\(\int (a+b \sin ^2(e+f x))^{3/2} \tan ^3(e+f x) \, dx\) [500]

Optimal result

Integrand size = 25, antiderivative size = 148 \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=-\frac {\sqrt {a+b} (2 a+5 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{2 f}+\frac {(2 a+5 b) \sqrt {a+b \sin ^2(e+f x)}}{2 f}+\frac {(2 a+5 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 (a+b) f}+\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 (a+b) f} \]

[Out]

1/6*(2*a+5*b)*(a+b*sin(f*x+e)^2)^(3/2)/(a+b)/f+1/2*sec(f*x+e)^2*(a+b*sin(f*x+e)^2)^(5/2)/(a+b)/f-1/2*(2*a+5*b)
*arctanh((a+b*sin(f*x+e)^2)^(1/2)/(a+b)^(1/2))*(a+b)^(1/2)/f+1/2*(2*a+5*b)*(a+b*sin(f*x+e)^2)^(1/2)/f

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3273, 79, 52, 65, 214} \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=-\frac {\sqrt {a+b} (2 a+5 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{2 f}+\frac {(2 a+5 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 f (a+b)}+\frac {(2 a+5 b) \sqrt {a+b \sin ^2(e+f x)}}{2 f}+\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 f (a+b)} \]

[In]

Int[(a + b*Sin[e + f*x]^2)^(3/2)*Tan[e + f*x]^3,x]

[Out]

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

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 3273

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

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x (a+b x)^{3/2}}{(1-x)^2} \, dx,x,\sin ^2(e+f x)\right )}{2 f} \\ & = \frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 (a+b) f}-\frac {(2 a+5 b) \text {Subst}\left (\int \frac {(a+b x)^{3/2}}{1-x} \, dx,x,\sin ^2(e+f x)\right )}{4 (a+b) f} \\ & = \frac {(2 a+5 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 (a+b) f}+\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 (a+b) f}-\frac {(2 a+5 b) \text {Subst}\left (\int \frac {\sqrt {a+b x}}{1-x} \, dx,x,\sin ^2(e+f x)\right )}{4 f} \\ & = \frac {(2 a+5 b) \sqrt {a+b \sin ^2(e+f x)}}{2 f}+\frac {(2 a+5 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 (a+b) f}+\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 (a+b) f}-\frac {((a+b) (2 a+5 b)) \text {Subst}\left (\int \frac {1}{(1-x) \sqrt {a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{4 f} \\ & = \frac {(2 a+5 b) \sqrt {a+b \sin ^2(e+f x)}}{2 f}+\frac {(2 a+5 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 (a+b) f}+\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 (a+b) f}-\frac {((a+b) (2 a+5 b)) \text {Subst}\left (\int \frac {1}{1+\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sin ^2(e+f x)}\right )}{2 b f} \\ & = -\frac {\sqrt {a+b} (2 a+5 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{2 f}+\frac {(2 a+5 b) \sqrt {a+b \sin ^2(e+f x)}}{2 f}+\frac {(2 a+5 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 (a+b) f}+\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 (a+b) f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.78 \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=\frac {3 \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}+(2 a+5 b) \left (-3 (a+b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )+\sqrt {a+b \sin ^2(e+f x)} \left (4 a+3 b+b \sin ^2(e+f x)\right )\right )}{6 (a+b) f} \]

[In]

Integrate[(a + b*Sin[e + f*x]^2)^(3/2)*Tan[e + f*x]^3,x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(425\) vs. \(2(128)=256\).

Time = 1.78 (sec) , antiderivative size = 426, normalized size of antiderivative = 2.88

method	result	size
default	\(\frac {-\frac {\sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (b \left (\cos ^{2}\left (f x +e \right )\right )+2 a -b \right )}{3}+2 b \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}+2 a \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}-\frac {\left (\frac {1}{2} a^{2}+2 a b +\frac {3}{2} b^{2}\right ) \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right )}{\sqrt {a +b}}-\frac {\left (\frac {1}{2} a^{2}+2 a b +\frac {3}{2} b^{2}\right ) \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right )}{\sqrt {a +b}}+\left (\frac {1}{4} a^{2}+\frac {1}{2} a b +\frac {1}{4} b^{2}\right ) \left (-\frac {\sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}}{\left (a +b \right ) \left (\sin \left (f x +e \right )-1\right )}+\frac {b \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right )}{\left (a +b \right )^{\frac {3}{2}}}\right )+\left (-\frac {1}{4} a^{2}-\frac {1}{2} a b -\frac {1}{4} b^{2}\right ) \left (-\frac {\sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}}{\left (a +b \right ) \left (1+\sin \left (f x +e \right )\right )}-\frac {b \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right )}{\left (a +b \right )^{\frac {3}{2}}}\right )}{f}\)	\(426\)

[In]

int((a+b*sin(f*x+e)^2)^(3/2)*tan(f*x+e)^3,x,method=_RETURNVERBOSE)

[Out]

