3.2.42 \(\int \frac {\sin (e+f x)}{(a+b \tan ^2(e+f x))^{5/2}} \, dx\) [142]

Optimal. Leaf size=118 \[ -\frac {\cos (e+f x)}{(a-b) f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac {4 b \sec (e+f x)}{3 (a-b)^2 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac {8 b \sec (e+f x)}{3 (a-b)^3 f \sqrt {a-b+b \sec ^2(e+f x)}} \]

[Out]

-cos(f*x+e)/(a-b)/f/(a-b+b*sec(f*x+e)^2)^(3/2)-4/3*b*sec(f*x+e)/(a-b)^2/f/(a-b+b*sec(f*x+e)^2)^(3/2)-8/3*b*sec
(f*x+e)/(a-b)^3/f/(a-b+b*sec(f*x+e)^2)^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3745, 277, 198, 197} \begin {gather*} -\frac {8 b \sec (e+f x)}{3 f (a-b)^3 \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {4 b \sec (e+f x)}{3 f (a-b)^2 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac {\cos (e+f x)}{f (a-b) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]/(a + b*Tan[e + f*x]^2)^(5/2),x]

[Out]

-(Cos[e + f*x]/((a - b)*f*(a - b + b*Sec[e + f*x]^2)^(3/2))) - (4*b*Sec[e + f*x])/(3*(a - b)^2*f*(a - b + b*Se
c[e + f*x]^2)^(3/2)) - (8*b*Sec[e + f*x])/(3*(a - b)^3*f*Sqrt[a - b + b*Sec[e + f*x]^2])

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 198

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p +
1], 0] && NeQ[p, -1]

Rule 277

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^n)^(p + 1)/(a*(m + 1))), x]
- Dist[b*((m + n*(p + 1) + 1)/(a*(m + 1))), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 3745

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a - b + b*ff^2*x^2)^p/x^(m
+ 1)), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \frac {\sin (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx &=\frac {\text {Subst}\left (\int \frac {1}{x^2 \left (a-b+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac {\cos (e+f x)}{(a-b) f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac {(4 b) \text {Subst}\left (\int \frac {1}{\left (a-b+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{(a-b) f}\\ &=-\frac {\cos (e+f x)}{(a-b) f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac {4 b \sec (e+f x)}{3 (a-b)^2 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac {(8 b) \text {Subst}\left (\int \frac {1}{\left (a-b+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{3 (a-b)^2 f}\\ &=-\frac {\cos (e+f x)}{(a-b) f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac {4 b \sec (e+f x)}{3 (a-b)^2 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac {8 b \sec (e+f x)}{3 (a-b)^3 f \sqrt {a-b+b \sec ^2(e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 1.31, size = 124, normalized size = 1.05 \begin {gather*} -\frac {\cos (e+f x) \left ((3 a+5 b)^2+12 \left (a^2+2 a b-3 b^2\right ) \cos (2 (e+f x))+3 (a-b)^2 \cos (4 (e+f x))\right ) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}}{6 \sqrt {2} (a-b)^3 f (a+b+(a-b) \cos (2 (e+f x)))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]/(a + b*Tan[e + f*x]^2)^(5/2),x]

[Out]

-1/6*(Cos[e + f*x]*((3*a + 5*b)^2 + 12*(a^2 + 2*a*b - 3*b^2)*Cos[2*(e + f*x)] + 3*(a - b)^2*Cos[4*(e + f*x)])*
Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])/(Sqrt[2]*(a - b)^3*f*(a + b + (a - b)*Cos[2*(e + f*x)
])^2)

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Maple [A]
time = 0.12, size = 147, normalized size = 1.25

method	result	size
default	\(-\frac {\left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right ) \left (3 \left (\cos ^{4}\left (f x +e \right )\right ) a^{2}-6 \left (\cos ^{4}\left (f x +e \right )\right ) a b +3 \left (\cos ^{4}\left (f x +e \right )\right ) b^{2}+12 \left (\cos ^{2}\left (f x +e \right )\right ) a b -12 \left (\cos ^{2}\left (f x +e \right )\right ) b^{2}+8 b^{2}\right )}{3 f \left (\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\cos \left (f x +e \right )^{2}}\right )^{\frac {5}{2}} \cos \left (f x +e \right )^{5} \left (a -b \right )^{3}}\)	\(147\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)/(a+b*tan(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/3/f*(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)*(3*cos(f*x+e)^4*a^2-6*cos(f*x+e)^4*a*b+3*cos(f*x+e)^4*b^2+12*cos(f*x+
e)^2*a*b-12*cos(f*x+e)^2*b^2+8*b^2)/((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/cos(f*x+e)^2)^(5/2)/cos(f*x+e)^5/(a-b)^
3

