3.1.50 \(\int \sqrt {c+d x} \cos ^2(a+b x) \, dx\) [50]

Optimal. Leaf size=158 \[ \frac {(c+d x)^{3/2}}{3 d}-\frac {\sqrt {d} \sqrt {\pi } \cos \left (2 a-\frac {2 b c}{d}\right ) S\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{8 b^{3/2}}-\frac {\sqrt {d} \sqrt {\pi } \text {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{8 b^{3/2}}+\frac {\sqrt {c+d x} \sin (2 a+2 b x)}{4 b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.18, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {3393, 3377, 3387, 3386, 3432, 3385, 3433} \begin {gather*} -\frac {\sqrt {\pi } \sqrt {d} \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 )}{8 b^{3/2}}-\frac {\sqrt {\pi } \sqrt {d} \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 )}{8 b^{3/2}}+\frac {\sqrt {c+d x} \sin (2 a+2 b x)}{4 b}+\frac {(c+d x)^{3/2}}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*x]*Cos[a + b*x]^2,x]

[Out]

(c + d*x)^(3/2)/(3*d) - (Sqrt[d]*Sqrt[Pi]*Cos[2*a - (2*b*c)/d]*FresnelS[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqr
t[Pi])])/(8*b^(3/2)) - (Sqrt[d]*Sqrt[Pi]*FresnelC[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])]*Sin[2*a - (2*b
*c)/d])/(8*b^(3/2)) + (Sqrt[c + d*x]*Sin[2*a + 2*b*x])/(4*b)

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3385

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 3386

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 3387

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 3393

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 3432

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 3433

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rubi steps

\begin {align*} \int \sqrt {c+d x} \cos ^2(a+b x) \, dx &=\int \left (\frac {1}{2} \sqrt {c+d x}+\frac {1}{2} \sqrt {c+d x} \cos (2 a+2 b x)\right ) \, dx\\ &=\frac {(c+d x)^{3/2}}{3 d}+\frac {1}{2} \int \sqrt {c+d x} \cos (2 a+2 b x) \, dx\\ &=\frac {(c+d x)^{3/2}}{3 d}+\frac {\sqrt {c+d x} \sin (2 a+2 b x)}{4 b}-\frac {d \int \frac {\sin (2 a+2 b x)}{\sqrt {c+d x}} \, dx}{8 b}\\ &=\frac {(c+d x)^{3/2}}{3 d}+\frac {\sqrt {c+d x} \sin (2 a+2 b x)}{4 b}-\frac {\left (d \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}{8 b}-\frac {\left (d \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}{8 b}\\ &=\frac {(c+d x)^{3/2}}{3 d}+\frac {\sqrt {c+d x} \sin (2 a+2 b x)}{4 b}-\frac {\cos \left (2 a-\frac {2 b c}{d}\right ) \text {Subst}\left (\int \sin \left (\frac {2 b x^2}{d}\right ) \, dx,x,\sqrt {c+d x}\right )}{4 b}-\frac {\sin \left (2 a-\frac {2 b c}{d}\right ) \text {Subst}\left (\int \cos \left (\frac {2 b x^2}{d}\right ) \, dx,x,\sqrt {c+d x}\right )}{4 b}\\ &=\frac {(c+d x)^{3/2}}{3 d}-\frac {\sqrt {d} \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 )}{8 b^{3/2}}-\frac {\sqrt {d} \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 )}{8 b^{3/2}}+\frac {\sqrt {c+d x} \sin (2 a+2 b x)}{4 b}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains complex when optimal does not.
time = 0.35, size = 146, normalized size = 0.92 \begin {gather*} \frac {1}{48} \sqrt {c+d x} \left (\frac {16 (c+d x)}{d}-\frac {3 i \sqrt {2} e^{2 i \left (a-\frac {b c}{d}\right )} \text {Gamma}\left (\frac {3}{2},-\frac {2 i b (c+d x)}{d}\right )}{b \sqrt {-\frac {i b (c+d x)}{d}}}+\frac {3 i \sqrt {2} e^{-2 i \left (a-\frac {b c}{d}\right )} \text {Gamma}\left (\frac {3}{2},\frac {2 i b (c+d x)}{d}\right )}{b \sqrt {\frac {i b (c+d x)}{d}}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x]*((16*(c + d*x))/d - ((3*I)*Sqrt[2]*E^((2*I)*(a - (b*c)/d))*Gamma[3/2, ((-2*I)*b*(c + d*x))/d])/
(b*Sqrt[((-I)*b*(c + d*x))/d]) + ((3*I)*Sqrt[2]*Gamma[3/2, ((2*I)*b*(c + d*x))/d])/(b*E^((2*I)*(a - (b*c)/d))*
Sqrt[(I*b*(c + d*x))/d])))/48

