\(\int \frac {\sin ^2(a+b x)}{(c+d x)^{7/2}} \, dx\) [51]

Optimal result

Integrand size = 18, antiderivative size = 216 \[ \int \frac {\sin ^2(a+b x)}{(c+d x)^{7/2}} \, dx=-\frac {16 b^2}{15 d^3 \sqrt {c+d x}}-\frac {32 b^{5/2} \sqrt {\pi } \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{15 d^{7/2}}-\frac {32 b^{5/2} \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{15 d^{7/2}}-\frac {8 b \cos (a+b x) \sin (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sin ^2(a+b x)}{5 d (c+d x)^{5/2}}+\frac {32 b^2 \sin ^2(a+b x)}{15 d^3 \sqrt {c+d x}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3395, 32, 3394, 12, 3387, 3386, 3432, 3385, 3433} \[ \int \frac {\sin ^2(a+b x)}{(c+d x)^{7/2}} \, dx=-\frac {32 \sqrt {\pi } b^{5/2} \sin \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{15 d^{7/2}}-\frac {32 \sqrt {\pi } b^{5/2} \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{15 d^{7/2}}+\frac {32 b^2 \sin ^2(a+b x)}{15 d^3 \sqrt {c+d x}}-\frac {8 b \sin (a+b x) \cos (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sin ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {16 b^2}{15 d^3 \sqrt {c+d x}} \]

[In]

Int[Sin[a + b*x]^2/(c + d*x)^(7/2),x]

[Out]

(-16*b^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)) - (2*Sin[a +
b*x]^2)/(5*d*(c + d*x)^(5/2)) + (32*b^2*Sin[a + b*x]^2)/(15*d^3*Sqrt[c + d*x])

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 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 3394

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 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*} \text {integral}& = -\frac {8 b \cos (a+b x) \sin (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sin ^2(a+b x)}{5 d (c+d x)^{5/2}}+\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 {\sin ^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 {8 b \cos (a+b x) \sin (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sin ^2(a+b x)}{5 d (c+d x)^{5/2}}+\frac {32 b^2 \sin ^2(a+b x)}{15 d^3 \sqrt {c+d x}}-\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 {8 b \cos (a+b x) \sin (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sin ^2(a+b x)}{5 d (c+d x)^{5/2}}+\frac {32 b^2 \sin ^2(a+b x)}{15 d^3 \sqrt {c+d x}}-\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 {8 b \cos (a+b x) \sin (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sin ^2(a+b x)}{5 d (c+d x)^{5/2}}+\frac {32 b^2 \sin ^2(a+b x)}{15 d^3 \sqrt {c+d x}}-\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 {8 b \cos (a+b x) \sin (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sin ^2(a+b x)}{5 d (c+d x)^{5/2}}+\frac {32 b^2 \sin ^2(a+b x)}{15 d^3 \sqrt {c+d x}}-\frac {\left (64 b^3 \cos \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 )}{15 d^4}-\frac {\left (64 b^3 \sin \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 )}{15 d^4} \\ & = -\frac {16 b^2}{15 d^3 \sqrt {c+d x}}-\frac {32 b^{5/2} \sqrt {\pi } \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelS}\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 } \operatorname {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 )}{15 d^{7/2}}-\frac {8 b \cos (a+b x) \sin (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sin ^2(a+b x)}{5 d (c+d x)^{5/2}}+\frac {32 b^2 \sin ^2(a+b x)}{15 d^3 \sqrt {c+d x}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.73 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.10 \[ \int \frac {\sin ^2(a+b x)}{(c+d x)^{7/2}} \, dx=\frac {-6 d^2+e^{2 i a} \left (3 d^2 e^{2 i b x}-4 b e^{-\frac {2 i b c}{d}} (c+d x) \left (e^{\frac {2 i b (c+d x)}{d}} (-i d+4 b (c+d x))-4 i \sqrt {2} d \left (-\frac {i b (c+d x)}{d}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {2 i b (c+d x)}{d}\right )\right )\right )+e^{-2 i (a+b x)} \left (3 d^2+2 i b (c+d x) \left (-2 d+8 i b (c+d x)-8 \sqrt {2} d e^{\frac {2 i b (c+d x)}{d}} \left (\frac {i b (c+d x)}{d}\right )^{3/2} \Gamma \left (\frac {1}{2},\frac {2 i b (c+d x)}{d}\right )\right )\right )}{30 d^3 (c+d x)^{5/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.06