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Maxima [A]
time = 0.28, size = 141, normalized size = 1.19 \begin {gather*} -\frac {\frac {3 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {6 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} b \cos \left (f x + e\right )^{2} - b^{2}}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3}}}{3 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

-1/3*(3*sqrt(a - b + b/cos(f*x + e)^2)*cos(f*x + e)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (6*(a - b + b/cos(f*x +
e)^2)*b*cos(f*x + e)^2 - b^2)/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*(a - b + b/cos(f*x + e)^2)^(3/2)*cos(f*x + e)^3
))/f

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Fricas [A]
time = 2.67, size = 209, normalized size = 1.77 \begin {gather*} -\frac {{\left (3 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} + 12 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{3} + 8 \, b^{2} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{3 \, {\left ({\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b - 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - 4 \, a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b^{2} - 3 \, a^{2} b^{3} + 3 \, a b^{4} - b^{5}\right )} f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

-1/3*(3*(a^2 - 2*a*b + b^2)*cos(f*x + e)^5 + 12*(a*b - b^2)*cos(f*x + e)^3 + 8*b^2*cos(f*x + e))*sqrt(((a - b)
*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*f*cos(f*x + e)
^4 + 2*(a^4*b - 4*a^3*b^2 + 6*a^2*b^3 - 4*a*b^4 + b^5)*f*cos(f*x + e)^2 + (a^3*b^2 - 3*a^2*b^3 + 3*a*b^4 - b^5
)*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)/(a+b*tan(f*x+e)**2)**(5/2),x)

[Out]

Integral(sin(e + f*x)/(a + b*tan(e + f*x)**2)**(5/2), x)

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Giac [A]
time = 2.56, size = 223, normalized size = 1.89 \begin {gather*} -\frac {f^{4} {\left (\frac {3 \, \sqrt {a \cos \left (f x + e\right )^{2} - b \cos \left (f x + e\right )^{2} + b}}{a {\left | f \right |} \mathrm {sgn}\left (f\right ) \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) - b {\left | f \right |} \mathrm {sgn}\left (f\right ) \mathrm {sgn}\left (\cos \left (f x + e\right )\right )} + \frac {6 \, {\left (a \cos \left (f x + e\right )^{2} - b \cos \left (f x + e\right )^{2} + b\right )} b - b^{2}}{{\left (a {\left | f \right |} \mathrm {sgn}\left (f\right ) \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) - b {\left | f \right |} \mathrm {sgn}\left (f\right ) \mathrm {sgn}\left (\cos \left (f x + e\right )\right )\right )} {\left (a \cos \left (f x + e\right )^{2} - b \cos \left (f x + e\right )^{2} + b\right )}^{\frac {3}{2}}}\right )}}{3 \, {\left (a f^{2} - b f^{2}\right )}^{2}} + \frac {8 \, \sqrt {b} \mathrm {sgn}\left (f\right ) \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{3 \, {\left (a^{3} {\left | f \right |} - 3 \, a^{2} b {\left | f \right |} + 3 \, a b^{2} {\left | f \right |} - b^{3} {\left | f \right |}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

-1/3*f^4*(3*sqrt(a*cos(f*x + e)^2 - b*cos(f*x + e)^2 + b)/(a*abs(f)*sgn(f)*sgn(cos(f*x + e)) - b*abs(f)*sgn(f)
*sgn(cos(f*x + e))) + (6*(a*cos(f*x + e)^2 - b*cos(f*x + e)^2 + b)*b - b^2)/((a*abs(f)*sgn(f)*sgn(cos(f*x + e)
) - b*abs(f)*sgn(f)*sgn(cos(f*x + e)))*(a*cos(f*x + e)^2 - b*cos(f*x + e)^2 + b)^(3/2)))/(a*f^2 - b*f^2)^2 + 8
/3*sqrt(b)*sgn(f)*sgn(cos(f*x + e))/(a^3*abs(f) - 3*a^2*b*abs(f) + 3*a*b^2*abs(f) - b^3*abs(f))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)/(a + b*tan(e + f*x)^2)^(5/2),x)

[Out]

int(sin(e + f*x)/(a + b*tan(e + f*x)^2)^(5/2), x)

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