3.45 \(\int (c+d x)^{5/2} \sinh ^2(a+b x) \, dx\)

Optimal. Leaf size=239 \[ \frac{15 \sqrt{\frac{\pi }{2}} d^{5/2} e^{\frac{2 b c}{d}-2 a} \text{Erf}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{256 b^{7/2}}-\frac{15 \sqrt{\frac{\pi }{2}} d^{5/2} e^{2 a-\frac{2 b c}{d}} \text{Erfi}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{256 b^{7/2}}+\frac{15 d^2 \sqrt{c+d x} \sinh (2 a+2 b x)}{64 b^3}-\frac{5 d (c+d x)^{3/2} \sinh ^2(a+b x)}{8 b^2}+\frac{(c+d x)^{5/2} \sinh (a+b x) \cosh (a+b x)}{2 b}-\frac{5 d (c+d x)^{3/2}}{16 b^2}-\frac{(c+d x)^{7/2}}{7 d} \]

[Out]

(-5*d*(c + d*x)^(3/2))/(16*b^2) - (c + d*x)^(7/2)/(7*d) + (15*d^(5/2)*E^(-2*a + (2*b*c)/d)*Sqrt[Pi/2]*Erf[(Sqr
t[2]*Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(256*b^(7/2)) - (15*d^(5/2)*E^(2*a - (2*b*c)/d)*Sqrt[Pi/2]*Erfi[(Sqrt[2]
*Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(256*b^(7/2)) + ((c + d*x)^(5/2)*Cosh[a + b*x]*Sinh[a + b*x])/(2*b) - (5*d*(
c + d*x)^(3/2)*Sinh[a + b*x]^2)/(8*b^2) + (15*d^2*Sqrt[c + d*x]*Sinh[2*a + 2*b*x])/(64*b^3)

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Rubi [A]  time = 0.448625, antiderivative size = 239, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {3311, 32, 3312, 3296, 3308, 2180, 2204, 2205} \[ \frac{15 \sqrt{\frac{\pi }{2}} d^{5/2} e^{\frac{2 b c}{d}-2 a} \text{Erf}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{256 b^{7/2}}-\frac{15 \sqrt{\frac{\pi }{2}} d^{5/2} e^{2 a-\frac{2 b c}{d}} \text{Erfi}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{256 b^{7/2}}+\frac{15 d^2 \sqrt{c+d x} \sinh (2 a+2 b x)}{64 b^3}-\frac{5 d (c+d x)^{3/2} \sinh ^2(a+b x)}{8 b^2}+\frac{(c+d x)^{5/2} \sinh (a+b x) \cosh (a+b x)}{2 b}-\frac{5 d (c+d x)^{3/2}}{16 b^2}-\frac{(c+d x)^{7/2}}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(5/2)*Sinh[a + b*x]^2,x]

[Out]

(-5*d*(c + d*x)^(3/2))/(16*b^2) - (c + d*x)^(7/2)/(7*d) + (15*d^(5/2)*E^(-2*a + (2*b*c)/d)*Sqrt[Pi/2]*Erf[(Sqr
t[2]*Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(256*b^(7/2)) - (15*d^(5/2)*E^(2*a - (2*b*c)/d)*Sqrt[Pi/2]*Erfi[(Sqrt[2]
*Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(256*b^(7/2)) + ((c + d*x)^(5/2)*Cosh[a + b*x]*Sinh[a + b*x])/(2*b) - (5*d*(
c + d*x)^(3/2)*Sinh[a + b*x]^2)/(8*b^2) + (15*d^2*Sqrt[c + d*x]*Sinh[2*a + 2*b*x])/(64*b^3)

Rule 3311

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*m*(c + d*x)^(m - 1)*(
b*Sin[e + f*x])^n)/(f^2*n^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[(d^2*m*(m - 1))/(f^2*n^2), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[(b*(c + d*x)^m*Cos[e +
f*x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3308

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/E^(I*(e + f*x))
, x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d, e, f, m}, x]

Rule 2180

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[F^(g*(e - (c*
f)/d) + (f*g*x^2)/d), x], x, Sqrt[c + d*x]], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2205

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erf[(c + d*x)*Rt[-(b*Log[F]),
2]])/(2*d*Rt[-(b*Log[F]), 2]), x] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rubi steps

\begin{align*} \int (c+d x)^{5/2} \sinh ^2(a+b x) \, dx &=\frac{(c+d x)^{5/2} \cosh (a+b x) \sinh (a+b x)}{2 b}-\frac{5 d (c+d x)^{3/2} \sinh ^2(a+b x)}{8 b^2}-\frac{1}{2} \int (c+d x)^{5/2} \, dx+\frac{\left (15 d^2\right ) \int \sqrt{c+d x} \sinh ^2(a+b x) \, dx}{16 b^2}\\ &=-\frac{(c+d x)^{7/2}}{7 d}+\frac{(c+d x)^{5/2} \cosh (a+b x) \sinh (a+b x)}{2 b}-\frac{5 d (c+d x)^{3/2} \sinh ^2(a+b x)}{8 b^2}-\frac{\left (15 d^2\right ) \int \left (\frac{1}{2} \sqrt{c+d x}-\frac{1}{2} \sqrt{c+d x} \cosh (2 a+2 b x)\right ) \, dx}{16 b^2}\\ &=-\frac{5 d (c+d x)^{3/2}}{16 b^2}-\frac{(c+d x)^{7/2}}{7 d}+\frac{(c+d x)^{5/2} \cosh (a+b x) \sinh (a+b x)}{2 b}-\frac{5 d (c+d x)^{3/2} \sinh ^2(a+b x)}{8 b^2}+\frac{\left (15 d^2\right ) \int \sqrt{c+d x} \cosh (2 a+2 b x) \, dx}{32 b^2}\\ &=-\frac{5 d (c+d x)^{3/2}}{16 b^2}-\frac{(c+d x)^{7/2}}{7 d}+\frac{(c+d x)^{5/2} \cosh (a+b x) \sinh (a+b x)}{2 b}-\frac{5 d (c+d x)^{3/2} \sinh ^2(a+b x)}{8 b^2}+\frac{15 d^2 \sqrt{c+d x} \sinh (2 a+2 b x)}{64 b^3}-\frac{\left (15 d^3\right ) \int \frac{\sinh (2 a+2 b x)}{\sqrt{c+d x}} \, dx}{128 b^3}\\ &=-\frac{5 d (c+d x)^{3/2}}{16 b^2}-\frac{(c+d x)^{7/2}}{7 d}+\frac{(c+d x)^{5/2} \cosh (a+b x) \sinh (a+b x)}{2 b}-\frac{5 d (c+d x)^{3/2} \sinh ^2(a+b x)}{8 b^2}+\frac{15 d^2 \sqrt{c+d x} \sinh (2 a+2 b x)}{64 b^3}-\frac{\left (15 d^3\right ) \int \frac{e^{-i (2 i a+2 i b x)}}{\sqrt{c+d x}} \, dx}{256 b^3}+\frac{\left (15 d^3\right ) \int \frac{e^{i (2 i a+2 i b x)}}{\sqrt{c+d x}} \, dx}{256 b^3}\\ &=-\frac{5 d (c+d x)^{3/2}}{16 b^2}-\frac{(c+d x)^{7/2}}{7 d}+\frac{(c+d x)^{5/2} \cosh (a+b x) \sinh (a+b x)}{2 b}-\frac{5 d (c+d x)^{3/2} \sinh ^2(a+b x)}{8 b^2}+\frac{15 d^2 \sqrt{c+d x} \sinh (2 a+2 b x)}{64 b^3}+\frac{\left (15 d^2\right ) \operatorname{Subst}\left (\int e^{i \left (2 i a-\frac{2 i b c}{d}\right )-\frac{2 b x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{128 b^3}-\frac{\left (15 d^2\right ) \operatorname{Subst}\left (\int e^{-i \left (2 i a-\frac{2 i b c}{d}\right )+\frac{2 b x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{128 b^3}\\ &=-\frac{5 d (c+d x)^{3/2}}{16 b^2}-\frac{(c+d x)^{7/2}}{7 d}+\frac{15 d^{5/2} e^{-2 a+\frac{2 b c}{d}} \sqrt{\frac{\pi }{2}} \text{erf}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{256 b^{7/2}}-\frac{15 d^{5/2} e^{2 a-\frac{2 b c}{d}} \sqrt{\frac{\pi }{2}} \text{erfi}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{256 b^{7/2}}+\frac{(c+d x)^{5/2} \cosh (a+b x) \sinh (a+b x)}{2 b}-\frac{5 d (c+d x)^{3/2} \sinh ^2(a+b x)}{8 b^2}+\frac{15 d^2 \sqrt{c+d x} \sinh (2 a+2 b x)}{64 b^3}\\ \end{align*}

Mathematica [A]  time = 6.59064, size = 190, normalized size = 0.79 \[ \frac{\sqrt{c+d x} \left (-7 \sqrt{2} d^4 \sqrt{-\frac{b^2 (c+d x)^2}{d^2}} \text{Gamma}\left (\frac{7}{2},-\frac{2 b (c+d x)}{d}\right ) \left (\sinh \left (2 a-\frac{2 b c}{d}\right )+\cosh \left (2 a-\frac{2 b c}{d}\right )\right )-b (c+d x) \left (7 \sqrt{2} d^3 \text{Gamma}\left (\frac{7}{2},\frac{2 b (c+d x)}{d}\right ) \left (\cosh \left (2 a-\frac{2 b c}{d}\right )-\sinh \left (2 a-\frac{2 b c}{d}\right )\right )+64 b^3 (c+d x)^3 \sqrt{\frac{b (c+d x)}{d}}\right )\right )}{448 b^3 d^2 \left (\frac{b (c+d x)}{d}\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*x)^(5/2)*Sinh[a + b*x]^2,x]

[Out]

(Sqrt[c + d*x]*(-(b*(c + d*x)*(64*b^3*(c + d*x)^3*Sqrt[(b*(c + d*x))/d] + 7*Sqrt[2]*d^3*Gamma[7/2, (2*b*(c + d
*x))/d]*(Cosh[2*a - (2*b*c)/d] - Sinh[2*a - (2*b*c)/d]))) - 7*Sqrt[2]*d^4*Sqrt[-((b^2*(c + d*x)^2)/d^2)]*Gamma
[7/2, (-2*b*(c + d*x))/d]*(Cosh[2*a - (2*b*c)/d] + Sinh[2*a - (2*b*c)/d])))/(448*b^3*d^2*((b*(c + d*x))/d)^(3/
2))

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Maple [F]  time = 0.062, size = 0, normalized size = 0. \begin{align*} \int \left ( dx+c \right ) ^{{\frac{5}{2}}} \left ( \sinh \left ( bx+a \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(5/2)*sinh(b*x+a)^2,x)

[Out]

int((d*x+c)^(5/2)*sinh(b*x+a)^2,x)

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Maxima [A]  time = 1.74935, size = 379, normalized size = 1.59 \begin{align*} -\frac{512 \,{\left (d x + c\right )}^{\frac{7}{2}} + \frac{105 \, \sqrt{2} \sqrt{\pi } d^{3} \operatorname{erf}\left (\sqrt{2} \sqrt{d x + c} \sqrt{-\frac{b}{d}}\right ) e^{\left (2 \, a - \frac{2 \, b c}{d}\right )}}{b^{3} \sqrt{-\frac{b}{d}}} - \frac{105 \, \sqrt{2} \sqrt{\pi } d^{3} \operatorname{erf}\left (\sqrt{2} \sqrt{d x + c} \sqrt{\frac{b}{d}}\right ) e^{\left (-2 \, a + \frac{2 \, b c}{d}\right )}}{b^{3} \sqrt{\frac{b}{d}}} + \frac{28 \,{\left (16 \,{\left (d x + c\right )}^{\frac{5}{2}} b^{2} d e^{\left (\frac{2 \, b c}{d}\right )} + 20 \,{\left (d x + c\right )}^{\frac{3}{2}} b d^{2} e^{\left (\frac{2 \, b c}{d}\right )} + 15 \, \sqrt{d x + c} d^{3} e^{\left (\frac{2 \, b c}{d}\right )}\right )} e^{\left (-2 \, a - \frac{2 \,{\left (d x + c\right )} b}{d}\right )}}{b^{3}} - \frac{28 \,{\left (16 \,{\left (d x + c\right )}^{\frac{5}{2}} b^{2} d e^{\left (2 \, a\right )} - 20 \,{\left (d x + c\right )}^{\frac{3}{2}} b d^{2} e^{\left (2 \, a\right )} + 15 \, \sqrt{d x + c} d^{3} e^{\left (2 \, a\right )}\right )} e^{\left (\frac{2 \,{\left (d x + c\right )} b}{d} - \frac{2 \, b c}{d}\right )}}{b^{3}}}{3584 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)*sinh(b*x+a)^2,x, algorithm="maxima")

[Out]

-1/3584*(512*(d*x + c)^(7/2) + 105*sqrt(2)*sqrt(pi)*d^3*erf(sqrt(2)*sqrt(d*x + c)*sqrt(-b/d))*e^(2*a - 2*b*c/d
)/(b^3*sqrt(-b/d)) - 105*sqrt(2)*sqrt(pi)*d^3*erf(sqrt(2)*sqrt(d*x + c)*sqrt(b/d))*e^(-2*a + 2*b*c/d)/(b^3*sqr
t(b/d)) + 28*(16*(d*x + c)^(5/2)*b^2*d*e^(2*b*c/d) + 20*(d*x + c)^(3/2)*b*d^2*e^(2*b*c/d) + 15*sqrt(d*x + c)*d
^3*e^(2*b*c/d))*e^(-2*a - 2*(d*x + c)*b/d)/b^3 - 28*(16*(d*x + c)^(5/2)*b^2*d*e^(2*a) - 20*(d*x + c)^(3/2)*b*d
^2*e^(2*a) + 15*sqrt(d*x + c)*d^3*e^(2*a))*e^(2*(d*x + c)*b/d - 2*b*c/d)/b^3)/d

