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

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

[Out]

-2/5*cos(b*x+a)^2/d/(d*x+c)^(5/2)+8/15*b*cos(b*x+a)*sin(b*x+a)/d^2/(d*x+c)^(3/2)+32/15*b^(5/2)*cos(2*a-2*b*c/d
)*FresnelS(2*b^(1/2)*(d*x+c)^(1/2)/d^(1/2)/Pi^(1/2))*Pi^(1/2)/d^(7/2)+32/15*b^(5/2)*FresnelC(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^(7/2)-16/15*b^2/d^3/(d*x+c)^(1/2)+32/15*b^2*cos(b*x+a)^2/
d^3/(d*x+c)^(1/2)

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Rubi [A]  time = 0.32, antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, 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 12

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

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Mathematica [A]  time = 1.27, size = 244, normalized size = 1.13 \[ \frac {16 b^2 c^2 \cos (2 (a+b x))+32 b^2 c d x \cos (2 (a+b x))+16 b^2 d^2 x^2 \cos (2 (a+b x))+32 \sqrt {\pi } b d \left (\frac {b}{d}\right )^{3/2} (c+d x)^{5/2} \sin \left (2 a-\frac {2 b c}{d}\right ) C\left (\frac {2 \sqrt {\frac {b}{d}} \sqrt {c+d x}}{\sqrt {\pi }}\right )+32 \sqrt {\pi } b d \left (\frac {b}{d}\right )^{3/2} (c+d x)^{5/2} \cos \left (2 a-\frac {2 b c}{d}\right ) S\left (\frac {2 \sqrt {\frac {b}{d}} \sqrt {c+d x}}{\sqrt {\pi }}\right )+4 b c d \sin (2 (a+b x))+4 b d^2 x \sin (2 (a+b x))-3 d^2 \cos (2 (a+b x))-3 d^2}{15 d^3 (c+d x)^{5/2}} \]

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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fricas [A]  time = 0.82, size = 323, normalized size = 1.50 \[ \frac {2 \, {\left (16 \, {\left (\pi b^{2} d^{3} x^{3} + 3 \, \pi b^{2} c d^{2} x^{2} + 3 \, \pi b^{2} c^{2} d x + \pi b^{2} c^{3}\right )} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {S}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) + 16 \, {\left (\pi b^{2} d^{3} x^{3} + 3 \, \pi b^{2} c d^{2} x^{2} + 3 \, \pi b^{2} c^{2} d x + \pi b^{2} c^{3}\right )} \sqrt {\frac {b}{\pi d}} \operatorname {C}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - {\left (8 \, b^{2} d^{2} x^{2} + 16 \, b^{2} c d x + 8 \, b^{2} c^{2} - {\left (16 \, b^{2} d^{2} x^{2} + 32 \, b^{2} c d x + 16 \, b^{2} c^{2} - 3 \, d^{2}\right )} \cos \left (b x + a\right )^{2} - 4 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right )\right )} \sqrt {d x + c}\right )}}{15 \, {\left (d^{6} x^{3} + 3 \, c d^{5} x^{2} + 3 \, c^{2} d^{4} x + c^{3} d^{3}\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (b x + a\right )^{2}}{{\left (d x + c\right )}^{\frac {7}{2}}}\,{d x} \]

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)

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maple [A]  time = 0.04, size = 230, normalized size = 1.06 \[ \frac {-\frac {1}{5 \left (d x +c \right )^{\frac {5}{2}}}-\frac {\cos \left (\frac {2 \left (d x +c \right ) b}{d}+\frac {2 d a -2 c b}{d}\right )}{5 \left (d x +c \right )^{\frac {5}{2}}}-\frac {4 b \left (-\frac {\sin \left (\frac {2 \left (d x +c \right ) b}{d}+\frac {2 d a -2 c b}{d}\right )}{3 \left (d x +c \right )^{\frac {3}{2}}}+\frac {4 b \left (-\frac {\cos \left (\frac {2 \left (d x +c \right ) b}{d}+\frac {2 d a -2 c b}{d}\right )}{\sqrt {d x +c}}-\frac {2 b \sqrt {\pi }\, \left (\cos \left (\frac {2 d a -2 c b}{d}\right ) \mathrm {S}\left (\frac {2 \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {2 d a -2 c b}{d}\right ) \FresnelC \left (\frac {2 \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{d \sqrt {\frac {b}{d}}}\right )}{3 d}\right )}{5 d}}{d} \]

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/d*b*(-1/3/(d*x+c)^(3/2)*sin(2
/d*(d*x+c)*b+2*(a*d-b*c)/d)+4/3/d*b*(-1/(d*x+c)^(1/2)*cos(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)*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/P
i^(1/2)/(1/d*b)^(1/2)*(d*x+c)^(1/2)*b/d)))))

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maxima [C]  time = 1.75, size = 135, normalized size = 0.62 \[ -\frac {\sqrt {2} {\left ({\left (-\left (5 i + 5\right ) \, \sqrt {2} \Gamma \left (-\frac {5}{2}, \frac {2 i \, {\left (d x + c\right )} b}{d}\right ) + \left (5 i - 5\right ) \, \sqrt {2} \Gamma \left (-\frac {5}{2}, -\frac {2 i \, {\left (d x + c\right )} b}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + {\left (\left (5 i - 5\right ) \, \sqrt {2} \Gamma \left (-\frac {5}{2}, \frac {2 i \, {\left (d x + c\right )} b}{d}\right ) - \left (5 i + 5\right ) \, \sqrt {2} \Gamma \left (-\frac {5}{2}, -\frac {2 i \, {\left (d x + c\right )} b}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \left (\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {5}{2}} + 2}{10 \, {\left (d x + c\right )}^{\frac {5}{2}} d} \]

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*I + 5)*sqrt(2)*gamma(-5/2, 2*I*(d*x + c)*b/d) + (5*I - 5)*sqrt(2)*gamma(-5/2, -2*I*(d*x +
c)*b/d))*cos(-2*(b*c - a*d)/d) + ((5*I - 5)*sqrt(2)*gamma(-5/2, 2*I*(d*x + c)*b/d) - (5*I + 5)*sqrt(2)*gamma(
-5/2, -2*I*(d*x + c)*b/d))*sin(-2*(b*c - a*d)/d))*((d*x + c)*b/d)^(5/2) + 2)/((d*x + c)^(5/2)*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (a+b\,x\right )}^2}{{\left (c+d\,x\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^2/(c + d*x)^(7/2),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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