3.52 \(\int \frac {\cos ^2(a+b x)}{(c+d x)^{3/2}} \, dx\)

Optimal. Leaf size=135 \[ -\frac {2 \sqrt {\pi } \sqrt {b} \sin \left (2 a-\frac {2 b c}{d}\right ) C\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{d^{3/2}}-\frac {2 \sqrt {\pi } \sqrt {b} \cos \left (2 a-\frac {2 b c}{d}\right ) S\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{d^{3/2}}-\frac {2 \cos ^2(a+b x)}{d \sqrt {c+d x}} \]

[Out]

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

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Rubi [A]  time = 0.26, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {3313, 12, 3306, 3305, 3351, 3304, 3352} \[ -\frac {2 \sqrt {\pi } \sqrt {b} \sin \left (2 a-\frac {2 b c}{d}\right ) \text {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {\pi } \sqrt {d}}\right )}{d^{3/2}}-\frac {2 \sqrt {\pi } \sqrt {b} \cos \left (2 a-\frac {2 b c}{d}\right ) S\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{d^{3/2}}-\frac {2 \cos ^2(a+b x)}{d \sqrt {c+d x}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^2/(c + d*x)^(3/2),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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 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 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 \frac {\cos ^2(a+b x)}{(c+d x)^{3/2}} \, dx &=-\frac {2 \cos ^2(a+b x)}{d \sqrt {c+d x}}+\frac {(4 b) \int -\frac {\sin (2 a+2 b x)}{2 \sqrt {c+d x}} \, dx}{d}\\ &=-\frac {2 \cos ^2(a+b x)}{d \sqrt {c+d x}}-\frac {(2 b) \int \frac {\sin (2 a+2 b x)}{\sqrt {c+d x}} \, dx}{d}\\ &=-\frac {2 \cos ^2(a+b x)}{d \sqrt {c+d x}}-\frac {\left (2 b \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}{d}-\frac {\left (2 b \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}{d}\\ &=-\frac {2 \cos ^2(a+b x)}{d \sqrt {c+d x}}-\frac {\left (4 b \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 )}{d^2}-\frac {\left (4 b \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 )}{d^2}\\ &=-\frac {2 \cos ^2(a+b x)}{d \sqrt {c+d x}}-\frac {2 \sqrt {b} \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 )}{d^{3/2}}-\frac {2 \sqrt {b} \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 )}{d^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.66, size = 133, normalized size = 0.99 \[ \frac {2 \left (-\sqrt {\pi } \sqrt {\frac {b}{d}} \sin \left (2 a-\frac {2 b c}{d}\right ) C\left (\frac {2 \sqrt {\frac {b}{d}} \sqrt {c+d x}}{\sqrt {\pi }}\right )-\sqrt {\pi } \sqrt {\frac {b}{d}} \cos \left (2 a-\frac {2 b c}{d}\right ) S\left (\frac {2 \sqrt {\frac {b}{d}} \sqrt {c+d x}}{\sqrt {\pi }}\right )-\frac {\cos ^2(a+b x)}{\sqrt {c+d x}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^2/(c + d*x)^(3/2),x]

[Out]

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

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fricas [A]  time = 0.55, size = 136, normalized size = 1.01 \[ -\frac {2 \, {\left ({\left (\pi d x + \pi c\right )} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {S}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) + {\left (\pi d x + \pi c\right )} \sqrt {\frac {b}{\pi d}} \operatorname {C}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + \sqrt {d x + c} \cos \left (b x + a\right )^{2}\right )}}{d^{2} x + c d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2/(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

-2*((pi*d*x + pi*c)*sqrt(b/(pi*d))*cos(-2*(b*c - a*d)/d)*fresnel_sin(2*sqrt(d*x + c)*sqrt(b/(pi*d))) + (pi*d*x
+ pi*c)*sqrt(b/(pi*d))*fresnel_cos(2*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-2*(b*c - a*d)/d) + sqrt(d*x + c)*cos(
b*x + a)^2)/(d^2*x + c*d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (b x + a\right )^{2}}{{\left (d x + c\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2/(d*x+c)^(3/2),x, algorithm="giac")

[Out]

integrate(cos(b*x + a)^2/(d*x + c)^(3/2), x)

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maple [A]  time = 0.05, size = 146, normalized size = 1.08 \[ \frac {-\frac {1}{\sqrt {d x +c}}-\frac {\cos \left (\frac {2 \left (d x +c \right ) b}{d}+\frac {2 d a -2 c b}{d}\right )}{\sqrt {d x +c}}-\frac {2 b \sqrt {\pi }\, \left (\cos \left (\frac {2 d a -2 c b}{d}\right ) \mathrm {S}\left (\frac {2 \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {2 d a -2 c b}{d}\right ) \FresnelC \left (\frac {2 \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{d \sqrt {\frac {b}{d}}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^2/(d*x+c)^(3/2),x)

[Out]

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

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maxima [C]  time = 1.82, size = 135, normalized size = 1.00 \[ \frac {\sqrt {2} {\left ({\left (-\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, \frac {2 i \, {\left (d x + c\right )} b}{d}\right ) + \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -\frac {2 i \, {\left (d x + c\right )} b}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + {\left (\left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, \frac {2 i \, {\left (d x + c\right )} b}{d}\right ) - \left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -\frac {2 i \, {\left (d x + c\right )} b}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \sqrt {\frac {{\left (d x + c\right )} b}{d}} - 8}{8 \, \sqrt {d x + c} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2/(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

1/8*(sqrt(2)*((-(I + 1)*sqrt(2)*gamma(-1/2, 2*I*(d*x + c)*b/d) + (I - 1)*sqrt(2)*gamma(-1/2, -2*I*(d*x + c)*b/
d))*cos(-2*(b*c - a*d)/d) + ((I - 1)*sqrt(2)*gamma(-1/2, 2*I*(d*x + c)*b/d) - (I + 1)*sqrt(2)*gamma(-1/2, -2*I
*(d*x + c)*b/d))*sin(-2*(b*c - a*d)/d))*sqrt((d*x + c)*b/d) - 8)/(sqrt(d*x + c)*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (a+b\,x\right )}^2}{{\left (c+d\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^2/(c + d*x)^(3/2),x)

[Out]

int(cos(a + b*x)^2/(c + d*x)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos ^{2}{\left (a + b x \right )}}{\left (c + d x\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**2/(d*x+c)**(3/2),x)

[Out]

Integral(cos(a + b*x)**2/(c + d*x)**(3/2), x)

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