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

Optimal. Leaf size=113 \[ \frac{3 b \sec (e+f x) \sqrt{a+b \sec ^2(e+f x)-b}}{2 f}-\frac{\cos (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{f}+\frac{3 \sqrt{b} (a-b) \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)-b}}\right )}{2 f} \]

[Out]

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

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Rubi [A]  time = 0.0819726, antiderivative size = 113, 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 = {3664, 277, 195, 217, 206} \[ \frac{3 b \sec (e+f x) \sqrt{a+b \sec ^2(e+f x)-b}}{2 f}-\frac{\cos (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{f}+\frac{3 \sqrt{b} (a-b) \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)-b}}\right )}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3664

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]

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

Mathematica [A]  time = 1.21397, size = 170, normalized size = 1.5 \[ \frac{\sec (e+f x) \sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (6 \sqrt{2} \sqrt{b} (a-b) \cos ^2(e+f x) \tanh ^{-1}\left (\frac{\sqrt{(a-b) \cos (2 (e+f x))+a+b}}{\sqrt{2} \sqrt{b}}\right )-2 ((a-b) \cos (2 (e+f x))+a-2 b) \sqrt{(a-b) \cos (2 (e+f x))+a+b}\right )}{4 \sqrt{2} f \sqrt{(a-b) \cos (2 (e+f x))+a+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.053, size = 359, normalized size = 3.2 \begin{align*} -{\frac{\cos \left ( fx+e \right ) }{2\,fb} \left ({\frac{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b}{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}} \right ) ^{{\frac{3}{2}}} \left ( 3\,{b}^{5/2}\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b}+b}{\cos \left ( fx+e \right ) }} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}-3\,{b}^{3/2}\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b}+b}{\cos \left ( fx+e \right ) }} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}a+ \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) ^{{\frac{3}{2}}} \left ( \cos \left ( fx+e \right ) \right ) ^{2}a- \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) ^{{\frac{3}{2}}} \left ( \cos \left ( fx+e \right ) \right ) ^{2}b- \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) ^{{\frac{5}{2}}}+3\,\sqrt{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b} \left ( \cos \left ( fx+e \right ) \right ) ^{2}ab-3\,\sqrt{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b} \left ( \cos \left ( fx+e \right ) \right ) ^{2}{b}^{2} \right ) \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.90307, size = 667, normalized size = 5.9 \begin{align*} \left [-\frac{3 \,{\left (a - b\right )} \sqrt{b} \cos \left (f x + e\right ) \log \left (-\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt{b} \sqrt{\frac{{\left (a - b\right )} \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 \,{\left (a - b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt{\frac{{\left (a - b\right )} \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 \,{\left (a - b\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} \sqrt{\frac{{\left (a - b\right )} \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 \,{\left (a - b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt{\frac{{\left (a - b\right )} \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(sin(f*x+e)*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError