3.1.50 \(\int \frac {\sin ^2(a+b x)}{(c+d x)^{5/2}} \, dx\) [50]

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

[Out]

-2/3*sin(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 = 0.22, 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 = {3395, 32, 3393, 3387, 3386, 3432, 3385, 3433} \begin {gather*} \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 \sin ^2(a+b x)}{3 d (c+d x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^2/(c + d*x)^(5/2),x]

[Out]

(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]*Sqrt[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]) - (2*Sin[a + b*x]^2)/(3*d*(c + d*x)^(3/2))

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 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 3395

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

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Mathematica [A]
time = 0.92, size = 158, normalized size = 0.93 \begin {gather*} \frac {2 \left (4 b \sqrt {\frac {b}{d}} \sqrt {\pi } \cos \left (2 a-\frac {2 b c}{d}\right ) C\left (\frac {2 \sqrt {\frac {b}{d}} \sqrt {c+d x}}{\sqrt {\pi }}\right )-4 b \sqrt {\frac {b}{d}} \sqrt {\pi } S\left (\frac {2 \sqrt {\frac {b}{d}} \sqrt {c+d x}}{\sqrt {\pi }}\right ) \sin \left (2 a-\frac {2 b c}{d}\right )-\frac {\sin (a+b x) (4 b (c+d x) \cos (a+b x)+d \sin (a+b x))}{(c+d x)^{3/2}}\right )}{3 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.05, size = 189, normalized size = 1.11

method	result	size
derivativedivides	\(\frac {-\frac {1}{3 \left (d x +c \right )^{\frac {3}{2}}}+\frac {\cos \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 d a -2 c b}{d}\right )}{3 \left (d x +c \right )^{\frac {3}{2}}}+\frac {4 b \left (-\frac {\sin \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 d a -2 c b}{d}\right )}{\sqrt {d x +c}}+\frac {2 b \sqrt {\pi }\, \left (\cos \left (\frac {2 d a -2 c b}{d}\right ) \FresnelC \left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {2 d a -2 c b}{d}\right ) \mathrm {S}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{d \sqrt {\frac {b}{d}}}\right )}{3 d}}{d}\)	\(189\)
default	\(\frac {-\frac {1}{3 \left (d x +c \right )^{\frac {3}{2}}}+\frac {\cos \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 d a -2 c b}{d}\right )}{3 \left (d x +c \right )^{\frac {3}{2}}}+\frac {4 b \left (-\frac {\sin \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 d a -2 c b}{d}\right )}{\sqrt {d x +c}}+\frac {2 b \sqrt {\pi }\, \left (\cos \left (\frac {2 d a -2 c b}{d}\right ) \FresnelC \left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {2 d a -2 c b}{d}\right ) \mathrm {S}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{d \sqrt {\frac {b}{d}}}\right )}{3 d}}{d}\)	\(189\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)^2/(d*x+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.61, size = 136, normalized size = 0.80 \begin {gather*} -\frac {3 \, \sqrt {2} {\left ({\left (-\left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, \frac {2 i \, {\left (d x + c\right )} b}{d}\right ) + \left (i + 1\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 (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, \frac {2 i \, {\left (d x + c\right )} b}{d}\right ) + \left (i - 1\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*sqrt(2)*((-(I - 1)*sqrt(2)*gamma(-3/2, 2*I*(d*x + c)*b/d) + (I + 1)*sqrt(2)*gamma(-3/2, -2*I*(d*x + c
)*b/d))*cos(-2*(b*c - a*d)/d) + (-(I + 1)*sqrt(2)*gamma(-3/2, 2*I*(d*x + c)*b/d) + (I - 1)*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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Fricas [A]
time = 0.37, size = 209, normalized size = 1.23 \begin {gather*} \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 ) - d\right )} \sqrt {d x + c}\right )}}{3 \, {\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(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)*sqr
t(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) - d)*sqrt(d
*x + c))/(d^4*x^2 + 2*c*d^3*x + c^2*d^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sin ^{2}{\left (a + b x \right )}}{\left (c + d x\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\sin \left (a+b\,x\right )}^2}{{\left (c+d\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^2/(c + d*x)^(5/2),x)

[Out]

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

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