________________________________________________________________________________________

Maple [A]
time = 0.06, size = 150, normalized size = 0.95

method	result	size
derivativedivides	\(\frac {\frac {\left (d x +c \right )^{\frac {3}{2}}}{3}+\frac {d \sqrt {d x +c}\, \sin \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 d a -2 b c}{d}\right )}{4 b}-\frac {d \sqrt {\pi }\, \left (\cos \left (\frac {2 d a -2 b c}{d}\right ) \mathrm {S}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {2 d a -2 b c}{d}\right ) \FresnelC \left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{8 b \sqrt {\frac {b}{d}}}}{d}\)	\(150\)
default	\(\frac {\frac {\left (d x +c \right )^{\frac {3}{2}}}{3}+\frac {d \sqrt {d x +c}\, \sin \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 d a -2 b c}{d}\right )}{4 b}-\frac {d \sqrt {\pi }\, \left (\cos \left (\frac {2 d a -2 b c}{d}\right ) \mathrm {S}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {2 d a -2 b c}{d}\right ) \FresnelC \left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{8 b \sqrt {\frac {b}{d}}}}{d}\)	\(150\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*(1/6*(d*x+c)^(3/2)+1/8/b*d*(d*x+c)^(1/2)*sin(2/d*b*(d*x+c)+2*(a*d-b*c)/d)-1/16/b*d*Pi^(1/2)/(b/d)^(1/2)*(c
os(2*(a*d-b*c)/d)*FresnelS(2/Pi^(1/2)/(b/d)^(1/2)*b*(d*x+c)^(1/2)/d)+sin(2*(a*d-b*c)/d)*FresnelC(2/Pi^(1/2)/(b
/d)^(1/2)*b*(d*x+c)^(1/2)/d)))

________________________________________________________________________________________

Maxima [C] Result contains complex when optimal does not.
time = 0.53, size = 229, normalized size = 1.45 \begin {gather*} \frac {\sqrt {2} {\left (\frac {32 \, \sqrt {2} {\left (d x + c\right )}^{\frac {3}{2}} b^{2}}{d} + 24 \, \sqrt {2} \sqrt {d x + c} b \sin \left (\frac {2 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right ) - 3 \, {\left (\left (i + 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d \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 } d \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 ) - 3 \, {\left (-\left (i - 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d \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 } d \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 )\right )}}{192 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*sqrt(2)*(32*sqrt(2)*(d*x + c)^(3/2)*b^2/d + 24*sqrt(2)*sqrt(d*x + c)*b*sin(2*((d*x + c)*b - b*c + a*d)/d
) - 3*((I + 1)*4^(1/4)*sqrt(pi)*d*(b^2/d^2)^(1/4)*cos(-2*(b*c - a*d)/d) - (I - 1)*4^(1/4)*sqrt(pi)*d*(b^2/d^2)
^(1/4)*sin(-2*(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(2*I*b/d)) - 3*(-(I - 1)*4^(1/4)*sqrt(pi)*d*(b^2/d^2)^(1/4
)*cos(-2*(b*c - a*d)/d) + (I + 1)*4^(1/4)*sqrt(pi)*d*(b^2/d^2)^(1/4)*sin(-2*(b*c - a*d)/d))*erf(sqrt(d*x + c)*
sqrt(-2*I*b/d)))/b^2

________________________________________________________________________________________

Fricas [A]
time = 0.44, size = 148, normalized size = 0.94 \begin {gather*} -\frac {3 \, \pi d^{2} \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 ) + 3 \, \pi d^{2} \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 ) - 4 \, {\left (2 \, b^{2} d x + 3 \, b d \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 2 \, b^{2} c\right )} \sqrt {d x + c}}{24 \, b^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(3*pi*d^2*sqrt(b/(pi*d))*cos(-2*(b*c - a*d)/d)*fresnel_sin(2*sqrt(d*x + c)*sqrt(b/(pi*d))) + 3*pi*d^2*sq
rt(b/(pi*d))*fresnel_cos(2*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-2*(b*c - a*d)/d) - 4*(2*b^2*d*x + 3*b*d*cos(b*x
+ a)*sin(b*x + a) + 2*b^2*c)*sqrt(d*x + c))/(b^2*d)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [C] Result contains complex when optimal does not.
time = 0.52, size = 434, normalized size = 2.75 \begin {gather*} -\frac {12 \, {\left (\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 \, {\left (i \, b c - i \, a d\right )}}{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 \, {\left (-i \, b c + i \, a d\right )}}{d}\right )}}{\sqrt {b d} {\left (\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}} - 4 \, \sqrt {d x + c}\right )} c - \frac {3 \, \sqrt {\pi } {\left (4 \, b c - i \, d\right )} 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 \, {\left (i \, b c - i \, a d\right )}}{d}\right )}}{\sqrt {b d} {\left (-\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )} b} - \frac {3 \, \sqrt {\pi } {\left (4 \, b c + i \, d\right )} 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 \, {\left (-i \, b c + i \, a d\right )}}{d}\right )}}{\sqrt {b d} {\left (\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )} b} - 16 \, {\left (d x + c\right )}^{\frac {3}{2}} + 48 \, \sqrt {d x + c} c - \frac {6 i \, \sqrt {d x + c} d e^{\left (-\frac {2 \, {\left (i \, {\left (d x + c\right )} b - i \, b c + i \, a d\right )}}{d}\right )}}{b} + \frac {6 i \, \sqrt {d x + c} d e^{\left (-\frac {2 \, {\left (-i \, {\left (d x + c\right )} b + i \, b c - i \, a d\right )}}{d}\right )}}{b}}{48 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(12*(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 - I*a*d)/d)/(sqr
t(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 + I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)) - 4*sqrt(d*x + c))*c - 3*sqrt(pi)*(4*b*c - I*d)*d*
erf(-sqrt(b*d)*sqrt(d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^(-2*(I*b*c - I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^
2*d^2) + 1)*b) - 3*sqrt(pi)*(4*b*c + I*d)*d*erf(-sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^(-2*(-
I*b*c + I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)*b) - 16*(d*x + c)^(3/2) + 48*sqrt(d*x + c)*c - 6*I*sqrt
(d*x + c)*d*e^(-2*(I*(d*x + c)*b - I*b*c + I*a*d)/d)/b + 6*I*sqrt(d*x + c)*d*e^(-2*(-I*(d*x + c)*b + I*b*c - I
*a*d)/d)/b)/d

________________________________________________________________________________________

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

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)

________________________________________________________________________________________