3.51 \(\int \frac {\cos ^2(a+b x)}{\sqrt {c+d x}} \, dx\)

Optimal. Leaf size=130 \[ \frac {\sqrt {\pi } \cos \left (2 a-\frac {2 b c}{d}\right ) C\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{2 \sqrt {b} \sqrt {d}}-\frac {\sqrt {\pi } \sin \left (2 a-\frac {2 b c}{d}\right ) S\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{2 \sqrt {b} \sqrt {d}}+\frac {\sqrt {c+d x}}{d} \]

[Out]

1/2*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)/b^(1/2)/d^(1/2)-1/2*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)/b^(1/2)/d^(1/2)+(d*x+c)^(1/2)/d

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Rubi [A]  time = 0.24, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3312, 3306, 3305, 3351, 3304, 3352} \[ \frac {\sqrt {\pi } \cos \left (2 a-\frac {2 b c}{d}\right ) \text {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {\pi } \sqrt {d}}\right )}{2 \sqrt {b} \sqrt {d}}-\frac {\sqrt {\pi } \sin \left (2 a-\frac {2 b c}{d}\right ) S\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{2 \sqrt {b} \sqrt {d}}+\frac {\sqrt {c+d x}}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^2/Sqrt[c + d*x],x]

[Out]

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

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 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)}{\sqrt {c+d x}} \, dx &=\int \left (\frac {1}{2 \sqrt {c+d x}}+\frac {\cos (2 a+2 b x)}{2 \sqrt {c+d x}}\right ) \, dx\\ &=\frac {\sqrt {c+d x}}{d}+\frac {1}{2} \int \frac {\cos (2 a+2 b x)}{\sqrt {c+d x}} \, dx\\ &=\frac {\sqrt {c+d x}}{d}+\frac {1}{2} \cos \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\cos \left (\frac {2 b c}{d}+2 b x\right )}{\sqrt {c+d x}} \, dx-\frac {1}{2} \sin \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x\right )}{\sqrt {c+d x}} \, dx\\ &=\frac {\sqrt {c+d x}}{d}+\frac {\cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {Subst}\left (\int \cos \left (\frac {2 b x^2}{d}\right ) \, dx,x,\sqrt {c+d x}\right )}{d}-\frac {\sin \left (2 a-\frac {2 b c}{d}\right ) \operatorname {Subst}\left (\int \sin \left (\frac {2 b x^2}{d}\right ) \, dx,x,\sqrt {c+d x}\right )}{d}\\ &=\frac {\sqrt {c+d x}}{d}+\frac {\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 )}{2 \sqrt {b} \sqrt {d}}-\frac {\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 )}{2 \sqrt {b} \sqrt {d}}\\ \end {align*}

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Mathematica [C]  time = 0.22, size = 145, normalized size = 1.12 \[ \frac {-\frac {i \sqrt {2} e^{2 i \left (a-\frac {b c}{d}\right )} \sqrt {-\frac {i b (c+d x)}{d}} \Gamma \left (\frac {1}{2},-\frac {2 i b (c+d x)}{d}\right )}{b}+\frac {i \sqrt {2} e^{-2 i \left (a-\frac {b c}{d}\right )} \sqrt {\frac {i b (c+d x)}{d}} \Gamma \left (\frac {1}{2},\frac {2 i b (c+d x)}{d}\right )}{b}+8 \left (\frac {c}{d}+x\right )}{8 \sqrt {c+d x}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^2/Sqrt[c + d*x],x]

[Out]

(8*(c/d + x) - (I*Sqrt[2]*E^((2*I)*(a - (b*c)/d))*Sqrt[((-I)*b*(c + d*x))/d]*Gamma[1/2, ((-2*I)*b*(c + d*x))/d
])/b + (I*Sqrt[2]*Sqrt[(I*b*(c + d*x))/d]*Gamma[1/2, ((2*I)*b*(c + d*x))/d])/(b*E^((2*I)*(a - (b*c)/d))))/(8*S
qrt[c + d*x])

