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

Optimal. Leaf size=170 \[ -\frac {8 \sqrt {\pi } b^{3/2} \cos \left (2 a-\frac {2 b c}{d}\right ) C\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{3 d^{5/2}}+\frac {8 \sqrt {\pi } b^{3/2} \sin \left (2 a-\frac {2 b c}{d}\right ) S\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{3 d^{5/2}}+\frac {8 b \sin (a+b x) \cos (a+b x)}{3 d^2 \sqrt {c+d x}}-\frac {2 \cos ^2(a+b x)}{3 d (c+d x)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.31, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3314, 32, 3312, 3306, 3305, 3351, 3304, 3352} \[ -\frac {8 \sqrt {\pi } b^{3/2} \cos \left (2 a-\frac {2 b c}{d}\right ) \text {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {\pi } \sqrt {d}}\right )}{3 d^{5/2}}+\frac {8 \sqrt {\pi } b^{3/2} \sin \left (2 a-\frac {2 b c}{d}\right ) S\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{3 d^{5/2}}+\frac {8 b \sin (a+b x) \cos (a+b x)}{3 d^2 \sqrt {c+d x}}-\frac {2 \cos ^2(a+b x)}{3 d (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^2/(c + d*x)^(5/2),x]

[Out]

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

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 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 3312

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

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

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Mathematica [C]  time = 1.39, size = 181, normalized size = 1.06 \[ \frac {e^{-\frac {2 i (a d+b c)}{d}} \left (-2 \sqrt {2} e^{4 i a} d \left (-\frac {i b (c+d x)}{d}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {2 i b (c+d x)}{d}\right )+2 e^{\frac {2 i (a d+b c)}{d}} \left (2 b (c+d x) \sin (2 (a+b x))-d \cos ^2(a+b x)\right )-2 \sqrt {2} d e^{\frac {4 i b c}{d}} \left (\frac {i b (c+d x)}{d}\right )^{3/2} \Gamma \left (\frac {1}{2},\frac {2 i b (c+d x)}{d}\right )\right )}{3 d^2 (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.68, size = 206, normalized size = 1.21 \[ -\frac {2 \, {\left (4 \, {\left (\pi b d^{2} x^{2} + 2 \, \pi b c d x + \pi b c^{2}\right )} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {C}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) - 4 \, {\left (\pi b d^{2} x^{2} + 2 \, \pi b c d x + \pi b c^{2}\right )} \sqrt {\frac {b}{\pi d}} \operatorname {S}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + {\left (d \cos \left (b x + a\right )^{2} - 4 \, {\left (b d x + b c\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right )\right )} \sqrt {d x + c}\right )}}{3 \, {\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 189, normalized size = 1.11 \[ \frac {-\frac {1}{3 \left (d x +c \right )^{\frac {3}{2}}}-\frac {\cos \left (\frac {2 \left (d x +c \right ) b}{d}+\frac {2 d a -2 c b}{d}\right )}{3 \left (d x +c \right )^{\frac {3}{2}}}-\frac {4 b \left (-\frac {\sin \left (\frac {2 \left (d x +c \right ) b}{d}+\frac {2 d a -2 c b}{d}\right )}{\sqrt {d x +c}}+\frac {2 b \sqrt {\pi }\, \left (\cos \left (\frac {2 d a -2 c b}{d}\right ) \FresnelC \left (\frac {2 \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {2 d a -2 c b}{d}\right ) \mathrm {S}\left (\frac {2 \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{d \sqrt {\frac {b}{d}}}\right )}{3 d}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*(-1/6/(d*x+c)^(3/2)-1/6/(d*x+c)^(3/2)*cos(2/d*(d*x+c)*b+2*(a*d-b*c)/d)-2/3/d*b*(-1/(d*x+c)^(1/2)*sin(2/d*(
d*x+c)*b+2*(a*d-b*c)/d)+2/d*b*Pi^(1/2)/(1/d*b)^(1/2)*(cos(2*(a*d-b*c)/d)*FresnelC(2/Pi^(1/2)/(1/d*b)^(1/2)*(d*
x+c)^(1/2)*b/d)-sin(2*(a*d-b*c)/d)*FresnelS(2/Pi^(1/2)/(1/d*b)^(1/2)*(d*x+c)^(1/2)*b/d))))

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maxima [C]  time = 2.21, size = 135, normalized size = 0.79 \[ -\frac {\sqrt {2} {\left ({\left (\left (3 i - 3\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, \frac {2 i \, {\left (d x + c\right )} b}{d}\right ) - \left (3 i + 3\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, -\frac {2 i \, {\left (d x + c\right )} b}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + {\left (\left (3 i + 3\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, \frac {2 i \, {\left (d x + c\right )} b}{d}\right ) - \left (3 i - 3\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, -\frac {2 i \, {\left (d x + c\right )} b}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \left (\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {3}{2}} + 4}{12 \, {\left (d x + c\right )}^{\frac {3}{2}} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(sqrt(2)*(((3*I - 3)*sqrt(2)*gamma(-3/2, 2*I*(d*x + c)*b/d) - (3*I + 3)*sqrt(2)*gamma(-3/2, -2*I*(d*x +
c)*b/d))*cos(-2*(b*c - a*d)/d) + ((3*I + 3)*sqrt(2)*gamma(-3/2, 2*I*(d*x + c)*b/d) - (3*I - 3)*sqrt(2)*gamma(-
3/2, -2*I*(d*x + c)*b/d))*sin(-2*(b*c - a*d)/d))*((d*x + c)*b/d)^(3/2) + 4)/((d*x + c)^(3/2)*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^2/(c + d*x)^(5/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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