(-1/3*(a+b-b*cos(f*x+e)^2)^(1/2)*(b*cos(f*x+e)^2+2*a-b)+2*b*(a+b*sin(f*x+e)^2)^(1/2)+2*a*(a+b*sin(f*x+e)^2)^(1
/2)-(1/2*a^2+2*a*b+3/2*b^2)/(a+b)^(1/2)*ln((2*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-2*b*sin(f*x+e)+2*a)/(1+si
n(f*x+e)))-(1/2*a^2+2*a*b+3/2*b^2)/(a+b)^(1/2)*ln((2*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+2*b*sin(f*x+e)+2*a
)/(sin(f*x+e)-1))+(1/4*a^2+1/2*a*b+1/4*b^2)*(-1/(a+b)/(sin(f*x+e)-1)*(a+b-b*cos(f*x+e)^2)^(1/2)+b/(a+b)^(3/2)*
ln((2*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+2*b*sin(f*x+e)+2*a)/(sin(f*x+e)-1)))+(-1/4*a^2-1/2*a*b-1/4*b^2)*(
-1/(a+b)/(1+sin(f*x+e))*(a+b-b*cos(f*x+e)^2)^(1/2)-b/(a+b)^(3/2)*ln((2*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-
2*b*sin(f*x+e)+2*a)/(1+sin(f*x+e)))))/f

Fricas [A] (verification not implemented)

none

Time = 0.53 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.79 \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=\left [\frac {3 \, {\left (2 \, a + 5 \, b\right )} \sqrt {a + b} \cos \left (f x + e\right )^{2} \log \left (\frac {b \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} - 2 \, a - 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) - 2 \, {\left (2 \, b \cos \left (f x + e\right )^{4} - 2 \, {\left (4 \, a + 7 \, b\right )} \cos \left (f x + e\right )^{2} - 3 \, a - 3 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{12 \, f \cos \left (f x + e\right )^{2}}, \frac {3 \, {\left (2 \, a + 5 \, b\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{a + b}\right ) \cos \left (f x + e\right )^{2} - {\left (2 \, b \cos \left (f x + e\right )^{4} - 2 \, {\left (4 \, a + 7 \, b\right )} \cos \left (f x + e\right )^{2} - 3 \, a - 3 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{6 \, f \cos \left (f x + e\right )^{2}}\right ] \]

[In]

integrate((a+b*sin(f*x+e)^2)^(3/2)*tan(f*x+e)^3,x, algorithm="fricas")

[Out]

[1/12*(3*(2*a + 5*b)*sqrt(a + b)*cos(f*x + e)^2*log((b*cos(f*x + e)^2 + 2*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt
(a + b) - 2*a - 2*b)/cos(f*x + e)^2) - 2*(2*b*cos(f*x + e)^4 - 2*(4*a + 7*b)*cos(f*x + e)^2 - 3*a - 3*b)*sqrt(
-b*cos(f*x + e)^2 + a + b))/(f*cos(f*x + e)^2), 1/6*(3*(2*a + 5*b)*sqrt(-a - b)*arctan(sqrt(-b*cos(f*x + e)^2
+ a + b)*sqrt(-a - b)/(a + b))*cos(f*x + e)^2 - (2*b*cos(f*x + e)^4 - 2*(4*a + 7*b)*cos(f*x + e)^2 - 3*a - 3*b
)*sqrt(-b*cos(f*x + e)^2 + a + b))/(f*cos(f*x + e)^2)]

Sympy [F(-1)]

Timed out. \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=\text {Timed out} \]

[In]

integrate((a+b*sin(f*x+e)**2)**(3/2)*tan(f*x+e)**3,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.14 \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=\frac {4 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} b^{2} + 12 \, {\left (a b^{2} + 2 \, b^{3}\right )} \sqrt {b \sin \left (f x + e\right )^{2} + a} + \frac {3 \, {\left (2 \, a^{2} b^{2} + 7 \, a b^{3} + 5 \, b^{4}\right )} \log \left (\frac {\sqrt {b \sin \left (f x + e\right )^{2} + a} - \sqrt {a + b}}{\sqrt {b \sin \left (f x + e\right )^{2} + a} + \sqrt {a + b}}\right )}{\sqrt {a + b}} - \frac {6 \, {\left (a b^{3} + b^{4}\right )} \sqrt {b \sin \left (f x + e\right )^{2} + a}}{b \sin \left (f x + e\right )^{2} - b}}{12 \, b^{2} f} \]

[In]

integrate((a+b*sin(f*x+e)^2)^(3/2)*tan(f*x+e)^3,x, algorithm="maxima")

[Out]

1/12*(4*(b*sin(f*x + e)^2 + a)^(3/2)*b^2 + 12*(a*b^2 + 2*b^3)*sqrt(b*sin(f*x + e)^2 + a) + 3*(2*a^2*b^2 + 7*a*
b^3 + 5*b^4)*log((sqrt(b*sin(f*x + e)^2 + a) - sqrt(a + b))/(sqrt(b*sin(f*x + e)^2 + a) + sqrt(a + b)))/sqrt(a
+ b) - 6*(a*b^3 + b^4)*sqrt(b*sin(f*x + e)^2 + a)/(b*sin(f*x + e)^2 - b))/(b^2*f)

Giac [F(-2)]

Exception generated. \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((a+b*sin(f*x+e)^2)^(3/2)*tan(f*x+e)^3,x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Error: Bad Argument Type

Mupad [F(-1)]

Timed out. \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x) \, dx=\int {\mathrm {tan}\left (e+f\,x\right )}^3\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2} \,d x \]

[In]

int(tan(e + f*x)^3*(a + b*sin(e + f*x)^2)^(3/2),x)

[Out]

int(tan(e + f*x)^3*(a + b*sin(e + f*x)^2)^(3/2), x)