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]

(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

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Rubi [A]  time = 0.0702479, antiderivative size = 100, normalized size of antiderivative = 1., 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 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])

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 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 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 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])

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 = 0.664805, size = 73, normalized size = 0.73 \[ -\frac{a \cos (e+f x) (a \cos (2 (e+f x))+a+2 b)^2 \sqrt{a+b \sec ^2(e+f x)} \text{Hypergeometric2F1}\left (2,\frac{5}{2},\frac{7}{2},\frac{a \cos ^2(e+f x)}{b}+1\right )}{20 b^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(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]*Sqr
t[a + b*Sec[e + f*x]^2])/(20*b^2*f)

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Maple [A]  time = 0.059, size = 121, normalized size = 1.2 \begin{align*} -{\frac{1}{fa\sec \left ( fx+e \right ) } \left ( a+b \left ( \sec \left ( fx+e \right ) \right ) ^{2} \right ) ^{{\frac{5}{2}}}}+{\frac{b\sec \left ( fx+e \right ) }{fa} \left ( a+b \left ( \sec \left ( fx+e \right ) \right ) ^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{3\,b\sec \left ( fx+e \right ) }{2\,f}\sqrt{a+b \left ( \sec \left ( fx+e \right ) \right ) ^{2}}}+{\frac{3\,a}{2\,f}\sqrt{b}\ln \left ( \sec \left ( fx+e \right ) \sqrt{b}+\sqrt{a+b \left ( \sec \left ( fx+e \right ) \right ) ^{2}} \right ) } \end{align*}

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*b/a*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*b^(1/2)*a*ln(sec(f*x+e)*b^(1/2)+(a+b*sec(f*x+e)^2)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.18364, size = 591, normalized size = 5.91 \begin{align*} \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 ] \end{align*}

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))]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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

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Giac [A]  time = 1.30538, size = 124, normalized size = 1.24 \begin{align*} -\frac{{\left (\frac{3 \, b \arctan \left (\frac{\sqrt{a \cos \left (f x + e\right )^{2} + b}}{\sqrt{-b}}\right )}{\sqrt{-b}} + 2 \, \sqrt{a \cos \left (f x + e\right )^{2} + b} - \frac{\sqrt{a \cos \left (f x + e\right )^{2} + b} b}{a \cos \left (f x + e\right )^{2}}\right )} a \mathrm{sgn}\left (\cos \left (f x + e\right )\right )}{2 \, f} \end{align*}

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]

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