3.82 \(\int (a+b \sec ^2(e+f x))^{3/2} \sin (e+f x) \, dx\)

Optimal. Leaf size=100 \[ \frac {3 b \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 f}-\frac {\cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{f}+\frac {3 a \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{2 f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {4134, 277, 195, 217, 206} \[ \frac {3 b \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 f}-\frac {\cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{f}+\frac {3 a \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*Sqrt[b]*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/(2*f) + (3*b*Sec[e + f*x]*Sqrt[a + b*
Sec[e + f*x]^2])/(2*f) - (Cos[e + f*x]*(a + b*Sec[e + f*x]^2)^(3/2))/f

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 277

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

Rule 4134

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Cos[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^
n)^p)/x^(m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (Gt
Q[m, 0] || EqQ[n, 2] || EqQ[n, 4])

Rubi steps

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

________________________________________________________________________________________

Mathematica [C]  time = 0.60, size = 73, normalized size = 0.73 \[ -\frac {a \cos (e+f x) (a \cos (2 (e+f x))+a+2 b)^2 \sqrt {a+b \sec ^2(e+f x)} \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};\frac {a \cos ^2(e+f x)}{b}+1\right )}{20 b^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/20*(a*Cos[e + f*x]*(a + 2*b + a*Cos[2*(e + f*x)])^2*Hypergeometric2F1[2, 5/2, 7/2, 1 + (a*Cos[e + f*x]^2)/b
]*Sqrt[a + b*Sec[e + f*x]^2])/(b^2*f)

________________________________________________________________________________________

fricas [A]  time = 0.87, size = 231, normalized size = 2.31 \[ \left [\frac {3 \, a \sqrt {b} \cos \left (f x + e\right ) \log \left (\frac {a \cos \left (f x + e\right )^{2} + 2 \, \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) - 2 \, {\left (2 \, a \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, f \cos \left (f x + e\right )}, -\frac {3 \, a \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{b}\right ) \cos \left (f x + e\right ) + {\left (2 \, a \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, f \cos \left (f x + e\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(3*a*sqrt(b)*cos(f*x + e)*log((a*cos(f*x + e)^2 + 2*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*c
os(f*x + e) + 2*b)/cos(f*x + e)^2) - 2*(2*a*cos(f*x + e)^2 - b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(
f*cos(f*x + e)), -1/2*(3*a*sqrt(-b)*arctan(sqrt(-b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b
)*cos(f*x + e) + (2*a*cos(f*x + e)^2 - b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(f*cos(f*x + e))]