method	result	size
derivativedivides	\(\frac {-\frac {1}{5 \left (d x +c \right )^{\frac {5}{2}}}+\frac {\cos \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 d a -2 c b}{d}\right )}{5 \left (d x +c \right )^{\frac {5}{2}}}+\frac {4 b \left (-\frac {\sin \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 d a -2 c b}{d}\right )}{3 \left (d x +c \right )^{\frac {3}{2}}}+\frac {4 b \left (-\frac {\cos \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 d a -2 c b}{d}\right )}{\sqrt {d x +c}}-\frac {2 b \sqrt {\pi }\, \left (\cos \left (\frac {2 d a -2 c b}{d}\right ) \operatorname {S}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {2 d a -2 c b}{d}\right ) \operatorname {C}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{d \sqrt {\frac {b}{d}}}\right )}{3 d}\right )}{5 d}}{d}\)	\(230\)
default	\(\frac {-\frac {1}{5 \left (d x +c \right )^{\frac {5}{2}}}+\frac {\cos \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 d a -2 c b}{d}\right )}{5 \left (d x +c \right )^{\frac {5}{2}}}+\frac {4 b \left (-\frac {\sin \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 d a -2 c b}{d}\right )}{3 \left (d x +c \right )^{\frac {3}{2}}}+\frac {4 b \left (-\frac {\cos \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 d a -2 c b}{d}\right )}{\sqrt {d x +c}}-\frac {2 b \sqrt {\pi }\, \left (\cos \left (\frac {2 d a -2 c b}{d}\right ) \operatorname {S}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {2 d a -2 c b}{d}\right ) \operatorname {C}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{d \sqrt {\frac {b}{d}}}\right )}{3 d}\right )}{5 d}}{d}\)	\(230\)

[In]

int(sin(b*x+a)^2/(d*x+c)^(7/2),x,method=_RETURNVERBOSE)

[Out]

2/d*(-1/10/(d*x+c)^(5/2)+1/10/(d*x+c)^(5/2)*cos(2*b*(d*x+c)/d+2*(a*d-b*c)/d)+2/5*b/d*(-1/3/(d*x+c)^(3/2)*sin(2
*b*(d*x+c)/d+2*(a*d-b*c)/d)+4/3*b/d*(-1/(d*x+c)^(1/2)*cos(2*b*(d*x+c)/d+2*(a*d-b*c)/d)-2*b/d*Pi^(1/2)/(b/d)^(1
/2)*(cos(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)))))

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.52 \[ \int \frac {\sin ^2(a+b x)}{(c+d x)^{7/2}} \, dx=-\frac {2 \, {\left (16 \, {\left (\pi b^{2} d^{3} x^{3} + 3 \, \pi b^{2} c d^{2} x^{2} + 3 \, \pi b^{2} c^{2} d x + \pi b^{2} c^{3}\right )} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {S}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) + 16 \, {\left (\pi b^{2} d^{3} x^{3} + 3 \, \pi b^{2} c d^{2} x^{2} + 3 \, \pi b^{2} c^{2} d x + \pi b^{2} c^{3}\right )} \sqrt {\frac {b}{\pi d}} \operatorname {C}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - {\left (8 \, b^{2} d^{2} x^{2} + 16 \, b^{2} c d x + 8 \, b^{2} c^{2} - {\left (16 \, b^{2} d^{2} x^{2} + 32 \, b^{2} c d x + 16 \, b^{2} c^{2} - 3 \, d^{2}\right )} \cos \left (b x + a\right )^{2} - 4 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 3 \, d^{2}\right )} \sqrt {d x + c}\right )}}{15 \, {\left (d^{6} x^{3} + 3 \, c d^{5} x^{2} + 3 \, c^{2} d^{4} x + c^{3} d^{3}\right )}} \]

[In]

integrate(sin(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) - 3*d^2)*sqrt(d*x + c))/(d^6*x^3 + 3*c*d^5*x^2 + 3*c^2*d^4*x + c^3*d^3)

Sympy [F]

\[ \int \frac {\sin ^2(a+b x)}{(c+d x)^{7/2}} \, dx=\int \frac {\sin ^{2}{\left (a + b x \right )}}{\left (c + d x\right )^{\frac {7}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.38 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.63 \[ \int \frac {\sin ^2(a+b x)}{(c+d x)^{7/2}} \, dx=-\frac {5 \, \sqrt {2} {\left ({\left (\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {5}{2}, \frac {2 i \, {\left (d x + c\right )} b}{d}\right ) - \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {5}{2}, -\frac {2 i \, {\left (d x + c\right )} b}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + {\left (-\left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {5}{2}, \frac {2 i \, {\left (d x + c\right )} b}{d}\right ) + \left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {5}{2}, -\frac {2 i \, {\left (d x + c\right )} b}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \left (\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {5}{2}} + 2}{10 \, {\left (d x + c\right )}^{\frac {5}{2}} d} \]

[In]

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

[Out]

-1/10*(5*sqrt(2)*(((I + 1)*sqrt(2)*gamma(-5/2, 2*I*(d*x + c)*b/d) - (I - 1)*sqrt(2)*gamma(-5/2, -2*I*(d*x + c)
*b/d))*cos(-2*(b*c - a*d)/d) + (-(I - 1)*sqrt(2)*gamma(-5/2, 2*I*(d*x + c)*b/d) + (I + 1)*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)

Giac [F]

\[ \int \frac {\sin ^2(a+b x)}{(c+d x)^{7/2}} \, dx=\int { \frac {\sin \left (b x + a\right )^{2}}{{\left (d x + c\right )}^{\frac {7}{2}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sin ^2(a+b x)}{(c+d x)^{7/2}} \, dx=\int \frac {{\sin \left (a+b\,x\right )}^2}{{\left (c+d\,x\right )}^{7/2}} \,d x \]

[In]

int(sin(a + b*x)^2/(c + d*x)^(7/2),x)

[Out]

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