Optimal. Leaf size=219 \[ \frac{2 \left (15 a^2-23 b^2\right ) \sqrt{a+b \tan (c+d x)}}{15 d \sqrt{\tan (c+d x)}}-\frac{2 a^2 \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{(-b+i a)^{5/2} \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{22 a b \sqrt{a+b \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{(b+i a)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d} \]
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Rubi [A] time = 0.994674, antiderivative size = 219, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {3565, 3649, 3616, 3615, 93, 203, 206} \[ \frac{2 \left (15 a^2-23 b^2\right ) \sqrt{a+b \tan (c+d x)}}{15 d \sqrt{\tan (c+d x)}}-\frac{2 a^2 \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{(-b+i a)^{5/2} \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{22 a b \sqrt{a+b \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{(b+i a)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d} \]
Antiderivative was successfully verified.
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Rule 3565
Rule 3649
Rule 3616
Rule 3615
Rule 93
Rule 203
Rule 206
Rubi steps
\begin{align*} \int \frac{(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac{7}{2}}(c+d x)} \, dx &=-\frac{2 a^2 \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2}{5} \int \frac{\frac{11 a^2 b}{2}-\frac{5}{2} a \left (a^2-3 b^2\right ) \tan (c+d x)-\frac{1}{2} b \left (4 a^2-5 b^2\right ) \tan ^2(c+d x)}{\tan ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx\\ &=-\frac{2 a^2 \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{22 a b \sqrt{a+b \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{4 \int \frac{\frac{1}{4} a^2 \left (15 a^2-23 b^2\right )+\frac{15}{4} a b \left (3 a^2-b^2\right ) \tan (c+d x)+\frac{11}{2} a^2 b^2 \tan ^2(c+d x)}{\tan ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx}{15 a}\\ &=-\frac{2 a^2 \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{22 a b \sqrt{a+b \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (15 a^2-23 b^2\right ) \sqrt{a+b \tan (c+d x)}}{15 d \sqrt{\tan (c+d x)}}+\frac{8 \int \frac{-\frac{15}{8} a^2 b \left (3 a^2-b^2\right )+\frac{15}{8} a^3 \left (a^2-3 b^2\right ) \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{15 a^2}\\ &=-\frac{2 a^2 \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{22 a b \sqrt{a+b \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (15 a^2-23 b^2\right ) \sqrt{a+b \tan (c+d x)}}{15 d \sqrt{\tan (c+d x)}}-\frac{1}{2} (i a-b)^3 \int \frac{1-i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx+\frac{1}{2} (i a+b)^3 \int \frac{1+i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx\\ &=-\frac{2 a^2 \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{22 a b \sqrt{a+b \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (15 a^2-23 b^2\right ) \sqrt{a+b \tan (c+d x)}}{15 d \sqrt{\tan (c+d x)}}-\frac{(i a-b)^3 \operatorname{Subst}\left (\int \frac{1}{(1+i x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{(i a+b)^3 \operatorname{Subst}\left (\int \frac{1}{(1-i x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=-\frac{2 a^2 \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{22 a b \sqrt{a+b \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (15 a^2-23 b^2\right ) \sqrt{a+b \tan (c+d x)}}{15 d \sqrt{\tan (c+d x)}}-\frac{(i a-b)^3 \operatorname{Subst}\left (\int \frac{1}{1-(-i a+b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{(i a+b)^3 \operatorname{Subst}\left (\int \frac{1}{1-(i a+b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}\\ &=-\frac{(i a-b)^{5/2} \tan ^{-1}\left (\frac{\sqrt{i a-b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{(i a+b)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{i a+b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 a^2 \sqrt{a+b \tan (c+d x)}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{22 a b \sqrt{a+b \tan (c+d x)}}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (15 a^2-23 b^2\right ) \sqrt{a+b \tan (c+d x)}}{15 d \sqrt{\tan (c+d x)}}\\ \end{align*}
Mathematica [A] time = 1.57302, size = 194, normalized size = 0.89 \[ \frac{\frac{2 \sqrt{a+b \tan (c+d x)} \left (\left (15 a^2-23 b^2\right ) \tan ^2(c+d x)-3 a^2-11 a b \tan (c+d x)\right )}{\tan ^{\frac{5}{2}}(c+d x)}-15 \sqrt [4]{-1} (-a-i b)^{5/2} \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )-15 \sqrt [4]{-1} (a-i b)^{5/2} \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{15 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.327, size = 1345351, normalized size = 6143.2 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}{\tan \left (d x + c\right )^{\frac{7}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [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.
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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.
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Giac [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.
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