\(\int \frac {\sin (a+b x)}{(d \cos (a+b x))^{7/2}} \, dx\) [192]

Optimal result

Integrand size = 19, antiderivative size = 22 \[ \int \frac {\sin (a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\frac {2}{5 b d (d \cos (a+b x))^{5/2}} \]

[Out]

2/5/b/d/(d*cos(b*x+a))^(5/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2645, 30} \[ \int \frac {\sin (a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\frac {2}{5 b d (d \cos (a+b x))^{5/2}} \]

[In]

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

[Out]

2/(5*b*d*(d*Cos[a + b*x])^(5/2))

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2645

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {1}{x^{7/2}} \, dx,x,d \cos (a+b x)\right )}{b d} \\ & = \frac {2}{5 b d (d \cos (a+b x))^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {\sin (a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\frac {2}{5 b d (d \cos (a+b x))^{5/2}} \]

[In]

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

[Out]

2/(5*b*d*(d*Cos[a + b*x])^(5/2))

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(\frac {2}{5 b d \left (d \cos \left (b x +a \right )\right )^{\frac {5}{2}}}\)	\(19\)
default	\(\frac {2}{5 b d \left (d \cos \left (b x +a \right )\right )^{\frac {5}{2}}}\)	\(19\)

[In]

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

[Out]

2/5/b/d/(d*cos(b*x+a))^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {\sin (a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\frac {2 \, \sqrt {d \cos \left (b x + a\right )}}{5 \, b d^{4} \cos \left (b x + a\right )^{3}} \]

[In]

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

[Out]

2/5*sqrt(d*cos(b*x + a))/(b*d^4*cos(b*x + a)^3)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 36 vs. \(2 (17) = 34\).

Time = 29.62 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.64 \[ \int \frac {\sin (a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\begin {cases} \frac {2 \cos {\left (a + b x \right )}}{5 b \left (d \cos {\left (a + b x \right )}\right )^{\frac {7}{2}}} & \text {for}\: b \neq 0 \\\frac {x \sin {\left (a \right )}}{\left (d \cos {\left (a \right )}\right )^{\frac {7}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*cos(a + b*x)/(5*b*(d*cos(a + b*x))**(7/2)), Ne(b, 0)), (x*sin(a)/(d*cos(a))**(7/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {\sin (a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\frac {2}{5 \, \left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}} b d} \]

[In]

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

[Out]

2/5/((d*cos(b*x + a))^(5/2)*b*d)

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.66 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.95 \[ \int \frac {\sin (a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\frac {16\,{\mathrm {e}}^{a\,3{}\mathrm {i}+b\,x\,3{}\mathrm {i}}\,\sqrt {d\,\left (\frac {{\mathrm {e}}^{-a\,1{}\mathrm {i}-b\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}}{2}\right )}}{5\,b\,d^4\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )}^3} \]

[In]

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

[Out]

(16*exp(a*3i + b*x*3i)*(d*(exp(- a*1i - b*x*1i)/2 + exp(a*1i + b*x*1i)/2))^(1/2))/(5*b*d^4*(exp(a*2i + b*x*2i)
+ 1)^3)