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

Optimal. Leaf size=193 \[ \frac {8 \sqrt {2 \pi } b^{5/2} \sin \left (a-\frac {b c}{d}\right ) C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{15 d^{7/2}}+\frac {8 \sqrt {2 \pi } b^{5/2} \cos \left (a-\frac {b c}{d}\right ) S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{15 d^{7/2}}+\frac {8 b^2 \cos (a+b x)}{15 d^3 \sqrt {c+d x}}+\frac {4 b \sin (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \cos (a+b x)}{5 d (c+d x)^{5/2}} \]

[Out]

-2/5*cos(b*x+a)/d/(d*x+c)^(5/2)+4/15*b*sin(b*x+a)/d^2/(d*x+c)^(3/2)+8/15*b^(5/2)*cos(a-b*c/d)*FresnelS(b^(1/2)
*2^(1/2)/Pi^(1/2)*(d*x+c)^(1/2)/d^(1/2))*2^(1/2)*Pi^(1/2)/d^(7/2)+8/15*b^(5/2)*FresnelC(b^(1/2)*2^(1/2)/Pi^(1/
2)*(d*x+c)^(1/2)/d^(1/2))*sin(a-b*c/d)*2^(1/2)*Pi^(1/2)/d^(7/2)+8/15*b^2*cos(b*x+a)/d^3/(d*x+c)^(1/2)

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Rubi [A]  time = 0.30, antiderivative size = 193, normalized size of antiderivative = 1.00, 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 = {3297, 3306, 3305, 3351, 3304, 3352} \[ \frac {8 \sqrt {2 \pi } b^{5/2} \sin \left (a-\frac {b c}{d}\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{15 d^{7/2}}+\frac {8 \sqrt {2 \pi } b^{5/2} \cos \left (a-\frac {b c}{d}\right ) S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{15 d^{7/2}}+\frac {8 b^2 \cos (a+b x)}{15 d^3 \sqrt {c+d x}}+\frac {4 b \sin (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \cos (a+b x)}{5 d (c+d x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]/(c + d*x)^(7/2),x]

[Out]

(-2*Cos[a + b*x])/(5*d*(c + d*x)^(5/2)) + (8*b^2*Cos[a + b*x])/(15*d^3*Sqrt[c + d*x]) + (8*b^(5/2)*Sqrt[2*Pi]*
Cos[a - (b*c)/d]*FresnelS[(Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(15*d^(7/2)) + (8*b^(5/2)*Sqrt[2*Pi]*Fr
esnelC[(Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[a - (b*c)/d])/(15*d^(7/2)) + (4*b*Sin[a + b*x])/(15*d^2
*(c + d*x)^(3/2))

Rule 3297

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[((c + d*x)^(m + 1)*Sin[e + f*x])/(d*(
m + 1)), x] - Dist[f/(d*(m + 1)), Int[(c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[
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 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 (a+b x)}{(c+d x)^{7/2}} \, dx &=-\frac {2 \cos (a+b x)}{5 d (c+d x)^{5/2}}-\frac {(2 b) \int \frac {\sin (a+b x)}{(c+d x)^{5/2}} \, dx}{5 d}\\ &=-\frac {2 \cos (a+b x)}{5 d (c+d x)^{5/2}}+\frac {4 b \sin (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {\left (4 b^2\right ) \int \frac {\cos (a+b x)}{(c+d x)^{3/2}} \, dx}{15 d^2}\\ &=-\frac {2 \cos (a+b x)}{5 d (c+d x)^{5/2}}+\frac {8 b^2 \cos (a+b x)}{15 d^3 \sqrt {c+d x}}+\frac {4 b \sin (a+b x)}{15 d^2 (c+d x)^{3/2}}+\frac {\left (8 b^3\right ) \int \frac {\sin (a+b x)}{\sqrt {c+d x}} \, dx}{15 d^3}\\ &=-\frac {2 \cos (a+b x)}{5 d (c+d x)^{5/2}}+\frac {8 b^2 \cos (a+b x)}{15 d^3 \sqrt {c+d x}}+\frac {4 b \sin (a+b x)}{15 d^2 (c+d x)^{3/2}}+\frac {\left (8 b^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}{15 d^3}+\frac {\left (8 b^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}{15 d^3}\\ &=-\frac {2 \cos (a+b x)}{5 d (c+d x)^{5/2}}+\frac {8 b^2 \cos (a+b x)}{15 d^3 \sqrt {c+d x}}+\frac {4 b \sin (a+b x)}{15 d^2 (c+d x)^{3/2}}+\frac {\left (16 b^3 \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 )}{15 d^4}+\frac {\left (16 b^3 \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 )}{15 d^4}\\ &=-\frac {2 \cos (a+b x)}{5 d (c+d x)^{5/2}}+\frac {8 b^2 \cos (a+b x)}{15 d^3 \sqrt {c+d x}}+\frac {8 b^{5/2} \sqrt {2 \pi } \cos \left (a-\frac {b c}{d}\right ) S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{15 d^{7/2}}+\frac {8 b^{5/2} \sqrt {2 \pi } C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right ) \sin \left (a-\frac {b c}{d}\right )}{15 d^{7/2}}+\frac {4 b \sin (a+b x)}{15 d^2 (c+d x)^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 0.34, size = 228, normalized size = 1.18 \[ \frac {e^{-i a} \left (2 e^{2 i a} \left (2 b e^{-\frac {i b c}{d}} (c+d x) \left (e^{\frac {i b (c+d x)}{d}} (2 b (c+d x)-i d)-2 i d \left (-\frac {i b (c+d x)}{d}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {i b (c+d x)}{d}\right )\right )-3 d^2 e^{i b x}\right )+e^{-i b x} \left (8 b^2 (c+d x)^2+8 d^2 e^{\frac {i b (c+d x)}{d}} \left (\frac {i b (c+d x)}{d}\right )^{5/2} \Gamma \left (\frac {1}{2},\frac {i b (c+d x)}{d}\right )+4 i b d (c+d x)-6 d^2\right )\right )}{30 d^3 (c+d x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*E^((2*I)*a)*(-3*d^2*E^(I*b*x) + (2*b*(c + d*x)*(E^((I*b*(c + d*x))/d)*((-I)*d + 2*b*(c + d*x)) - (2*I)*d*((
(-I)*b*(c + d*x))/d)^(3/2)*Gamma[1/2, ((-I)*b*(c + d*x))/d]))/E^((I*b*c)/d)) + (-6*d^2 + (4*I)*b*d*(c + d*x) +
8*b^2*(c + d*x)^2 + 8*d^2*E^((I*b*(c + d*x))/d)*((I*b*(c + d*x))/d)^(5/2)*Gamma[1/2, (I*b*(c + d*x))/d])/E^(I
*b*x))/(30*d^3*E^(I*a)*(c + d*x)^(5/2))

