin (2 (a+b x))+4 b d^2 x \sin (2 (a+b x))-3 d^2 \cos (2 (a+b x))-3 d^2}{15 d^3 (c+d x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^2/(c + d*x)^(7/2),x]

[Out]

(-3*d^2 + 16*b^2*c^2*Cos[2*(a + b*x)] - 3*d^2*Cos[2*(a + b*x)] + 32*b^2*c*d*x*Cos[2*(a + b*x)] + 16*b^2*d^2*x^
2*Cos[2*(a + b*x)] + 32*b*(b/d)^(3/2)*d*Sqrt[Pi]*(c + d*x)^(5/2)*Cos[2*a - (2*b*c)/d]*FresnelS[(2*Sqrt[b/d]*Sq
rt[c + d*x])/Sqrt[Pi]] + 32*b*(b/d)^(3/2)*d*Sqrt[Pi]*(c + d*x)^(5/2)*FresnelC[(2*Sqrt[b/d]*Sqrt[c + d*x])/Sqrt
[Pi]]*Sin[2*a - (2*b*c)/d] + 4*b*c*d*Sin[2*(a + b*x)] + 4*b*d^2*x*Sin[2*(a + b*x)])/(15*d^3*(c + d*x)^(5/2))

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Maple [A]  time = 0.037, size = 230, normalized size = 1.1 \begin{align*} 2\,{\frac{1}{d} \left ( -1/10\, \left ( dx+c \right ) ^{-5/2}-1/10\,{\frac{1}{ \left ( dx+c \right ) ^{5/2}}\cos \left ( 2\,{\frac{ \left ( dx+c \right ) b}{d}}+2\,{\frac{da-cb}{d}} \right ) }-2/5\,{\frac{b}{d} \left ( -1/3\,{\frac{1}{ \left ( dx+c \right ) ^{3/2}}\sin \left ( 2\,{\frac{ \left ( dx+c \right ) b}{d}}+2\,{\frac{da-cb}{d}} \right ) }+4/3\,{\frac{b}{d} \left ( -{\frac{1}{\sqrt{dx+c}}\cos \left ( 2\,{\frac{ \left ( dx+c \right ) b}{d}}+2\,{\frac{da-cb}{d}} \right ) }-2\,{\frac{b\sqrt{\pi }}{d} \left ( \cos \left ( 2\,{\frac{da-cb}{d}} \right ){\it FresnelS} \left ( 2\,{\frac{b\sqrt{dx+c}}{\sqrt{\pi }d}{\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) +\sin \left ( 2\,{\frac{da-cb}{d}} \right ){\it FresnelC} \left ( 2\,{\frac{b\sqrt{dx+c}}{\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(cos(b*x+a)^2/(d*x+c)^(7/2),x)

[Out]

2/d*(-1/10/(d*x+c)^(5/2)-1/10/(d*x+c)^(5/2)*cos(2/d*(d*x+c)*b+2*(a*d-b*c)/d)-2/5*b/d*(-1/3/(d*x+c)^(3/2)*sin(2
/d*(d*x+c)*b+2*(a*d-b*c)/d)+4/3*b/d*(-1/(d*x+c)^(1/2)*cos(2/d*(d*x+c)*b+2*(a*d-b*c)/d)-2*b/d*Pi^(1/2)/(b/d)^(1
/2)*(cos(2*(a*d-b*c)/d)*FresnelS(2/Pi^(1/2)/(b/d)^(1/2)*(d*x+c)^(1/2)*b/d)+sin(2*(a*d-b*c)/d)*FresnelC(2/Pi^(1
/2)/(b/d)^(1/2)*(d*x+c)^(1/2)*b/d)))))

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Maxima [C]  time = 1.65191, size = 644, normalized size = 2.98 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2/(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