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check si
gn: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)4*((a^2*sqrt(a+b)*sign(cos(f*x+exp(1)))+a^2*(-sqrt(a+b)*tan(1/
2*(f*x+exp(1)))^2+sqrt(a*tan(1/2*(f*x+exp(1)))^4+b*tan(1/2*(f*x+exp(1)))^4-2*a*tan(1/2*(f*x+exp(1)))^2+2*b*tan
(1/2*(f*x+exp(1)))^2+a+b))*sign(cos(f*x+exp(1))))/(-2*sqrt(a+b)*(-sqrt(a+b)*tan(1/2*(f*x+exp(1)))^2+sqrt(a*tan
(1/2*(f*x+exp(1)))^4+b*tan(1/2*(f*x+exp(1)))^4-2*a*tan(1/2*(f*x+exp(1)))^2+2*b*tan(1/2*(f*x+exp(1)))^2+a+b))+(
-sqrt(a+b)*tan(1/2*(f*x+exp(1)))^2+sqrt(a*tan(1/2*(f*x+exp(1)))^4+b*tan(1/2*(f*x+exp(1)))^4-2*a*tan(1/2*(f*x+e
xp(1)))^2+2*b*tan(1/2*(f*x+exp(1)))^2+a+b))^2-3*a+b)+1/2*(sqrt(a+b)*(2*b^3-a*b^2+a^2*b)*sign(cos(f*x+exp(1)))+
(2*b^2+a*b)*(-sqrt(a+b)*tan(1/2*(f*x+exp(1)))^2+sqrt(a*tan(1/2*(f*x+exp(1)))^4+b*tan(1/2*(f*x+exp(1)))^4-2*a*t
an(1/2*(f*x+exp(1)))^2+2*b*tan(1/2*(f*x+exp(1)))^2+a+b))^3*sign(cos(f*x+exp(1)))+(-sqrt(a+b)*tan(1/2*(f*x+exp(
1)))^2+sqrt(a*tan(1/2*(f*x+exp(1)))^4+b*tan(1/2*(f*x+exp(1)))^4-2*a*tan(1/2*(f*x+exp(1)))^2+2*b*tan(1/2*(f*x+e
xp(1)))^2+a+b))*(-2*b^3-3*a*b^2+3*a^2*b)*sign(cos(f*x+exp(1)))+sqrt(a+b)*(-2*b^2+3*a*b)*(-sqrt(a+b)*tan(1/2*(f
*x+exp(1)))^2+sqrt(a*tan(1/2*(f*x+exp(1)))^4+b*tan(1/2*(f*x+exp(1)))^4-2*a*tan(1/2*(f*x+exp(1)))^2+2*b*tan(1/2
*(f*x+exp(1)))^2+a+b))^2*sign(cos(f*x+exp(1))))/(-2*sqrt(a+b)*(-sqrt(a+b)*tan(1/2*(f*x+exp(1)))^2+sqrt(a*tan(1
/2*(f*x+exp(1)))^4+b*tan(1/2*(f*x+exp(1)))^4-2*a*tan(1/2*(f*x+exp(1)))^2+2*b*tan(1/2*(f*x+exp(1)))^2+a+b))-(-s
qrt(a+b)*tan(1/2*(f*x+exp(1)))^2+sqrt(a*tan(1/2*(f*x+exp(1)))^4+b*tan(1/2*(f*x+exp(1)))^4-2*a*tan(1/2*(f*x+exp
(1)))^2+2*b*tan(1/2*(f*x+exp(1)))^2+a+b))^2-a+3*b)^2-3/4*a*b*sign(cos(f*x+exp(1)))*atan(1/2*(-sqrt(a+b)*tan(1/
2*(f*x+exp(1)))^2+sqrt(a+b)+sqrt(a*tan(1/2*(f*x+exp(1)))^4+b*tan(1/2*(f*x+exp(1)))^4-2*a*tan(1/2*(f*x+exp(1)))
^2+2*b*tan(1/2*(f*x+exp(1)))^2+a+b))/sqrt(-b))/sqrt(-b))/f

________________________________________________________________________________________

maple [A]  time = 0.36, size = 121, normalized size = 1.21 \[ -\frac {\left (a +b \left (\sec ^{2}\left (f x +e \right )\right )\right )^{\frac {5}{2}}}{f a \sec \left (f x +e \right )}+\frac {b \sec \left (f x +e \right ) \left (a +b \left (\sec ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}}}{f a}+\frac {3 b \sec \left (f x +e \right ) \sqrt {a +b \left (\sec ^{2}\left (f x +e \right )\right )}}{2 f}+\frac {3 a \sqrt {b}\, \ln \left (\sec \left (f x +e \right ) \sqrt {b}+\sqrt {a +b \left (\sec ^{2}\left (f x +e \right )\right )}\right )}{2 f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/f/a/sec(f*x+e)*(a+b*sec(f*x+e)^2)^(5/2)+1/f/a*b*sec(f*x+e)*(a+b*sec(f*x+e)^2)^(3/2)+3/2*b*sec(f*x+e)*(a+b*s
ec(f*x+e)^2)^(1/2)/f+3/2/f*a*b^(1/2)*ln(sec(f*x+e)*b^(1/2)+(a+b*sec(f*x+e)^2)^(1/2))

________________________________________________________________________________________

maxima [A]  time = 0.43, size = 142, normalized size = 1.42 \[ -\frac {4 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} a \cos \left (f x + e\right ) - \frac {2 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} a b \cos \left (f x + e\right )}{{\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} \cos \left (f x + e\right )^{2} - b} + 3 \, a \sqrt {b} \log \left (\frac {\sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - \sqrt {b}}{\sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + \sqrt {b}}\right )}{4 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 6.11, size = 61, normalized size = 0.61 \[ -\frac {\cos \left (e+f\,x\right )\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},-\frac {1}{2};\ \frac {1}{2};\ -\frac {b}{a\,{\cos \left (e+f\,x\right )}^2}\right )}{f\,{\left (\frac {b}{a\,{\cos \left (e+f\,x\right )}^2}+1\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(cos(e + f*x)*(a + b/cos(e + f*x)^2)^(3/2)*hypergeom([-3/2, -1/2], 1/2, -b/(a*cos(e + f*x)^2)))/(f*(b/(a*cos(
e + f*x)^2) + 1)^(3/2))

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________