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fricas [A]  time = 0.66, size = 296, normalized size = 1.53 \[ \frac {2 \, {\left (4 \, \sqrt {2} {\left (\pi b^{2} d^{3} x^{3} + 3 \, \pi b^{2} c d^{2} x^{2} + 3 \, \pi b^{2} c^{2} d x + \pi b^{2} c^{3}\right )} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {b c - a d}{d}\right ) \operatorname {S}\left (\sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) + 4 \, \sqrt {2} {\left (\pi b^{2} d^{3} x^{3} + 3 \, \pi b^{2} c d^{2} x^{2} + 3 \, \pi b^{2} c^{2} d x + \pi b^{2} c^{3}\right )} \sqrt {\frac {b}{\pi d}} \operatorname {C}\left (\sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {b c - a d}{d}\right ) + \sqrt {d x + c} {\left ({\left (4 \, b^{2} d^{2} x^{2} + 8 \, b^{2} c d x + 4 \, b^{2} c^{2} - 3 \, d^{2}\right )} \cos \left (b x + a\right ) + 2 \, {\left (b d^{2} x + b c d\right )} \sin \left (b x + a\right )\right )}\right )}}{15 \, {\left (d^{6} x^{3} + 3 \, c d^{5} x^{2} + 3 \, c^{2} d^{4} x + c^{3} d^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(4*sqrt(2)*(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(-(b*c
- a*d)/d)*fresnel_sin(sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d))) + 4*sqrt(2)*(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(sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-(b*c -
a*d)/d) + sqrt(d*x + c)*((4*b^2*d^2*x^2 + 8*b^2*c*d*x + 4*b^2*c^2 - 3*d^2)*cos(b*x + a) + 2*(b*d^2*x + b*c*d)*
sin(b*x + a)))/(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 \[ \int \frac {\cos \left (b x + a\right )}{{\left (d x + c\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 220, normalized size = 1.14 \[ \frac {-\frac {2 \cos \left (\frac {\left (d x +c \right ) b}{d}+\frac {d a -c b}{d}\right )}{5 \left (d x +c \right )^{\frac {5}{2}}}-\frac {4 b \left (-\frac {\sin \left (\frac {\left (d x +c \right ) b}{d}+\frac {d a -c b}{d}\right )}{3 \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 b \left (-\frac {\cos \left (\frac {\left (d x +c \right ) b}{d}+\frac {d a -c b}{d}\right )}{\sqrt {d x +c}}-\frac {b \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {d a -c b}{d}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {d a -c b}{d}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{d \sqrt {\frac {b}{d}}}\right )}{3 d}\right )}{5 d}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 1.80, size = 129, normalized size = 0.67 \[ -\frac {{\left ({\left (-\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {5}{2}, \frac {i \, {\left (d x + c\right )} b}{d}\right ) + \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {5}{2}, -\frac {i \, {\left (d x + c\right )} b}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) + {\left (\left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {5}{2}, \frac {i \, {\left (d x + c\right )} b}{d}\right ) - \left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {5}{2}, -\frac {i \, {\left (d x + c\right )} b}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right )\right )} \left (\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {5}{2}}}{4 \, {\left (d x + c\right )}^{\frac {5}{2}} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*((-(I + 1)*sqrt(2)*gamma(-5/2, I*(d*x + c)*b/d) + (I - 1)*sqrt(2)*gamma(-5/2, -I*(d*x + c)*b/d))*cos(-(b*
c - a*d)/d) + ((I - 1)*sqrt(2)*gamma(-5/2, I*(d*x + c)*b/d) - (I + 1)*sqrt(2)*gamma(-5/2, -I*(d*x + c)*b/d))*s
in(-(b*c - a*d)/d))*((d*x + c)*b/d)^(5/2)/((d*x + c)^(5/2)*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\cos \left (a+b\,x\right )}{{\left (c+d\,x\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)/(c + d*x)^(7/2),x)

[Out]

int(cos(a + b*x)/(c + d*x)^(7/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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