-1/10*(sqrt(2)*((5*(gamma(-5/2, 2*I*(d*x + c)*b/d) + gamma(-5/2, -2*I*(d*x + c)*b/d))*cos(5/4*pi + 5/2*arctan2
(0, b) + 5/2*arctan2(0, d/sqrt(d^2))) + 5*(gamma(-5/2, 2*I*(d*x + c)*b/d) + gamma(-5/2, -2*I*(d*x + c)*b/d))*c
os(-5/4*pi + 5/2*arctan2(0, b) + 5/2*arctan2(0, d/sqrt(d^2))) + (5*I*gamma(-5/2, 2*I*(d*x + c)*b/d) - 5*I*gamm
a(-5/2, -2*I*(d*x + c)*b/d))*sin(5/4*pi + 5/2*arctan2(0, b) + 5/2*arctan2(0, d/sqrt(d^2))) + (-5*I*gamma(-5/2,
2*I*(d*x + c)*b/d) + 5*I*gamma(-5/2, -2*I*(d*x + c)*b/d))*sin(-5/4*pi + 5/2*arctan2(0, b) + 5/2*arctan2(0, d/
sqrt(d^2))))*cos(-2*(b*c - a*d)/d) + ((-5*I*gamma(-5/2, 2*I*(d*x + c)*b/d) + 5*I*gamma(-5/2, -2*I*(d*x + c)*b/
d))*cos(5/4*pi + 5/2*arctan2(0, b) + 5/2*arctan2(0, d/sqrt(d^2))) + (-5*I*gamma(-5/2, 2*I*(d*x + c)*b/d) + 5*I
*gamma(-5/2, -2*I*(d*x + c)*b/d))*cos(-5/4*pi + 5/2*arctan2(0, b) + 5/2*arctan2(0, d/sqrt(d^2))) + 5*(gamma(-5
/2, 2*I*(d*x + c)*b/d) + gamma(-5/2, -2*I*(d*x + c)*b/d))*sin(5/4*pi + 5/2*arctan2(0, b) + 5/2*arctan2(0, d/sq
rt(d^2))) - 5*(gamma(-5/2, 2*I*(d*x + c)*b/d) + gamma(-5/2, -2*I*(d*x + c)*b/d))*sin(-5/4*pi + 5/2*arctan2(0,
b) + 5/2*arctan2(0, d/sqrt(d^2))))*sin(-2*(b*c - a*d)/d))*((d*x + c)*abs(b)/abs(d))^(5/2) + 2)/((d*x + c)^(5/2
)*d)

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Fricas [A]  time = 2.07002, size = 733, normalized size = 3.39 \begin{align*} \frac{2 \,{\left (16 \,{\left (\pi b^{2} d^{3} x^{3} + 3 \, \pi b^{2} c d^{2} x^{2} + 3 \, \pi b^{2} c^{2} d x + \pi b^{2} c^{3}\right )} \sqrt{\frac{b}{\pi d}} \cos \left (-\frac{2 \,{\left (b c - a d\right )}}{d}\right ) \operatorname{S}\left (2 \, \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) + 16 \,{\left (\pi b^{2} d^{3} x^{3} + 3 \, \pi b^{2} c d^{2} x^{2} + 3 \, \pi b^{2} c^{2} d x + \pi b^{2} c^{3}\right )} \sqrt{\frac{b}{\pi d}} \operatorname{C}\left (2 \, \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) \sin \left (-\frac{2 \,{\left (b c - a d\right )}}{d}\right ) -{\left (8 \, b^{2} d^{2} x^{2} + 16 \, b^{2} c d x + 8 \, b^{2} c^{2} -{\left (16 \, b^{2} d^{2} x^{2} + 32 \, b^{2} c d x + 16 \, b^{2} c^{2} - 3 \, d^{2}\right )} \cos \left (b x + a\right )^{2} - 4 \,{\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right )\right )} \sqrt{d x + c}\right )}}{15 \,{\left (d^{6} x^{3} + 3 \, c d^{5} x^{2} + 3 \, c^{2} d^{4} x + c^{3} d^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2/(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

2/15*(16*(pi*b^2*d^3*x^3 + 3*pi*b^2*c*d^2*x^2 + 3*pi*b^2*c^2*d*x + pi*b^2*c^3)*sqrt(b/(pi*d))*cos(-2*(b*c - a*
d)/d)*fresnel_sin(2*sqrt(d*x + c)*sqrt(b/(pi*d))) + 16*(pi*b^2*d^3*x^3 + 3*pi*b^2*c*d^2*x^2 + 3*pi*b^2*c^2*d*x
+ pi*b^2*c^3)*sqrt(b/(pi*d))*fresnel_cos(2*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-2*(b*c - a*d)/d) - (8*b^2*d^2*x
^2 + 16*b^2*c*d*x + 8*b^2*c^2 - (16*b^2*d^2*x^2 + 32*b^2*c*d*x + 16*b^2*c^2 - 3*d^2)*cos(b*x + a)^2 - 4*(b*d^2
*x + b*c*d)*cos(b*x + a)*sin(b*x + a))*sqrt(d*x + c))/(d^6*x^3 + 3*c*d^5*x^2 + 3*c^2*d^4*x + c^3*d^3)

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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(cos(b*x+a)**2/(d*x+c)**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (b x + a\right )^{2}}{{\left (d x + c\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2/(d*x+c)^(7/2),x, algorithm="giac")

[Out]

integrate(cos(b*x + a)^2/(d*x + c)^(7/2), x)