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Fricas [B]  time = 3.19787, size = 2319, normalized size = 9.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)*sinh(b*x+a)^2,x, algorithm="fricas")

[Out]

1/3584*(105*sqrt(2)*sqrt(pi)*(d^4*cosh(b*x + a)^2*cosh(-2*(b*c - a*d)/d) - d^4*cosh(b*x + a)^2*sinh(-2*(b*c -
a*d)/d) + (d^4*cosh(-2*(b*c - a*d)/d) - d^4*sinh(-2*(b*c - a*d)/d))*sinh(b*x + a)^2 + 2*(d^4*cosh(b*x + a)*cos
h(-2*(b*c - a*d)/d) - d^4*cosh(b*x + a)*sinh(-2*(b*c - a*d)/d))*sinh(b*x + a))*sqrt(b/d)*erf(sqrt(2)*sqrt(d*x
+ c)*sqrt(b/d)) + 105*sqrt(2)*sqrt(pi)*(d^4*cosh(b*x + a)^2*cosh(-2*(b*c - a*d)/d) + d^4*cosh(b*x + a)^2*sinh(
-2*(b*c - a*d)/d) + (d^4*cosh(-2*(b*c - a*d)/d) + d^4*sinh(-2*(b*c - a*d)/d))*sinh(b*x + a)^2 + 2*(d^4*cosh(b*
x + a)*cosh(-2*(b*c - a*d)/d) + d^4*cosh(b*x + a)*sinh(-2*(b*c - a*d)/d))*sinh(b*x + a))*sqrt(-b/d)*erf(sqrt(2
)*sqrt(d*x + c)*sqrt(-b/d)) - 4*(112*b^3*d^3*x^2 + 112*b^3*c^2*d + 140*b^2*c*d^2 - 7*(16*b^3*d^3*x^2 + 16*b^3*
c^2*d - 20*b^2*c*d^2 + 15*b*d^3 + 4*(8*b^3*c*d^2 - 5*b^2*d^3)*x)*cosh(b*x + a)^4 - 28*(16*b^3*d^3*x^2 + 16*b^3
*c^2*d - 20*b^2*c*d^2 + 15*b*d^3 + 4*(8*b^3*c*d^2 - 5*b^2*d^3)*x)*cosh(b*x + a)*sinh(b*x + a)^3 - 7*(16*b^3*d^
3*x^2 + 16*b^3*c^2*d - 20*b^2*c*d^2 + 15*b*d^3 + 4*(8*b^3*c*d^2 - 5*b^2*d^3)*x)*sinh(b*x + a)^4 + 105*b*d^3 +
128*(b^4*d^3*x^3 + 3*b^4*c*d^2*x^2 + 3*b^4*c^2*d*x + b^4*c^3)*cosh(b*x + a)^2 + 2*(64*b^4*d^3*x^3 + 192*b^4*c*
d^2*x^2 + 192*b^4*c^2*d*x + 64*b^4*c^3 - 21*(16*b^3*d^3*x^2 + 16*b^3*c^2*d - 20*b^2*c*d^2 + 15*b*d^3 + 4*(8*b^
3*c*d^2 - 5*b^2*d^3)*x)*cosh(b*x + a)^2)*sinh(b*x + a)^2 + 28*(8*b^3*c*d^2 + 5*b^2*d^3)*x - 4*(7*(16*b^3*d^3*x
^2 + 16*b^3*c^2*d - 20*b^2*c*d^2 + 15*b*d^3 + 4*(8*b^3*c*d^2 - 5*b^2*d^3)*x)*cosh(b*x + a)^3 - 64*(b^4*d^3*x^3
+ 3*b^4*c*d^2*x^2 + 3*b^4*c^2*d*x + b^4*c^3)*cosh(b*x + a))*sinh(b*x + a))*sqrt(d*x + c))/(b^4*d*cosh(b*x + a
)^2 + 2*b^4*d*cosh(b*x + a)*sinh(b*x + a) + b^4*d*sinh(b*x + a)^2)

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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((d*x+c)**(5/2)*sinh(b*x+a)**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )}^{\frac{5}{2}} \sinh \left (b x + a\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)*sinh(b*x+a)^2,x, algorithm="giac")

[Out]

integrate((d*x + c)^(5/2)*sinh(b*x + a)^2, x)