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fricas [A]  time = 0.50, size = 114, normalized size = 0.88 \[ \frac {\pi d \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {C}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) - \pi d \sqrt {\frac {b}{\pi d}} \operatorname {S}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + 2 \, \sqrt {d x + c} b}{2 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2/(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

1/2*(pi*d*sqrt(b/(pi*d))*cos(-2*(b*c - a*d)/d)*fresnel_cos(2*sqrt(d*x + c)*sqrt(b/(pi*d))) - pi*d*sqrt(b/(pi*d
))*fresnel_sin(2*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-2*(b*c - a*d)/d) + 2*sqrt(d*x + c)*b)/(b*d)

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giac [C]  time = 0.49, size = 163, normalized size = 1.25 \[ -\frac {\frac {\sqrt {\pi } d \operatorname {erf}\left (-\frac {\sqrt {b d} \sqrt {d x + c} {\left (\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}{d}\right ) e^{\left (\frac {2 i \, b c - 2 i \, a d}{d}\right )}}{\sqrt {b d} {\left (\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}} + \frac {\sqrt {\pi } d \operatorname {erf}\left (-\frac {\sqrt {b d} \sqrt {d x + c} {\left (-\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}{d}\right ) e^{\left (\frac {-2 i \, b c + 2 i \, a d}{d}\right )}}{\sqrt {b d} {\left (-\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}} - 4 \, \sqrt {d x + c}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2/(d*x+c)^(1/2),x, algorithm="giac")

[Out]

-1/4*(sqrt(pi)*d*erf(-sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((2*I*b*c - 2*I*a*d)/d)/(sqrt(b*d
)*(I*b*d/sqrt(b^2*d^2) + 1)) + sqrt(pi)*d*erf(-sqrt(b*d)*sqrt(d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((-2*I*
b*c + 2*I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) + 1)) - 4*sqrt(d*x + c))/d

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maple [A]  time = 0.06, size = 108, normalized size = 0.83 \[ \frac {\sqrt {d x +c}+\frac {\sqrt {\pi }\, \left (\cos \left (\frac {2 d a -2 c b}{d}\right ) \FresnelC \left (\frac {2 \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {2 d a -2 c b}{d}\right ) \mathrm {S}\left (\frac {2 \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{2 \sqrt {\frac {b}{d}}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^2/(d*x+c)^(1/2),x)

[Out]

2/d*(1/2*(d*x+c)^(1/2)+1/4*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 = 1.44, size = 187, normalized size = 1.44 \[ -\frac {\sqrt {2} {\left ({\left (\left (i - 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + \left (i + 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {2 i \, b}{d}}\right ) + {\left (-\left (i + 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - \left (i - 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {2 i \, b}{d}}\right ) - \frac {8 \, \sqrt {2} \sqrt {d x + c} b}{d}\right )}}{16 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2/(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

-1/16*sqrt(2)*(((I - 1)*4^(1/4)*sqrt(pi)*(b^2/d^2)^(1/4)*cos(-2*(b*c - a*d)/d) + (I + 1)*4^(1/4)*sqrt(pi)*(b^2
/d^2)^(1/4)*sin(-2*(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(2*I*b/d)) + (-(I + 1)*4^(1/4)*sqrt(pi)*(b^2/d^2)^(1/
4)*cos(-2*(b*c - a*d)/d) - (I - 1)*4^(1/4)*sqrt(pi)*(b^2/d^2)^(1/4)*sin(-2*(b*c - a*d)/d))*erf(sqrt(d*x + c)*s
qrt(-2*I*b/d)) - 8*sqrt(2)*sqrt(d*x + c)*b/d)/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^2/(c + d*x)^(1/2),x)

[Out]

int(cos(a + b*x)^2/(c + d*x)^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**2/(d*x+c)**(1/2),x)

[Out]

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

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